Optimal. Leaf size=97 \[ 2 \sinh ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{2 \sinh ^{-1}(a x)}\right )-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]
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Rubi [A] time = 0.122107, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5659, 3716, 2190, 2531, 6609, 2282, 6589} \[ 2 \sinh ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{2 \sinh ^{-1}(a x)}\right )-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]
Antiderivative was successfully verified.
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Rule 5659
Rule 3716
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\sinh ^{-1}(a x)^4}{x} \, dx &=\operatorname{Subst}\left (\int x^4 \coth (x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5-2 \operatorname{Subst}\left (\int \frac{e^{2 x} x^4}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-4 \operatorname{Subst}\left (\int x^3 \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-6 \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+6 \operatorname{Subst}\left (\int x \text{Li}_3\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \operatorname{Subst}\left (\int \text{Li}_4\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac{3}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_4(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )\\ &=-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac{3}{2} \text{Li}_5\left (e^{2 \sinh ^{-1}(a x)}\right )\\ \end{align*}
Mathematica [A] time = 0.0077906, size = 97, normalized size = 1. \[ 2 \sinh ^{-1}(a x)^3 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(a x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{2 \sinh ^{-1}(a x)}\right )-\frac{1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.036, size = 257, normalized size = 2.7 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{5}}{5}}+ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}\ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) +4\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}{\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) -12\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 3,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +24\,{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 4,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) -24\,{\it polylog} \left ( 5,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) + \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}\ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +4\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}{\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -12\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 3,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) +24\,{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 4,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -24\,{\it polylog} \left ( 5,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*} \int \frac{\operatorname{arsinh}\left (a x\right )^{4}}{x}\,{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}, 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}\, 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}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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