Optimal. Leaf size=100 \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]
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Rubi [A] time = 0.464066, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6032, 6028, 5966, 6034, 5448, 12, 3298} \[ \frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 6032
Rule 6028
Rule 5966
Rule 6034
Rule 5448
Rule 12
Rule 3298
Rubi steps
\begin{align*} \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{\int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}+\frac{1}{2} (3 a) \int \frac{x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+2 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+\frac{3 \int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}-\frac{3 \int \frac{1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-3 \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+6 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac{6 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac{6 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}\\ \end{align*}
Mathematica [A] time = 0.189889, size = 96, normalized size = 0.96 \[ -\frac{-\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )-2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )+3 a^2 x^2 \tanh ^{-1}(a x)+a x+\tanh ^{-1}(a x)}{2 a^2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.062, size = 82, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{8\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\frac{{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2}}-{\frac{\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 4\,{\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*} -\frac{2 \, a x +{\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) -{\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}{{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) +{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-a x + 1\right )^{2}} + \int -\frac{2 \,{\left (3 \, a^{2} x^{3} + 5 \, x\right )}}{{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) -{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.99856, size = 606, normalized size = 6.06 \begin{align*} \frac{{\left (2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) +{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) -{\left (a^{4} x^{4} - 2 \, 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 )^{2} - 8 \, a x - 4 \,{\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{4 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2}} \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{x}{a^{6} x^{6} \operatorname{atanh}^{3}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname{atanh}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atanh}^{3}{\left (a x \right )} - \operatorname{atanh}^{3}{\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{x}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname{artanh}\left (a x\right )^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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