Optimal. Leaf size=133 \[ \frac{1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )+\frac{\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+3)}-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+1)} \]
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Rubi [A] time = 0.18136, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6153, 6148, 764, 364, 266, 43} \[ \frac{1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )+\frac{\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+3)}-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+1)} \]
Antiderivative was successfully verified.
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Rule 6153
Rule 6148
Rule 764
Rule 364
Rule 266
Rule 43
Rubi steps
\begin{align*} \int e^{\tanh ^{-1}(a x)} x^2 \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int e^{\tanh ^{-1}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 (1+a x) \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx+\left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^3 \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=\frac{1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )+\frac{1}{2} \left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \operatorname{Subst}\left (\int x \left (1-a^2 x\right )^{-\frac{1}{2}+p} \, dx,x,x^2\right )\\ &=\frac{1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )+\frac{1}{2} \left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \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{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (1+2 p)}+\frac{\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (3+2 p)}+\frac{1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{3}{2},\frac{1}{2}-p;\frac{5}{2};a^2 x^2\right )\\ \end{align*}
Mathematica [A] time = 0.0837518, size = 102, normalized size = 0.77 \[ \frac{1}{3} \left (c-a^2 c x^2\right )^p \left (x^3 \left (1-a^2 x^2\right )^{-p} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{1}{2}-p,\frac{5}{2},a^2 x^2\right )-\frac{3 \sqrt{1-a^2 x^2} \left (a^2 (2 p+1) x^2+2\right )}{a^3 \left (4 p^2+8 p+3\right )}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.318, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ){x}^{2} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, 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 (a x + 1\right )}{\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{\sqrt{-a^{2} x^{2} + 1}}\,{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{\sqrt{-a^{2} x^{2} + 1}{\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{a x - 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 20.0578, size = 269, normalized size = 2.02 \begin{align*} - \frac{a^{2 p} c^{p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} \frac{1}{2}, 1, p + \frac{3}{2} \\ p + 1, p + \frac{5}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + 1\right )} - \frac{a^{2 p} c^{p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac{3}{2}\right ) \Gamma \left (p + \frac{1}{2}\right ){{}_{3}F_{2}\left (\begin{matrix} 1, - p, - p - \frac{3}{2} \\ \frac{1}{2}, - p - \frac{1}{2} \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 \sqrt{\pi } \Gamma \left (- p - \frac{1}{2}\right ) \Gamma \left (p + 1\right )} - \frac{c^{p}{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^{3}} - \frac{c^{p}{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^{3} \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} c x^{2} + c\right )}^{p} x^{2}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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