Optimal. Leaf size=156 \[ 12 i a \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-24 a \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+24 a \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-24 i a \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )+24 i a \text{PolyLog}\left (4,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]
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Rubi [A] time = 0.193764, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4627, 4709, 4183, 2531, 6609, 2282, 6589} \[ 12 i a \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-24 a \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+24 a \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-24 i a \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )+24 i a \text{PolyLog}\left (4,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]
Antiderivative was successfully verified.
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Rule 4627
Rule 4709
Rule 4183
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\sin ^{-1}(a x)^4}{x^2} \, dx &=-\frac{\sin ^{-1}(a x)^4}{x}+(4 a) \int \frac{\sin ^{-1}(a x)^3}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sin ^{-1}(a x)^4}{x}+(4 a) \operatorname{Subst}\left (\int x^3 \csc (x) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-(12 a) \operatorname{Subst}\left (\int x^2 \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+(12 a) \operatorname{Subst}\left (\int x^2 \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-(24 i a) \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+(24 i a) \operatorname{Subst}\left (\int x \text{Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-24 a \sin ^{-1}(a x) \text{Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+24 a \sin ^{-1}(a x) \text{Li}_3\left (e^{i \sin ^{-1}(a x)}\right )+(24 a) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )-(24 a) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-24 a \sin ^{-1}(a x) \text{Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+24 a \sin ^{-1}(a x) \text{Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-(24 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )+(24 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )\\ &=-\frac{\sin ^{-1}(a x)^4}{x}-8 a \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-12 i a \sin ^{-1}(a x)^2 \text{Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-24 a \sin ^{-1}(a x) \text{Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+24 a \sin ^{-1}(a x) \text{Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-24 i a \text{Li}_4\left (-e^{i \sin ^{-1}(a x)}\right )+24 i a \text{Li}_4\left (e^{i \sin ^{-1}(a x)}\right )\\ \end{align*}
Mathematica [A] time = 0.266898, size = 198, normalized size = 1.27 \[ a \left (12 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{-i \sin ^{-1}(a x)}\right )+12 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )+24 \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{-i \sin ^{-1}(a x)}\right )-24 \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )-24 i \text{PolyLog}\left (4,e^{-i \sin ^{-1}(a x)}\right )-24 i \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^4}{a x}+i \sin ^{-1}(a x)^4+4 \sin ^{-1}(a x)^3 \log \left (1-e^{-i \sin ^{-1}(a x)}\right )-4 \sin ^{-1}(a x)^3 \log \left (1+e^{i \sin ^{-1}(a x)}\right )-\frac{i \pi ^4}{2}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.077, size = 241, normalized size = 1.5 \begin{align*} -{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{4}}{x}}+4\,a \left ( \arcsin \left ( ax \right ) \right ) ^{3}\ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -4\,a \left ( \arcsin \left ( ax \right ) \right ) ^{3}\ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) -24\,a\arcsin \left ( ax \right ){\it polylog} \left ( 3,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +24\,a\arcsin \left ( ax \right ){\it polylog} \left ( 3,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +12\,ia \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -12\,ia \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +24\,ia{\it polylog} \left ( 4,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) -24\,ia{\it polylog} \left ( 4,-iax-\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{\arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{4} + 4 \, a x \int \frac{\sqrt{-a x + 1} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3}}{\sqrt{a x + 1}{\left (a x - 1\right )} x}\,{d x}}{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{\arcsin \left (a x\right )^{4}}{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{asin}^{4}{\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{\arcsin \left (a x\right )^{4}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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