Optimal. Leaf size=136 \[ \frac {1}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac {a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{2} \text {Li}_3\left (\frac {2}{a x+1}-1\right )-\text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{3} \tanh ^{-1}(a x)^3-\frac {1}{4} \tanh ^{-1}(a x)^2+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
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Rubi [A] time = 0.29, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6030, 5988, 5932, 5948, 6056, 6610, 5994, 5956, 261} \[ -\frac {1}{2} \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )-\tanh ^{-1}(a x) \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )+\frac {1}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac {a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{3} \tanh ^{-1}(a x)^3-\frac {1}{4} \tanh ^{-1}(a x)^2+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
Antiderivative was successfully verified.
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Rule 261
Rule 5932
Rule 5948
Rule 5956
Rule 5988
Rule 5994
Rule 6030
Rule 6056
Rule 6610
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )^2} \, dx &=a^2 \int \frac {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx\\ &=\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {1}{3} \tanh ^{-1}(a x)^3-a \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=-\frac {a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{4} \tanh ^{-1}(a x)^2+\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-(2 a) \int \frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx+\frac {1}{2} a^2 \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {1}{4 \left (1-a^2 x^2\right )}-\frac {a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{4} \tanh ^{-1}(a x)^2+\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+a \int \frac {\text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {1}{4 \left (1-a^2 x^2\right )}-\frac {a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{4} \tanh ^{-1}(a x)^2+\frac {\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {1}{2} \text {Li}_3\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}
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Mathematica [C] time = 0.18, size = 106, normalized size = 0.78 \[ \frac {1}{24} \left (24 \tanh ^{-1}(a x) \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )-12 \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )-8 \tanh ^{-1}(a x)^3+24 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \sinh \left (2 \tanh ^{-1}(a x)\right )+6 \tanh ^{-1}(a x)^2 \cosh \left (2 \tanh ^{-1}(a x)\right )+3 \cosh \left (2 \tanh ^{-1}(a x)\right )+i \pi ^3\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )^{2}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}, x\right ) \]
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 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{2} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.73, size = 1290, normalized size = 9.49 \[ \text {result too large to display} \]
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 \[ \frac {1}{4} \, a^{4} \int \frac {x^{4} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} + \frac {1}{4} \, a^{3} \int \frac {x^{3} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac {1}{32} \, {\left (a {\left (\frac {2}{a^{4} x - a^{3}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} + \frac {4 \, \log \left (-a x + 1\right )}{a^{4} x^{2} - a^{2}}\right )} a^{2} - \frac {1}{4} \, a^{2} \int \frac {x^{2} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac {1}{4} \, a \int \frac {x \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} + \frac {1}{4} \, a \int \frac {x \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac {{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )^{3} + 3 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{2}}{24 \, {\left (a^{2} x^{2} - 1\right )}} + \frac {1}{4} \, \int \frac {\log \left (a x + 1\right )^{2}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac {1}{2} \, \int \frac {\log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} \]
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 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{x\,{\left (a^2\,x^2-1\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{x \left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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