Optimal. Leaf size=35 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a}-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]
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Rubi [A] time = 0.0945495, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {5966, 6034, 5448, 12, 3298} \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a}-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 5966
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)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+(2 a) \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a}\\ \end{align*}
Mathematica [A] time = 0.0912557, size = 30, normalized size = 0.86 \[ \frac{\frac{1}{\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)}+\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.066, size = 36, normalized size = 1. \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{2\,{\it Artanh} \left ( ax \right ) }}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) \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*} -4 \, a \int -\frac{x}{{\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 )}\,{d x} + \frac{2}{{\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) -{\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.99044, size = 244, normalized size = 6.97 \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 ) + 4}{2 \,{\left (a^{3} x^{2} - a\right )} \log \left (-\frac{a x + 1}{a x - 1}\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 \frac{1}{\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 \frac{1}{{\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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