Optimal. Leaf size=85 \[ \frac{1}{5} a x^5 \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-p,\frac{7}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^4 (2 p+1)}+\frac{\left (1-a^2 x^2\right )^{p+\frac{3}{2}}}{a^4 (2 p+3)} \]
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Rubi [A] time = 0.113711, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6148, 764, 266, 43, 364} \[ \frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}}}{a^4 (2 p+1)}+\frac{\left (1-a^2 x^2\right )^{p+\frac{3}{2}}}{a^4 (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 6148
Rule 764
Rule 266
Rule 43
Rule 364
Rubi steps
\begin{align*} \int e^{\tanh ^{-1}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx &=\int x^3 (1+a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=a \int x^4 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx+\int x^3 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int x \left (1-a^2 x\right )^{-\frac{1}{2}+p} \, dx,x,x^2\right )\\ &=\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{\left (1-a^2 x\right )^{-\frac{1}{2}+p}}{a^2}-\frac{\left (1-a^2 x\right )^{\frac{1}{2}+p}}{a^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{\left (1-a^2 x^2\right )^{\frac{1}{2}+p}}{a^4 (1+2 p)}+\frac{\left (1-a^2 x^2\right )^{\frac{3}{2}+p}}{a^4 (3+2 p)}+\frac{1}{5} a x^5 \, _2F_1\left (\frac{5}{2},\frac{1}{2}-p;\frac{7}{2};a^2 x^2\right )\\ \end{align*}
Mathematica [A] time = 0.0695191, size = 77, normalized size = 0.91 \[ \frac{1}{5} a x^5 \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-p,\frac{7}{2},a^2 x^2\right )-\frac{\left (1-a^2 x^2\right )^{p+\frac{1}{2}} \left (a^2 (2 p+1) x^2+2\right )}{a^4 \left (4 p^2+8 p+3\right )} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.357, size = 47, normalized size = 0.6 \begin{align*}{\frac{a{x}^{5}}{5}{\mbox{$_2$F$_1$}({\frac{5}{2}},{\frac{1}{2}}-p;\,{\frac{7}{2}};\,{a}^{2}{x}^{2})}}+{\frac{{x}^{4}}{4}{\mbox{$_2$F$_1$}(2,{\frac{1}{2}}-p;\,3;\,{a}^{2}{x}^{2})}} \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*} a \int \frac{x^{4} e^{\left (p \log \left (a x + 1\right ) + p \log \left (-a x + 1\right )\right )}}{\sqrt{a x + 1} \sqrt{-a x + 1}}\,{d x} + \frac{{\left (a^{4}{\left (2 \, p + 1\right )} x^{4} - a^{2}{\left (2 \, p - 1\right )} x^{2} - 2\right )}{\left (-a^{2} x^{2} + 1\right )}^{p}}{\sqrt{-a^{2} x^{2} + 1}{\left (4 \, p^{2} + 8 \, p + 3\right )} a^{4}} \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{\sqrt{-a^{2} x^{2} + 1}{\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{a x - 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 29.6525, size = 258, normalized size = 3.04 \begin{align*} - \frac{a a^{2 p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{5}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p + \frac{5}{2} \\ p + 1, p + \frac{7}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + 1\right )} - \frac{a a^{2 p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{5}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, - p - \frac{5}{2} \\ \frac{1}{2}, - p - \frac{3}{2} \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + 1\right )} - \frac{{G_{3, 3}^{2, 2}\left (\begin{matrix} - p - 1, 1 & -1 \\- p - \frac{3}{2}, - p - 1 & 0 \end{matrix} \middle |{\frac{e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac{1}{2}\right )}{2 \pi a^{4}} - \frac{{G_{3, 3}^{1, 3}\left (\begin{matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac{3}{2}, 0 \end{matrix} \middle |{\frac{e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac{1}{2}\right )}{2 a^{4} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \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 (a x + 1\right )}{\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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