Optimal. Leaf size=115 \[ \frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}-\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {2 \tanh ^{-1}(a x)^2}{3 a}-\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{3 a}-\frac {x}{3} \]
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Rubi [A] time = 0.10, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5944, 5910, 5984, 5918, 2402, 2315, 8} \[ -\frac {2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{3 a}+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {2 \tanh ^{-1}(a x)^2}{3 a}-\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{3 a}-\frac {x}{3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2315
Rule 2402
Rule 5910
Rule 5918
Rule 5944
Rule 5984
Rubi steps
\begin {align*} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx &=\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {\int 1 \, dx}{3}+\frac {2}{3} \int \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {1}{3} (4 a) \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4}{3} \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}+\frac {4}{3} \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{3 a}\\ &=-\frac {x}{3}+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{3 a}+\frac {2 \tanh ^{-1}(a x)^2}{3 a}+\frac {2}{3} x \tanh ^{-1}(a x)^2+\frac {1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {4 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{3 a}-\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{3 a}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 71, normalized size = 0.62 \[ -\frac {\tanh ^{-1}(a x) \left (a^2 x^2+4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-1\right )-2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+a x+(a x-1)^2 (a x+2) \tanh ^{-1}(a x)^2}{3 a} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]
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 \[ \int -{\left (a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 182, normalized size = 1.58 \[ -\frac {a^{2} \arctanh \left (a x \right )^{2} x^{3}}{3}+x \arctanh \left (a x \right )^{2}-\frac {a \arctanh \left (a x \right ) x^{2}}{3}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{3 a}+\frac {2 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{3 a}-\frac {x}{3}-\frac {\ln \left (a x -1\right )}{6 a}+\frac {\ln \left (a x +1\right )}{6 a}+\frac {\ln \left (a x -1\right )^{2}}{6 a}-\frac {2 \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{3 a}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{3 a}-\frac {\ln \left (a x +1\right )^{2}}{6 a}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{3 a}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{3 a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 144, normalized size = 1.25 \[ -\frac {1}{6} \, a^{2} {\left (\frac {2 \, a x + \log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2} + \log \left (a x - 1\right )}{a^{3}} + \frac {4 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a^{3}} - \frac {\log \left (a x + 1\right )}{a^{3}}\right )} - \frac {1}{3} \, {\left (x^{2} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a \operatorname {artanh}\left (a x\right ) - \frac {1}{3} \, {\left (a^{2} x^{3} - 3 \, x\right )} \operatorname {artanh}\left (a x\right )^{2} \]
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 \[ -\int {\mathrm {atanh}\left (a\,x\right )}^2\,\left (a^2\,x^2-1\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \int a^{2} x^{2} \operatorname {atanh}^{2}{\left (a x \right )}\, dx - \int \left (- \operatorname {atanh}^{2}{\left (a x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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