Optimal. Leaf size=85 \[ \frac {1}{5} a x^5 \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}+\frac {\left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)} \]
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Rubi [A] time = 0.11, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 43
Rule 266
Rule 364
Rule 764
Rule 6148
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*}
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Mathematica [A] time = 0.07, size = 77, normalized size = 0.91 \[ \frac {1}{5} a x^5 \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}} \left (a^2 (2 p+1) x^2+2\right )}{a^4 \left (4 p^2+8 p+3\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{a x - 1}, x\right ) \]
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 \[ \int \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 47, normalized size = 0.55 \[ \frac {a \,x^{5} \hypergeom \left (\left [\frac {5}{2}, \frac {1}{2}-p \right ], \left [\frac {7}{2}\right ], a^{2} x^{2}\right )}{5}+\frac {x^{4} \hypergeom \left (\left [2, \frac {1}{2}-p \right ], \relax [3], a^{2} x^{2}\right )}{4} \]
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 \[ a \int \frac {x^{4} e^{\left (p \log \left (a x + 1\right ) + p \log \left (-a x + 1\right )\right )}}{\sqrt {a x + 1} \sqrt {-a x + 1}}\,{d x} + \frac {{\left (a^{4} {\left (2 \, p + 1\right )} x^{4} - a^{2} {\left (2 \, p - 1\right )} x^{2} - 2\right )} {\left (-a^{2} x^{2} + 1\right )}^{p}}{\sqrt {-a^{2} x^{2} + 1} {\left (4 \, p^{2} + 8 \, p + 3\right )} a^{4}} \]
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 \[ \int \frac {x^3\,{\left (1-a^2\,x^2\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 19.10, size = 258, normalized size = 3.04 \[ - \frac {a a^{2 p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {5}{2} \\ p + 1, p + \frac {7}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {5}{2} \\ \frac {1}{2}, - p - \frac {3}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} - p - 1, 1 & -1 \\- p - \frac {3}{2}, - p - 1 & 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 \pi a^{4}} - \frac {{G_{3, 3}^{1, 3}\left (\begin {matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a^{4} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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