Optimal. Leaf size=70 \[ \frac {x^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]
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Rubi [A] time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6150, 82, 73, 364} \[ \frac {x^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]
Antiderivative was successfully verified.
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Rule 73
Rule 82
Rule 364
Rule 6150
Rubi steps
\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx\\ &=a \int \frac {x^{1+m}}{(1-a x)^2 (1+a x)^2} \, dx+\int \frac {x^m}{(1-a x)^2 (1+a x)^2} \, dx\\ &=a \int \frac {x^{1+m}}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 67, normalized size = 0.96 \[ x^{m+1} \left (\frac {a x \, _2F_1\left (2,\frac {m}{2}+1;\frac {m}{2}+2;a^2 x^2\right )}{m+2}+\frac {\, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m}}{a^{3} x^{3} - a^{2} x^{2} - a x + 1}, x\right ) \]
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 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.26, size = 177, normalized size = 2.53 \[ -\frac {\left (-a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (-m -2\right )}{\left (2+m \right ) \left (-a^{2} x^{2}+1\right )}+\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} m \Phi \left (a^{2} x^{2}, 1, \frac {m}{2}\right )}{2}\right )}{2 a}+\frac {\left (-a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (-\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-1-m \right )}{\left (1+m \right ) \left (-2 a^{2} x^{2}+2\right )}+\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-\frac {m^{2}}{4}+\frac {1}{4}\right ) \Phi \left (a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}\right )}{2} \]
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 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]
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 \[ \int \frac {x^m\,\left (a\,x+1\right )}{{\left (a^2\,x^2-1\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 29.32, size = 673, normalized size = 9.61 \[ - \frac {a^{2} m^{2} x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{2} x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + a \left (- \frac {a^{2} m^{2} x^{4} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {2 a^{2} m x^{4} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {m^{2} x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {2 m x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {2 m x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {4 x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )}\right ) + \frac {m^{2} x x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {2 m x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {x x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {2 x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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