Optimal. Leaf size=219 \[ -\frac{5 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{5 i \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac{\sin ^{-1}(a x)}{a^4}+\frac{5 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4} \]
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Rubi [A] time = 0.569595, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6016, 5994, 216, 5952, 4180, 2531, 2282, 6589} \[ -\frac{5 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{5 i \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac{\sin ^{-1}(a x)}{a^4}+\frac{5 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^4} \]
Antiderivative was successfully verified.
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Rule 6016
Rule 5994
Rule 216
Rule 5952
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{x^3 \tanh ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx &=-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac{2 \int \frac{x \tanh ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx}{3 a^2}+\frac{\int \frac{x^2 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{a}\\ &=-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac{\int \frac{\tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{2 a^3}+\frac{2 \int \frac{\tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{a^3}+\frac{\int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}+\frac{\operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^4}+\frac{2 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a^3}\\ &=\frac{\sin ^{-1}(a x)}{a^4}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac{5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{i \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{i \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{(4 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{(4 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac{\sin ^{-1}(a x)}{a^4}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac{5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{(4 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{(4 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac{\sin ^{-1}(a x)}{a^4}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac{5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{(4 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{(4 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ &=\frac{\sin ^{-1}(a x)}{a^4}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}+\frac{5 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac{2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^4}-\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^3}{3 a^2}-\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}+\frac{5 i \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^4}-\frac{5 i \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^4}\\ \end{align*}
Mathematica [A] time = 0.886576, size = 215, normalized size = 0.98 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{3 i \left (10 \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-10 \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+10 \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-10 \text{PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )+5 \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-5 \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+4 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right )\right )}{\sqrt{1-a^2 x^2}}+2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-3 a x \tanh ^{-1}(a x)^2-6 \left (\tanh ^{-1}(a x)^2+1\right ) \tanh ^{-1}(a x)\right )}{6 a^4} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.3, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \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{x^{3} \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} x^{3} \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{x^{3} \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{x^{3} \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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