Optimal. Leaf size=161 \[ \frac{x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,1;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d}-\frac{d^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e (p+1) \left (a e^2+b d^2\right )}+\frac{\left (a+b x^2\right )^{p+1}}{2 b e (p+1)} \]
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Rubi [A] time = 0.148172, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {959, 511, 510, 446, 80, 68} \[ \frac{x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,1;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d}-\frac{d^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e (p+1) \left (a e^2+b d^2\right )}+\frac{\left (a+b x^2\right )^{p+1}}{2 b e (p+1)} \]
Antiderivative was successfully verified.
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Rule 959
Rule 511
Rule 510
Rule 446
Rule 80
Rule 68
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b x^2\right )^p}{d+e x} \, dx &=d \int \frac{x^2 \left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx-e \int \frac{x^3 \left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx\\ &=-\left (\frac{1}{2} e \operatorname{Subst}\left (\int \frac{x (a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )\right )+\left (d \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{x^2 \left (1+\frac{b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx\\ &=\frac{\left (a+b x^2\right )^{1+p}}{2 b e (1+p)}+\frac{x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,1;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d}-\frac{d^2 \operatorname{Subst}\left (\int \frac{(a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )}{2 e}\\ &=\frac{\left (a+b x^2\right )^{1+p}}{2 b e (1+p)}+\frac{x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,1;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d}-\frac{d^2 \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e \left (b d^2+a e^2\right ) (1+p)}\\ \end{align*}
Mathematica [A] time = 0.253906, size = 227, normalized size = 1.41 \[ \frac{\left (a+b x^2\right )^p \left (b d^2 (p+1) \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )-2 b d e p (p+1) x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-a e^2 p \left (\frac{b x^2}{a}+1\right )^{-p}+a e^2 p+b e^2 p x^2\right )}{2 b e^3 p (p+1)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.652, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( b{x}^{2}+a \right ) ^{p}}{ex+d}}\, 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*} \int \frac{{\left (b x^{2} + a\right )}^{p} x^{2}}{e x + d}\,{d x} \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 x^{2} + a\right )}^{p} x^{2}}{e x + d}, 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^{2} + a\right )}^{p} x^{2}}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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