Optimal. Leaf size=322 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) (b c (a B (m+1)-A b (m-2 n+1)) (a d (m+1)-b c (m-n+1))-a d (A b (m+1)-a B (m+2 n+1)) (b c (m+1)-a d (m+n+1)))}{2 a^3 b^3 e (m+1) n^2}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)-A b (m-2 n+1))-d x^n (A b (m+1)-a B (m+2 n+1))\right )}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+2 n+1)) (b c (m+1)-a d (m+n+1))}{2 a^2 b^3 e (m+1) n^2}+\frac{(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.543598, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {594, 459, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) (b c (a B (m+1)-A b (m-2 n+1)) (a d (m+1)-b c (m-n+1))-a d (A b (m+1)-a B (m+2 n+1)) (b c (m+1)-a d (m+n+1)))}{2 a^3 b^3 e (m+1) n^2}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)-A b (m-2 n+1))-d x^n (A b (m+1)-a B (m+2 n+1))\right )}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+2 n+1)) (b c (m+1)-a d (m+n+1))}{2 a^2 b^3 e (m+1) n^2}+\frac{(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 594
Rule 459
Rule 364
Rubi steps
\begin{align*} \int \frac{(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^3} \, dx &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2}-\frac{\int \frac{(e x)^m \left (c+d x^n\right ) \left (-c (a B (1+m)-A b (1+m-2 n))+d (A b (1+m)-a B (1+m+2 n)) x^n\right )}{\left (a+b x^n\right )^2} \, dx}{2 a b n}\\ &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (a B (1+m)-A b (1+m-2 n))-d (A b (1+m)-a B (1+m+2 n)) x^n\right )}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac{\int \frac{(e x)^m \left (c (a B (1+m)-A b (1+m-2 n)) (a d (1+m)-b c (1+m-n))+d (b c (1+m)-a d (1+m+n)) (A b (1+m)-a B (1+m+2 n)) x^n\right )}{a+b x^n} \, dx}{2 a^2 b^2 n^2}\\ &=\frac{d (b c (1+m)-a d (1+m+n)) (A b (1+m)-a B (1+m+2 n)) (e x)^{1+m}}{2 a^2 b^3 e (1+m) n^2}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (a B (1+m)-A b (1+m-2 n))-d (A b (1+m)-a B (1+m+2 n)) x^n\right )}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac{\left (c (a B (1+m)-A b (1+m-2 n)) (a d (1+m)-b c (1+m-n))-\frac{a d (b c (1+m)-a d (1+m+n)) (A b (1+m)-a B (1+m+2 n))}{b}\right ) \int \frac{(e x)^m}{a+b x^n} \, dx}{2 a^2 b^2 n^2}\\ &=\frac{d (b c (1+m)-a d (1+m+n)) (A b (1+m)-a B (1+m+2 n)) (e x)^{1+m}}{2 a^2 b^3 e (1+m) n^2}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{2 a b e n \left (a+b x^n\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (a B (1+m)-A b (1+m-2 n))-d (A b (1+m)-a B (1+m+2 n)) x^n\right )}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac{\left (c (a B (1+m)-A b (1+m-2 n)) (a d (1+m)-b c (1+m-n))-\frac{a d (b c (1+m)-a d (1+m+n)) (A b (1+m)-a B (1+m+2 n))}{b}\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{b x^n}{a}\right )}{2 a^3 b^2 e (1+m) n^2}\\ \end{align*}
Mathematica [A] time = 0.239841, size = 168, normalized size = 0.52 \[ \frac{x (e x)^m \left (\frac{(b c-a d) (-3 a B d+2 A b d+b B c) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2}+\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^3}+\frac{d (-3 a B d+A b d+2 b B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a}+B d^2\right )}{b^3 (m+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.498, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( A+B{x}^{n} \right ) \left ( c+d{x}^{n} \right ) ^{2}}{ \left ( a+b{x}^{n} \right ) ^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B d^{2} x^{3 \, n} + A c^{2} +{\left (2 \, B c d + A d^{2}\right )} x^{2 \, n} +{\left (B c^{2} + 2 \, A c d\right )} x^{n}\right )} \left (e x\right )^{m}}{b^{3} x^{3 \, n} + 3 \, a b^{2} x^{2 \, n} + 3 \, a^{2} b x^{n} + a^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{n} + A\right )}{\left (d x^{n} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]