Optimal. Leaf size=267 \[ -\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m+n+1)-B c (m+2 n+1)))}{c^2 d^3 e (m+1) n}-\frac{b (e x)^{m+1} (2 a d (A d (m+1)-B c (m+n+1))-b c (A d (m+n+1)-B c (m+2 n+1)))}{c d^3 e (m+1) n}-\frac{(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{c d e n \left (c+d x^n\right )}-\frac{b^2 x^{n+1} (e x)^m (A d (m+n+1)-B c (m+2 n+1))}{c d^2 n (m+n+1)} \]
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Rubi [A] time = 0.675441, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {594, 570, 20, 30, 364} \[ -\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m+n+1)-B c (m+2 n+1)))}{c^2 d^3 e (m+1) n}-\frac{b (e x)^{m+1} (2 a d (A d (m+1)-B c (m+n+1))-b c (A d (m+n+1)-B c (m+2 n+1)))}{c d^3 e (m+1) n}-\frac{(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{c d e n \left (c+d x^n\right )}-\frac{b^2 x^{n+1} (e x)^m (A d (m+n+1)-B c (m+2 n+1))}{c d^2 n (m+n+1)} \]
Antiderivative was successfully verified.
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Rule 594
Rule 570
Rule 20
Rule 30
Rule 364
Rubi steps
\begin{align*} \int \frac{(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac{\int \frac{(e x)^m \left (a+b x^n\right ) \left (-a (B c (1+m)-A d (1+m-n))+b (A d (1+m+n)-B c (1+m+2 n)) x^n\right )}{c+d x^n} \, dx}{c d n}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac{\int \left (\frac{b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^m}{d^2}+\frac{b^2 (A d (1+m+n)-B c (1+m+2 n)) x^n (e x)^m}{d}+\frac{(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^m}{d^2 \left (c+d x^n\right )}\right ) \, dx}{c d n}\\ &=-\frac{b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac{\left (b^2 (A d (1+m+n)-B c (1+m+2 n))\right ) \int x^n (e x)^m \, dx}{c d^2 n}-\frac{((b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n)))) \int \frac{(e x)^m}{c+d x^n} \, dx}{c d^3 n}\\ &=-\frac{b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac{(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{c^2 d^3 e (1+m) n}-\frac{\left (b^2 (A d (1+m+n)-B c (1+m+2 n)) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{c d^2 n}\\ &=-\frac{b^2 (A d (1+m+n)-B c (1+m+2 n)) x^{1+n} (e x)^m}{c d^2 n (1+m+n)}-\frac{b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac{(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{c^2 d^3 e (1+m) n}\\ \end{align*}
Mathematica [A] time = 0.30429, size = 161, normalized size = 0.6 \[ \frac{x (e x)^m \left (-\frac{(b c-a d)^2 (B c-A d) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c^2 (m+1)}+\frac{(b c-a d) (-a B d-2 A b d+3 b B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c (m+1)}+\frac{b (2 a B d+A b d-2 b B c)}{m+1}+\frac{b^2 B d x^n}{m+n+1}\right )}{d^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.54, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( a+b{x}^{n} \right ) ^{2} \left ( A+B{x}^{n} \right ) }{ \left ( c+d{x}^{n} \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -{\left ({\left (b^{2} c^{2} d e^{m}{\left (m + n + 1\right )} + a^{2} d^{3} e^{m}{\left (m - n + 1\right )} - 2 \, a b c d^{2} e^{m}{\left (m + 1\right )}\right )} A -{\left (b^{2} c^{3} e^{m}{\left (m + 2 \, n + 1\right )} - 2 \, a b c^{2} d e^{m}{\left (m + n + 1\right )} + a^{2} c d^{2} e^{m}{\left (m + 1\right )}\right )} B\right )} \int \frac{x^{m}}{c d^{4} n x^{n} + c^{2} d^{3} n}\,{d x} + \frac{{\left (m n + n\right )} B b^{2} c d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )} +{\left ({\left ({\left (m^{2} + 2 \, m{\left (n + 1\right )} + n^{2} + 2 \, n + 1\right )} b^{2} c^{2} d e^{m} - 2 \,{\left (m^{2} + m{\left (n + 2\right )} + n + 1\right )} a b c d^{2} e^{m} +{\left (m^{2} + m{\left (n + 2\right )} + n + 1\right )} a^{2} d^{3} e^{m}\right )} A -{\left ({\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} b^{2} c^{3} e^{m} - 2 \,{\left (m^{2} + 2 \, m{\left (n + 1\right )} + n^{2} + 2 \, n + 1\right )} a b c^{2} d e^{m} +{\left (m^{2} + m{\left (n + 2\right )} + n + 1\right )} a^{2} c d^{2} e^{m}\right )} B\right )} x x^{m} +{\left ({\left (m n + n^{2} + n\right )} A b^{2} c d^{2} e^{m} -{\left ({\left (m n + 2 \, n^{2} + n\right )} b^{2} c^{2} d e^{m} - 2 \,{\left (m n + n^{2} + n\right )} a b c d^{2} e^{m}\right )} B\right )} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{{\left (m^{2} n +{\left (n^{2} + 2 \, n\right )} m + n^{2} + n\right )} c d^{4} x^{n} +{\left (m^{2} n +{\left (n^{2} + 2 \, n\right )} m + n^{2} + n\right )} c^{2} d^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{2} x^{3 \, n} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} x^{2 \, n} +{\left (B a^{2} + 2 \, A a b\right )} x^{n}\right )} \left (e x\right )^{m}}{d^{2} x^{2 \, n} + 2 \, c d x^{n} + c^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{n} + A\right )}{\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{n} + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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