Optimal. Leaf size=43 \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2} \]
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Rubi [A] time = 0.112312, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6034, 5448, 3298} \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2} \]
Antiderivative was successfully verified.
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Rule 6034
Rule 5448
Rule 3298
Rubi steps
\begin{align*} \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{5 \sinh (2 x)}{32 x}+\frac{\sinh (4 x)}{8 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}\\ &=\frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2}\\ \end{align*}
Mathematica [A] time = 0.198329, size = 43, normalized size = 1. \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.063, size = 33, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{5\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32}}+{\frac{{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8}}+{\frac{{\it Shi} \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32}} \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*} \int \frac{x}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.91039, size = 516, normalized size = 12. \begin{align*} \frac{\logintegral \left (-\frac{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - \logintegral \left (-\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 4 \, \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) - 5 \, \logintegral \left (-\frac{a x - 1}{a x + 1}\right )}{64 \, a^{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}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname{atanh}{\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 )}^{4} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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