Optimal. Leaf size=136 \[ \frac{x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{m+1}{2},\frac{1}{2}-p,\frac{m+3}{2},a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{m+2}{2},\frac{1}{2}-p,\frac{m+4}{2},a^2 x^2\right )}{m+2} \]
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Rubi [A] time = 0.149491, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6153, 6148, 808, 364} \[ \frac{x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{m+1}{2},\frac{1}{2}-p;\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{m+2}{2},\frac{1}{2}-p;\frac{m+4}{2};a^2 x^2\right )}{m+2} \]
Antiderivative was successfully verified.
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Rule 6153
Rule 6148
Rule 808
Rule 364
Rubi steps
\begin{align*} \int e^{\tanh ^{-1}(a x)} x^m \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^m \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^m (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^m \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^{1+m} \left (1-a^2 x^2\right )^{-\frac{1}{2}+p} \, dx\\ &=\frac{x^{1+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{1+m}{2},\frac{1}{2}-p;\frac{3+m}{2};a^2 x^2\right )}{1+m}+\frac{a x^{2+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{2+m}{2},\frac{1}{2}-p;\frac{4+m}{2};a^2 x^2\right )}{2+m}\\ \end{align*}
Mathematica [A] time = 0.041569, size = 114, normalized size = 0.84 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac{x^{m+1} \text{Hypergeometric2F1}\left (\frac{m+1}{2},\frac{1}{2}-p,\frac{m+1}{2}+1,a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{m+2}{2},\frac{1}{2}-p,\frac{m+2}{2}+1,a^2 x^2\right )}{m+2}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.329, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ){x}^{m} \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^{m}}{\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^{m}}{a x - 1}, 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 (a x + 1\right )}{\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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