Optimal. Leaf size=246 \[ \frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \text{PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^3} \]
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Rubi [A] time = 0.317475, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6028, 5952, 4180, 2531, 6609, 2282, 6589, 5962, 5958} \[ \frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \text{PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^3} \]
Antiderivative was successfully verified.
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Rule 6028
Rule 5952
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 5962
Rule 5958
Rubi steps
\begin{align*} \int \frac{x^2 \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac{\int \frac{\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}-\frac{\int \frac{\tanh ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac{6 \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}\\ &=-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac{(3 i) \operatorname{Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac{(3 i) \operatorname{Subst}\left (\int x^2 \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{(6 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac{(6 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_3\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_3\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=-\frac{6}{a^3 \sqrt{1-a^2 x^2}}+\frac{6 x \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \tanh ^{-1}(a x)^2}{a^3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{3 i \tanh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \tanh ^{-1}(a x) \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{6 i \text{Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{6 i \text{Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ \end{align*}
Mathematica [B] time = 0.936025, size = 541, normalized size = 2.2 \[ \frac{192 i \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-192 \pi \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )+384 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-384 i \tanh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )-48 i \left (\pi -2 i \tanh ^{-1}(a x)\right )^2 \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-48 i \pi ^2 \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-192 \pi \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )+192 \pi \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )+384 i \text{PolyLog}\left (4,-i e^{-\tanh ^{-1}(a x)}\right )+384 i \text{PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )-\frac{384}{\sqrt{1-a^2 x^2}}+\frac{64 a x \tanh ^{-1}(a x)^3}{\sqrt{1-a^2 x^2}}-\frac{192 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}}+\frac{384 a x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}-16 i \tanh ^{-1}(a x)^4+32 \pi \tanh ^{-1}(a x)^3+24 i \pi ^2 \tanh ^{-1}(a x)^2-8 \pi ^3 \tanh ^{-1}(a x)-64 i \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+64 i \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+96 \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-96 \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+48 i \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 i \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )-8 \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+8 \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )-8 \pi ^3 \log \left (\tan \left (\frac{1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )+7 i \pi ^4}{64 a^3} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.276, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{3} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}\, 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^{2} \operatorname{artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{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^{2} \operatorname{artanh}\left (a x\right )^{3}}{a^{4} x^{4} - 2 \, 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^{2} \operatorname{atanh}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{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{x^{2} \operatorname{artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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