Optimal. Leaf size=108 \[ 6 a^2 \sinh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 a^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )-\frac{2 a \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{x}-2 a^2 \sinh ^{-1}(a x)^3+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^4}{2 x^2} \]
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Rubi [A] time = 0.20714, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5661, 5723, 5659, 3716, 2190, 2531, 2282, 6589} \[ 6 a^2 \sinh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 a^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )-\frac{2 a \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{x}-2 a^2 \sinh ^{-1}(a x)^3+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\frac{\sinh ^{-1}(a x)^4}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 5661
Rule 5723
Rule 5659
Rule 3716
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\sinh ^{-1}(a x)^4}{x^3} \, dx &=-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+(2 a) \int \frac{\sinh ^{-1}(a x)^3}{x^2 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+\left (6 a^2\right ) \int \frac{\sinh ^{-1}(a x)^2}{x} \, dx\\ &=-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+\left (6 a^2\right ) \operatorname{Subst}\left (\int x^2 \coth (x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 a^2 \sinh ^{-1}(a x)^3-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}-\left (12 a^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} x^2}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 a^2 \sinh ^{-1}(a x)^3-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\left (12 a^2\right ) \operatorname{Subst}\left (\int x \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 a^2 \sinh ^{-1}(a x)^3-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+6 a^2 \sinh ^{-1}(a x) \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-\left (6 a^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 a^2 \sinh ^{-1}(a x)^3-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+6 a^2 \sinh ^{-1}(a x) \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )\\ &=-2 a^2 \sinh ^{-1}(a x)^3-\frac{2 a \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{x}-\frac{\sinh ^{-1}(a x)^4}{2 x^2}+6 a^2 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+6 a^2 \sinh ^{-1}(a x) \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 a^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )\\ \end{align*}
Mathematica [C] time = 0.259094, size = 113, normalized size = 1.05 \[ -\frac{\sinh ^{-1}(a x)^4}{2 x^2}+\frac{1}{4} a^2 \left (24 \sinh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-12 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )-\frac{8 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{a x}-8 \sinh ^{-1}(a x)^3+24 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+i \pi ^3\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 208, normalized size = 1.9 \begin{align*} -2\,{a}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}-2\,{\frac{a \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}\sqrt{{a}^{2}{x}^{2}+1}}{x}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}}{2\,{x}^{2}}}+6\,{a}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}\ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) +12\,{a}^{2}{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) -12\,{a}^{2}{\it polylog} \left ( 3,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +6\,{a}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}\ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +12\,{a}^{2}{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -12\,{a}^{2}{\it polylog} \left ( 3,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 )^{4}}{2 \, x^{2}} + \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 )^{3}}{a^{3} x^{5} + a x^{3} +{\left (a^{2} x^{4} + x^{2}\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 )^{4}}{x^{3}}, 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}^{4}{\left (a x \right )}}{x^{3}}\, 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 )^{4}}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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