Optimal. Leaf size=62 \[ -\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{15 a^3}-\frac {a x^4}{20}+\frac {1}{3} x^3 \tanh ^{-1}(a x)+\frac {x^2}{15 a} \]
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Rubi [A] time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6014, 5916, 266, 43} \[ \frac {\log \left (1-a^2 x^2\right )}{15 a^3}-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac {a x^4}{20}+\frac {x^2}{15 a}+\frac {1}{3} x^3 \tanh ^{-1}(a x) \]
Antiderivative was successfully verified.
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Rule 43
Rule 266
Rule 5916
Rule 6014
Rubi steps
\begin {align*} \int x^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^4 \tanh ^{-1}(a x) \, dx\right )+\int x^2 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac {1}{3} a \int \frac {x^3}{1-a^2 x^2} \, dx+\frac {1}{5} a^3 \int \frac {x^5}{1-a^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )+\frac {1}{10} a^3 \operatorname {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {1}{10} a^3 \operatorname {Subst}\left (\int \left (-\frac {1}{a^4}-\frac {x}{a^2}-\frac {1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{15 a}-\frac {a x^4}{20}+\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{15 a^3}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 62, normalized size = 1.00 \[ -\frac {1}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{15 a^3}-\frac {a x^4}{20}+\frac {1}{3} x^3 \tanh ^{-1}(a x)+\frac {x^2}{15 a} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 68, normalized size = 1.10 \[ -\frac {3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 2 \, {\left (3 \, a^{5} x^{5} - 5 \, a^{3} x^{3}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 4 \, \log \left (a^{2} x^{2} - 1\right )}{60 \, a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 268, normalized size = 4.32 \[ \frac {2}{15} \, a {\left (\frac {\log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {\log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{4}} - \frac {\frac {{\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {4 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {a x + 1}{a x - 1}}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}} - \frac {{\left (\frac {15 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {5 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 59, normalized size = 0.95 \[ -\frac {a^{2} x^{5} \arctanh \left (a x \right )}{5}+\frac {x^{3} \arctanh \left (a x \right )}{3}-\frac {x^{4} a}{20}+\frac {x^{2}}{15 a}+\frac {\ln \left (a x -1\right )}{15 a^{3}}+\frac {\ln \left (a x +1\right )}{15 a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 65, normalized size = 1.05 \[ -\frac {1}{60} \, a {\left (\frac {3 \, a^{2} x^{4} - 4 \, x^{2}}{a^{2}} - \frac {4 \, \log \left (a x + 1\right )}{a^{4}} - \frac {4 \, \log \left (a x - 1\right )}{a^{4}}\right )} - \frac {1}{15} \, {\left (3 \, a^{2} x^{5} - 5 \, x^{3}\right )} \operatorname {artanh}\left (a x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.91, size = 53, normalized size = 0.85 \[ \frac {\frac {\ln \left (a^2\,x^2-1\right )}{15}+\frac {a^2\,x^2}{15}}{a^3}-\frac {a\,x^4}{20}+\frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{3}-\frac {a^2\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.22, size = 63, normalized size = 1.02 \[ \begin {cases} - \frac {a^{2} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} - \frac {a x^{4}}{20} + \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {x^{2}}{15 a} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{15 a^{3}} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{15 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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