Optimal. Leaf size=50 \[ -2 a \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 a \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]
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Rubi [A] time = 0.0997795, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5760, 4182, 2279, 2391} \[ -2 a \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 a \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]
Antiderivative was successfully verified.
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Rule 5661
Rule 5760
Rule 4182
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\sinh ^{-1}(a x)^2}{x^2} \, dx &=-\frac{\sinh ^{-1}(a x)^2}{x}+(2 a) \int \frac{\sinh ^{-1}(a x)}{x \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}+(2 a) \operatorname{Subst}\left (\int x \text{csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+(2 a) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+(2 a) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )\\ &=-\frac{\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-2 a \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 a \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )\\ \end{align*}
Mathematica [A] time = 0.191734, size = 75, normalized size = 1.5 \[ a \left (2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )-2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )-\sinh ^{-1}(a x) \left (\frac{\sinh ^{-1}(a x)}{a x}-2 \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+2 \log \left (e^{-\sinh ^{-1}(a x)}+1\right )\right )\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.049, size = 107, normalized size = 2.1 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{x}}-2\,a{\it Arcsinh} \left ( ax \right ) \ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -2\,a{\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +2\,a{\it Arcsinh} \left ( ax \right ) \ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +2\,a{\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \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{\log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{x} + \int \frac{2 \,{\left (a^{3} x^{2} + \sqrt{a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{a^{3} x^{4} + a x^{2} +{\left (a^{2} x^{3} + x\right )} \sqrt{a^{2} x^{2} + 1}}\,{d x} \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 (\frac{\operatorname{arsinh}\left (a x\right )^{2}}{x^{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 \frac{\operatorname{asinh}^{2}{\left (a x \right )}}{x^{2}}\, 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{\operatorname{arsinh}\left (a x\right )^{2}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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