Optimal. Leaf size=153 \[ -\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \text{PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]
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Rubi [A] time = 0.1331, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5952, 4180, 2531, 6609, 2282, 6589} \[ -\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \text{PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]
Antiderivative was successfully verified.
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Rule 5952
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac{(3 i) \operatorname{Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac{(3 i) \operatorname{Subst}\left (\int x^2 \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_3\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_3\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}\\ &=\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac{6 i \text{Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac{6 i \text{Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a}\\ \end{align*}
Mathematica [B] time = 0.449595, size = 451, normalized size = 2.95 \[ -\frac{i \left (192 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )+384 \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )-48 \left (\pi -2 i \tanh ^{-1}(a x)\right )^2 \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-192 i \pi \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )+384 \text{PolyLog}\left (4,-i e^{-\tanh ^{-1}(a x)}\right )+384 \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )-16 \tanh ^{-1}(a x)^4-32 i \pi \tanh ^{-1}(a x)^3+24 \pi ^2 \tanh ^{-1}(a x)^2+8 i \pi ^3 \tanh ^{-1}(a x)-64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )-96 i \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+96 i \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+48 \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-8 i \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (\tan \left (\frac{1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )+7 \pi ^4\right )}{64 a} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.377, size = 0, normalized size = 0. \begin{align*} \int{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, dx \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{artanh}\left (a x\right )^{3}}{\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{\sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, 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{atanh}^{3}{\left (a x \right )}}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\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{\operatorname{artanh}\left (a x\right )^{3}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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