The following is a list of integrals (antiderivative functions) of rational functions. Any rational function can be integrated by partial fraction decomposition of the function into a sum of functions of the form:
![{\displaystyle {\frac {a}{(x-b)^{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da948f1f4e347058d6e34eac1bfb6617a70f1428)
, and
![{\displaystyle {\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/277e9eff8e906c97a0ea9b8d3dcf69566a5aa4d7)
which can then be integrated term by term.
For other types of functions, see lists of integrals.
Miscellaneous integrands
![{\displaystyle \int {\frac {f'(x)}{f(x)}}\,dx=\ln \left|f(x)\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68ff9dbfd7660c62e29ea8147600248c4ff07238)
![{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}\,dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1acafa96f0cd8a2d9f05b62f34fef88a73225d3)
![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C={\begin{cases}\displaystyle -{\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f707d02dbc04ceb2d02ddb5bfd60ab31f45b6b55)
![{\displaystyle \int {\frac {1}{a^{2}-x^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {a+x}{a-x}}\right|+C={\begin{cases}\displaystyle {\frac {1}{a}}\,\operatorname {artanh} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {a+x}{a-x}}+C&{\text{(for }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{a}}\,\operatorname {arcoth} {\frac {x}{a}}+C={\frac {1}{2a}}\ln {\frac {x+a}{x-a}}+C&{\text{(for }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d41f30fbfce8f00f5e1503b29b5e0b8415fadec)
![{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}={\frac {1}{2^{n-1}}}\sum _{k=1}^{2^{n-1}}\sin \left({\frac {2k-1}{2^{n}}}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n}}}\pi \right)\right]-{\frac {1}{2}}\cos \left({\frac {2k-1}{2^{n}}}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n}}}\pi \right)+1\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5086de865f6047c3f115bfbc3bd5ffde147a645c)
Integrands of the form xm(a x + b)n
Many of the following antiderivatives have a term of the form ln |ax + b|. Because this is undefined when x = −b / a, the most general form of the antiderivative replaces the constant of integration with a locally constant function.[1] However, it is conventional to omit this from the notation. For example,
is usually abbreviated as
where C is to be understood as notation for a locally constant function of x. This convention will be adhered to in the following.
(Cavalieri's quadrature formula) ![{\displaystyle \int {\frac {x}{ax+b}}\,dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0044cf33077ed2838244a1f49755bfda52d1b5c2)
![{\displaystyle \int {\frac {mx+n}{ax+b}}\,dx={\frac {m}{a}}x+{\frac {an-bm}{a^{2}}}\ln \left|ax+b\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b191362b12ea43318e6bec514eeea3b80dd1d9eb)
![{\displaystyle \int {\frac {x}{(ax+b)^{2}}}\,dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fca8db2b5e2c57c5991f42012f3fe019e82d6db8)
![{\displaystyle \int {\frac {x}{(ax+b)^{n}}}\,dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}+C\qquad {\text{(for }}n\not \in \{1,2\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6c32c87074e94fd830bbf325284f542fd4ab0f3)
![{\displaystyle \int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}+C\qquad {\text{(for }}n\not \in \{-1,-2\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3005bc8299d9e0f8b4de337c934017d6e6353064)
![{\displaystyle \int {\frac {x^{2}}{ax+b}}\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3}}}+{\frac {ax^{2}-2bx}{2a^{2}}}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52efc74406ece758b265c368e08dd9ee16c8a8fc)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}\,dx={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bbcfb06f106d8fe208d68fc69d74c426a18ce8c)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}\,dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d9738490df1b57b94860a05b22b0ff7c61237d)
![{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}\,dx={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(ax+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)+C\qquad {\text{(for }}n\not \in \{1,2,3\}{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4814f0ed2c13d5860225e30b78c3b08044d11551)
![{\displaystyle \int {\frac {1}{x(ax+b)}}\,dx=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eb20d2f8f53fa28110b76a629615131429bbdea)
![{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}\,dx=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d446865683df020ec83ce8838cdcae2f1d2bd5f)
![{\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}\,dx=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e03e60c23681f5186b4a95a5e9ea921732e122a)
Integrands of the form xm / (a x2 + b x + c)n
For
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c6e45e8f485cc92285459242e5edc389b0a4b3c)
![{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}\,dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2741cb28c134c0cfe27162393974e3c56475461)
![{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{2a{\sqrt {b^{2}-4ac}}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C={\begin{cases}\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(for }}|2ax+b|<{\sqrt {b^{2}-4ac}}{\mbox{)}}\\[6pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(else)}}\end{cases}}&{\text{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C={\frac {m}{a}}\ln \left|x+{\frac {b}{2a}}\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\text{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/137aeb719faa0d412412ce2afb21f694747e79af)
![{\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70aa274c4d40db21d09cced28142d167f4fd1aab)
![{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}\,dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9596eeb7ecf42279db0fe8ddbff66b7aa5ffca3)
![{\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}\,dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}\,dx+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce3a0fffb5a4379f6c63480665bb09f9d6d1f61b)
Integrands of the form xm (a + b xn)p
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+n\,p+1}}\,+\,{\frac {a\,n\,p}{m+n\,p+1}}\int x^{m}\left(a+b\,x^{n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4790301459e72262aafa18b525a4f5e3bdf47c07)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx=-{\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a\,n(p+1)}}\,+\,{\frac {m+n(p+1)+1}{a\,n(p+1)}}\int x^{m}\left(a+b\,x^{n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2c8816cb99165f737c15e4d99c66d493f720137)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+1}}\,-\,{\frac {b\,n\,p}{m+1}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96bba069ebcdb45e98e0062bb4b07008b0d35fc2)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b\,n(p+1)}}\,-\,{\frac {m-n+1}{b\,n(p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3422de38d52f4da42c1bf9968ce5768120c7086)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b(m+n\,p+1)}}\,-\,{\frac {a(m-n+1)}{b(m+n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26dfd412da178d54770054a6acc2eb5bfe05fa4e)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a(m+1)}}\,-\,{\frac {b(m+n(p+1)+1)}{a(m+1)}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/853386481e1c31cde5b48c5df4f22b78d7b39324)
Integrands of the form (A + B x) (a + b x)m (c + d x)n (e + f x)p
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, n and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
by setting B to 0.
![{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx=-{\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p+1}}{b(m+1)(a\,f-b\,e)}}\,+\,{\frac {1}{b(m+1)(a\,f-b\,e)}}\,\cdot \\&\qquad \int (b\,c(m+1)(A\,f-B\,e)+(A\,b-a\,B)(n\,d\,e+c\,f(p+1))+d(b(m+1)(A\,f-B\,e)+f(n+p+1)(A\,b-a\,B))x)(a+b\,x)^{m+1}(c+d\,x)^{n-1}(e+f\,x)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a656bfa04081bcc75754b278a90aaa126c6979f0)
![{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {B(a+b\,x)^{m}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{d\,f(m+n+p+2)}}\,+\,{\frac {1}{d\,f(m+n+p+2)}}\,\cdot \\&\qquad \int (A\,a\,d\,f(m+n+p+2)-B(b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B(a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1))))x)(a+b\,x)^{m-1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/691ba1b24823df8e3d20a2d1df3cb6bb3f183a15)
![{\displaystyle {\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n+1}(e+f\,x)^{p+1}}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,+\,{\frac {1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}}\,\cdot \\&\qquad \int ((m+1)(A(a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B)(d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3)(A\,b-a\,B)x)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a289e2385b566eef40cfb38b33e21459512699e7)
Integrands of the form xm (A + B xn) (a + b xn)p (c + d xn)q
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
and
by setting m and/or B to 0.
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a\,b\,n(p+1)}}\,+\,{\frac {1}{a\,b\,n(p+1)}}\,\cdot \\&\qquad \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d879f60ce2644ee846af896e0e8958aaee47fac)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{b(m+n(p+q+1)+1)}}\,+\,{\frac {1}{b(m+n(p+q+1)+1)}}\,\cdot \\&\qquad \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e4c59cd60abc07ec0902e6aeb407a623413d84a)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,n(b\,c-a\,d)(p+1)}}\,+\,{\frac {1}{a\,n(b\,c-a\,d)(p+1)}}\,\cdot \\&\qquad \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baef71373d1818f465b45ecb06da846ae8811997)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,d(m+n(p+q+1)+1)}}\,-\,{\frac {1}{b\,d(m+n(p+q+1)+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a985b092ec350d39f68f9cef5d82749f13e333c0)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,c(m+1)}}\,+\,{\frac {1}{a\,c(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77654f111b0e2002f94c9c4058abb79897ff344c)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a(m+1)}}\,-\,{\frac {1}{a(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95017c34adb3b380849bd99e1b5989e0efbc0d78)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {(A\,b-a\,B)x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,n(b\,c-a\,d)(p+1)}}\,-\,{\frac {1}{b\,n(b\,c-a\,d)(p+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5522c5813f569e417b3311767af9043f019b615)
Integrands of the form (d + e x)m (a + b x + c x2)p when b2 − 4 a c = 0
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
when
by setting m to 0.
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)}{e^{2}(m+1)(m+2p+1)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81d4c261f463293ee05481519bb1634bf765009b)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+1)}}\,-\,{\frac {p(d+e\,x)^{m+2}(b+2\,c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{e^{2}(m+1)(m+2)}}\,+\,{\frac {2\,c\,p\,(2\,p-1)}{e^{2}(m+1)(m+2)}}\int (d+e\,x)^{m+2}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52be3971d00f1b06cbb659d213cb6e875c7beb23)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e(m+2p+2)(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {e^{2}m(m+2p+2)}{(p+1)(2p+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd7c90e7a6da4a559c53a66731a2d51e3533e7af)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e\,m(d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{2c(p+1)(2p+1)}}\,+\,{\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(2p+1)}}\,+\,{\frac {e^{2}m(m-1)}{2c(p+1)(2p+1)}}\int (d+e\,x)^{m-2}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48f0d694e4cffcb65184a89244a10abaf0b9b346)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}}{e(m+2p+1)}}\,-\,{\frac {p(2c\,d-b\,e)(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1}}{2c\,e^{2}(m+2p)(m+2p+1)}}\,+\,{\frac {p(2p-1)(2c\,d-b\,e)^{2}}{2c\,e^{2}(m+2p)(m+2p+1)}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5639f373ae3a1cbf3900954d75e024c850b57be5)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {2c\,e(m+2p+2)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(2p+1)(2c\,d-b\,e)}}\,+\,{\frac {2c\,e^{2}(m+2p+2)(m+2p+3)}{(p+1)(2p+1)(2c\,d-b\,e)^{2}}}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cfc86b744b2f318e9e0a51ec13b7576e7b53917)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{2c(m+2p+1)}}\,+\,{\frac {m(2c\,d-b\,e)}{2c(m+2p+1)}}\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9a0ad7c5a71f6bcb324373ed772deb8ec0a6e1a)
![{\displaystyle \int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p}}{(m+1)(2c\,d-b\,e)}}\,+\,{\frac {2c(m+2p+2)}{(m+1)(2c\,d-b\,e)}}\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f81c419c1332e643d09afb2f927a42c10c468bb)
Integrands of the form (d + e x)m (A + B x) (a + b x + c x2)p
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
and
by setting m and/or B to 0.
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,e(m+2p+2)-B\,d(2p+1)+e\,B(m+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{e^{2}(m+1)(m+2p+2)}}\,+\,{\frac {1}{e^{2}(m+1)(m+2p+2)}}p\,\cdot \\&\qquad \int (d+e\,x)^{m+1}(B(b\,d+2a\,e+2a\,e\,m+2b\,d\,p)-A\,b\,e(m+2p+2)+(B(2c\,d+b\,e+b\,em+4c\,d\,p)-2A\,c\,e(m+2p+2))x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0956840d0dd30a387bcab57ab22725c26002bc9d)
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(A\,b-2a\,B-(b\,B-2A\,c)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(B(2a\,e\,m+b\,d(2p+3))-A(b\,e\,m+2c\,d(2p+3))+e(b\,B-2A\,c)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e7e950b265895e208e449680e4f5f33b5002141)
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,c\,e(m+2p+2)-B(c\,d+2c\,d\,p-b\,e\,p)+B\,c\,e(m+2p+1)x)\left(a+b\,x+c\,x^{2}\right)^{p}}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,-\,{\frac {p}{c\,e^{2}(m+2p+1)(m+2p+2)}}\,\cdot \\&\qquad \int (d+e\,x)^{m}(A\,c\,e(b\,d-2a\,e)(m+2p+2)+B(a\,e(b\,e-2c\,d\,m+b\,e\,m)+b\,d(b\,e\,p-c\,d-2c\,d\,p))+\\&\qquad \qquad \left(A\,c\,e(2c\,d-b\,e)(m+2p+2)-B\left(-b^{2}e^{2}(m+p+1)+2c^{2}d^{2}(1+2p)+c\,e(b\,d(m-2p)+2a\,e(m+2p+1))\right)\right)x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c33436c8438bd88f632412429060a9773eae514c)
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(A\left(b\,c\,d-b^{2}e+2a\,c\,e\right)-a\,B(2c\,d-b\,e)+c(A(2c\,d-b\,e)-B(b\,d-2a\,e))x\right)\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+\\&\qquad {\frac {1}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot \\&\qquad \qquad \int (d+e\,x)^{m}(A\left(b\,c\,d\,e(2p-m+2)+b^{2}e^{2}(m+p+2)-2c^{2}d^{2}(3+2p)-2a\,c\,e^{2}(m+2p+3)\right)-\\&\qquad \qquad \qquad B(a\,e(b\,e-2c\,dm+b\,e\,m)+b\,d(-3c\,d+b\,e-2c\,d\,p+b\,e\,p))+c\,e(B(b\,d-2a\,e)-A(2c\,d-b\,e))(m+2p+4)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35c6fec327101fa7948382930075e2e0d79c8e18)
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {B(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{c(m+2p+2)}}\,+\,{\frac {1}{c(m+2p+2)}}\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(m(A\,c\,d-a\,B\,e)-d(b\,B-2A\,c)(p+1)+((B\,c\,d-b\,B\,e+A\,c\,e)m-e(b\,B-2A\,c)(p+1))x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee370ed6be53099c2bd7594087a26ef9348656b5)
![{\displaystyle {\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(B\,d-A\,e)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1}}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,+\,{\frac {1}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)}}\,\cdot \\&\qquad \int (d+e\,x)^{m+1}((A\,c\,d-A\,b\,e+a\,B\,e)(m+1)+b(B\,d-A\,e)(p+1)+c(B\,d-A\,e)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52808a2af103a45429e2363edced3c5a83733ea2)
Integrands of the form xm (a + b xn + c x2n)p when b2 − 4 a c = 0
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
when
by setting m to 0.
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+2n\,p+1)}}\,-\,{\frac {b\,n^{2}p(2p-1)}{(m+1)(m+2n\,p+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8b1f53bf9a1baa716e5af791b71798ab7c7409d)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p-1)+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n+1)}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+1)(m+n+1)}}\,+\,{\frac {2c\,p\,n^{2}(2p-1)}{(m+1)(m+n+1)}}\int x^{m+2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daf94843c8208eb4a740395f5c20467bd18f7890)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p+1)+1)x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{b\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b\,n(2p+1)}}\,-\,{\frac {(m-n+1)(m+n(2p+1)+1)}{b\,n^{2}(p+1)(2p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/021f1e8e9598be88cfa1607774fcee9206cf2e6c)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m-3n-2n\,p+1)x^{m-2n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2c\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m-2n+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c\,n(2p+1)}}\,+\,{\frac {(m-n+1)(m-2n+1)}{2c\,n^{2}(p+1)(2p+1)}}\int x^{m-2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5519f9ce654c5b38b15c36e913ab34ed398b7a4e)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{m+2n\,p+1}}\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}}{(m+2n\,p+1)(m+n(2p-1)+1)}}\,+\,{\frac {2a\,n^{2}p(2p-1)}{(m+2n\,p+1)(m+n(2p-1)+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f56db1ed0a596ae938014d966e70cec21caf6147)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m+n+2n\,p+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{2a\,n^{2}(p+1)(2p+1)}}\,-\,{\frac {x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2a\,n(2p+1)}}\,+\,{\frac {(m+n(2p+1)+1)(m+2n(p+1)+1)}{2a\,n^{2}(p+1)(2p+1)}}\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38eababe79b99a4cd22d8a14cd9c524f14bd2c1d)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{2c(m+2n\,p+1)}}\,-\,{\frac {b(m-n+1)}{2c(m+2n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25d28b779eace5a71ad73fb2672f516197d71004)
![{\displaystyle \int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{b(m+1)}}\,-\,{\frac {2c(m+n(2p+1)+1)}{b(m+1)}}\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08f3bc0cc9c22df9b31c927fae0a91b906d41216)
Integrands of the form xm (A + B xn) (a + b xn + c x2n)p
- The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
- These reduction formulas can be used for integrands having integer and/or fractional exponents.
- Special cases of these reductions formulas can be used for integrands of the form
and
by setting m and/or B to 0.
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(A(m+n(2p+1)+1)+B(m+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{(m+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{(m+1)(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(2a\,B(m+1)-A\,b(m+n(2p+1)+1)+(b\,B(m+1)-2\,A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea81d2fb0c4d264a7382e987e7f0e5f4be69c556)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(A\,b-2a\,B-(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int x^{m-n}\left((m-n+1)(2a\,B-A\,b)+(m+2n(p+1)+1)(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/705cc3ecfaeec2f85d5a019ed635bad265ffd03c)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b\,B\,n\,p+A\,c(m+n(2p+1)+1)+B\,c(m+2n\,p+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,+\,{\frac {n\,p}{c(m+2n\,p+1)(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m}\left(2a\,A\,c(m+n(2p+1)+1)-a\,b\,B(m+1)+\left(2a\,B\,c(m+2n\,p+1)+A\,b\,c(m+n(2p+1)+1)-b^{2}B(m+n\,p+1)\right)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb90ba92d73dfd52ad585180d75324bdfa970525)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {x^{m+1}\left(A\,b^{2}-a\,b\,B-2a\,A\,c+(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,+\,{\frac {1}{a\,n(p+1)\left(b^{2}-4a\,c\right)}}\,\cdot \\&\qquad \int x^{m}\left((m+n(p+1)+1)A\,b^{2}-a\,b\,B(m+1)-2(m+2n(p+1)+1)a\,A\,c+(m+n(2p+3)+1)(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fa5c5301bea384110229218bf5a9be1713db74b)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{c(m+n(2p+1)+1)}}\,-\,{\frac {1}{c(m+n(2p+1)+1)}}\,\cdot \\&\qquad \int x^{m-n}\left(a\,B(m-n+1)+(b\,B(m+n\,p+1)-A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48e6ee601195e9a8ec6ee3bbfec7fae0aab8206e)
![{\displaystyle {\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}}{a(m+1)}}\,+\,{\frac {1}{a(m+1)}}\,\cdot \\&\qquad \int x^{m+n}\left(a\,B(m+1)-A\,b(m+n(p+1)+1)-A\,c(m+2n(p+1)+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c27f89e406a34f04f4dd70bb1196076d1dadbf04)
References
- ^ "Reader Survey: log|x| + C", Tom Leinster, The n-category Café, March 19, 2012