Optimal. Leaf size=133 \[ \frac {1}{3} x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+3)}-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+1)} \]
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Rubi [A] time = 0.18, antiderivative size = 133, normalized size of antiderivative = 1.00, 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 43
Rule 266
Rule 364
Rule 764
Rule 6148
Rule 6153
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*}
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Mathematica [A] time = 0.09, size = 102, normalized size = 0.77 \[ \frac {1}{3} \left (c-a^2 c x^2\right )^p \left (x^3 \left (1-a^2 x^2\right )^{-p} \, _2F_1\left (\frac {3}{2},\frac {1}{2}-p;\frac {5}{2};a^2 x^2\right )-\frac {3 \sqrt {1-a^2 x^2} \left (a^2 (2 p+1) x^2+2\right )}{a^3 \left (4 p^2+8 p+3\right )}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{a x - 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{2} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\left (c-a^2\,c\,x^2\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 16.70, size = 269, normalized size = 2.02 \[ - \frac {a^{2 p} c^{p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {3}{2} \\ p + 1, p + \frac {5}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} c^{p} x^{3} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {3}{2} \\ \frac {1}{2}, - p - \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {c^{p} {G_{3, 3}^{2, 2}\left (\begin {matrix} - p - 1, 1 & -1 \\- p - \frac {3}{2}, - p - 1 & 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 \pi a^{3}} - \frac {c^{p} {G_{3, 3}^{1, 3}\left (\begin {matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a^{3} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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