Optimal. Leaf size=171 \[ -\frac{8 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{15 a}+\frac{a^2 x^3}{30}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{8 \tanh ^{-1}(a x)^2}{15 a}-\frac{16 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a}-\frac{11 x}{30} \]
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Rubi [A] time = 0.133453, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5944, 5910, 5984, 5918, 2402, 2315, 8} \[ -\frac{8 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{15 a}+\frac{a^2 x^3}{30}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{8 \tanh ^{-1}(a x)^2}{15 a}-\frac{16 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{15 a}-\frac{11 x}{30} \]
Antiderivative was successfully verified.
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Rule 5944
Rule 5910
Rule 5984
Rule 5918
Rule 2402
Rule 2315
Rule 8
Rubi steps
\begin{align*} \int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2 \, dx &=\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{1}{10} \int \left (1-a^2 x^2\right ) \, dx+\frac{4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac{x}{10}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{4 \int 1 \, dx}{15}+\frac{8}{15} \int \tanh ^{-1}(a x)^2 \, dx\\ &=-\frac{11 x}{30}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{1}{15} (16 a) \int \frac{x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-\frac{11 x}{30}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{8 \tanh ^{-1}(a x)^2}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{16}{15} \int \frac{\tanh ^{-1}(a x)}{1-a x} \, dx\\ &=-\frac{11 x}{30}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{8 \tanh ^{-1}(a x)^2}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{16 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{15 a}+\frac{16}{15} \int \frac{\log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac{11 x}{30}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{8 \tanh ^{-1}(a x)^2}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{16 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{15 a}-\frac{16 \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a x}\right )}{15 a}\\ &=-\frac{11 x}{30}+\frac{a^2 x^3}{30}+\frac{4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}{15 a}+\frac{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}{10 a}+\frac{8 \tanh ^{-1}(a x)^2}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^2+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2-\frac{16 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{15 a}-\frac{8 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{15 a}\\ \end{align*}
Mathematica [A] time = 0.653835, size = 99, normalized size = 0.58 \[ \frac{16 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+a x \left (a^2 x^2-11\right )+2 \left (3 a^2 x^2+9 a x+8\right ) (a x-1)^3 \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \left (3 a^4 x^4-14 a^2 x^2-32 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )+11\right )}{30 a} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.047, size = 216, normalized size = 1.3 \begin{align*}{\frac{{a}^{4} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{5}}{5}}-{\frac{2\,{a}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{3}}{3}}+x \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+{\frac{{a}^{3}{\it Artanh} \left ( ax \right ){x}^{4}}{10}}-{\frac{7\,a{\it Artanh} \left ( ax \right ){x}^{2}}{15}}+{\frac{8\,{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{15\,a}}+{\frac{8\,{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{15\,a}}+{\frac{{x}^{3}{a}^{2}}{30}}-{\frac{11\,x}{30}}-{\frac{11\,\ln \left ( ax-1 \right ) }{60\,a}}+{\frac{11\,\ln \left ( ax+1 \right ) }{60\,a}}+{\frac{2\, \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{15\,a}}-{\frac{8}{15\,a}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{4\,\ln \left ( ax-1 \right ) }{15\,a}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{4}{15\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{4\,\ln \left ( ax+1 \right ) }{15\,a}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{2\, \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{15\,a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.963282, size = 236, normalized size = 1.38 \begin{align*} \frac{1}{60} \, a^{2}{\left (\frac{2 \, a^{3} x^{3} - 22 \, a x - 8 \, \log \left (a x + 1\right )^{2} + 16 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + 8 \, \log \left (a x - 1\right )^{2} - 11 \, \log \left (a x - 1\right )}{a^{3}} - \frac{32 \,{\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{11 \, \log \left (a x + 1\right )}{a^{3}}\right )} + \frac{1}{30} \,{\left (3 \, a^{2} x^{4} - 14 \, x^{2} + \frac{16 \, \log \left (a x + 1\right )}{a^{2}} + \frac{16 \, \log \left (a x - 1\right )}{a^{2}}\right )} a \operatorname{artanh}\left (a x\right ) + \frac{1}{15} \,{\left (3 \, a^{4} x^{5} - 10 \, a^{2} x^{3} + 15 \, x\right )} \operatorname{artanh}\left (a x\right )^{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname{atanh}^{2}{\left (a x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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