Optimal. Leaf size=154 \[ -\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{a^2 x^2+1}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{a^2 x^2+1}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac{2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a} \]
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Rubi [A] time = 0.195573, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {5966, 5996, 6034, 5448, 12, 3298} \[ -\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{a^2 x^2+1}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{a^2 x^2+1}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac{2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a} \]
Antiderivative was successfully verified.
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Rule 5966
Rule 5996
Rule 6034
Rule 5448
Rule 12
Rule 3298
Rubi steps
\begin{align*} \int \frac{1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^6} \, dx &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac{1}{5} (2 a) \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5} \, dx\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}+\frac{1}{15} (2 a) \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1}{15} (4 a) \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}\\ &=-\frac{1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac{x}{10 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac{1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac{x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}\\ \end{align*}
Mathematica [A] time = 0.123685, size = 101, normalized size = 0.66 \[ \frac{4 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+2 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^4+\left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2+2 a x \tanh ^{-1}(a x)^3+3 a x \tanh ^{-1}(a x)+6}{30 a \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.067, size = 98, normalized size = 0.6 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{10\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{5}}}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{10\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{5}}}-{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{20\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{30\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}}-{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{30\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{15\,{\it Artanh} \left ( ax \right ) }}+{\frac{2\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{15}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -8 \, a \int -\frac{x}{15 \,{\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )\right )}}\,{d x} + \frac{2 \,{\left (2 \, a x \log \left (a x + 1\right )^{3} +{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{4} +{\left (a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )^{4} - 2 \,{\left (a x + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3} + 12 \, a x \log \left (a x + 1\right ) + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 2 \,{\left (a^{2} x^{2} + 3 \, a x \log \left (a x + 1\right ) + 3 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 1\right )} \log \left (-a x + 1\right )^{2} - 2 \,{\left (3 \, a x \log \left (a x + 1\right )^{2} + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 6 \, a x + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right ) + 48\right )}}{15 \,{\left ({\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{5} - 5 \,{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{4} \log \left (-a x + 1\right ) + 10 \,{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right )^{2} - 10 \,{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{3} + 5 \,{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{4} -{\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )^{5}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.01265, size = 473, normalized size = 3.07 \begin{align*} \frac{{\left ({\left (a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) -{\left (a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x - 1}{a x + 1}\right )\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{5} + 4 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right )^{3} + 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{4} + 24 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) + 4 \,{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 96}{15 \,{\left (a^{3} x^{2} - a\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{5}} \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 \frac{1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname{atanh}^{6}{\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 \frac{1}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname{artanh}\left (a x\right )^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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