Optimal. Leaf size=254 \[ -\frac{i \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac{i \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{x \sqrt{1-a^2 x^2}}{12 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{4 a^3} \]
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Rubi [A] time = 0.852707, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 29, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6014, 6016, 5994, 216, 5952, 4180, 2531, 2282, 6589, 321} \[ -\frac{i \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac{i \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{x \sqrt{1-a^2 x^2}}{12 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{4 a^3} \]
Antiderivative was successfully verified.
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Rule 6014
Rule 6016
Rule 5994
Rule 216
Rule 5952
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rule 321
Rubi steps
\begin{align*} \int x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int \frac{x^4 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx\right )+\int \frac{x^2 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{3}{4} \int \frac{x^2 \tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx+\frac{\int \frac{\tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{2 a^2}+\frac{\int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{a}-\frac{1}{2} a \int \frac{x^3 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac{1}{6} \int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx+\frac{\operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^3}-\frac{3 \int \frac{\tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{8 a^2}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a^2}-\frac{\int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{3 a}-\frac{3 \int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{4 a}\\ &=\frac{x \sqrt{1-a^2 x^2}}{12 a^2}+\frac{\sin ^{-1}(a x)}{a^3}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^3}-\frac{i \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac{i \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^3}-\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{12 a^2}-\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{3 a^2}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{4 a^2}\\ &=\frac{x \sqrt{1-a^2 x^2}}{12 a^2}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac{(3 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}-\frac{(3 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=\frac{x \sqrt{1-a^2 x^2}}{12 a^2}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac{(3 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac{(3 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^3}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=\frac{x \sqrt{1-a^2 x^2}}{12 a^2}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{i \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^3}\\ &=\frac{x \sqrt{1-a^2 x^2}}{12 a^2}-\frac{\sin ^{-1}(a x)}{6 a^3}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac{x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{6 a}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac{\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^3}-\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}+\frac{i \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}-\frac{i \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^3}\\ \end{align*}
Mathematica [A] time = 1.10596, size = 228, normalized size = 0.9 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{i \left (6 \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+6 \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-6 \text{PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-8 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right )\right )}{\sqrt{1-a^2 x^2}}-6 a x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+6 \tanh ^{-1}(a x)+a x \left (3 \tanh ^{-1}(a x)^2+2\right )\right )}{24 a^3} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.303, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}\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 \sqrt{-a^{2} x^{2} + 1} x^{2} \operatorname{artanh}\left (a x\right )^{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 (\sqrt{-a^{2} x^{2} + 1} x^{2} \operatorname{artanh}\left (a x\right )^{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 x^{2} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname{atanh}^{2}{\left (a x \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 \sqrt{-a^{2} x^{2} + 1} x^{2} \operatorname{artanh}\left (a x\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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