Optimal. Leaf size=197 \[ 2 a \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-2 a \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+2 i a \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )+2 i a \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-4 a \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)-2 a \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \]
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Rubi [A] time = 0.382518, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6014, 6008, 6018, 5952, 4180, 2531, 2282, 6589} \[ 2 a \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-2 a \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+2 i a \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )+2 i a \text{PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-4 a \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)-2 a \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \]
Antiderivative was successfully verified.
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Rule 6014
Rule 6008
Rule 6018
Rule 5952
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x^2} \, dx &=-\left (a^2 \int \frac{\tanh ^{-1}(a x)^2}{\sqrt{1-a^2 x^2}} \, dx\right )+\int \frac{\tanh ^{-1}(a x)^2}{x^2 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-a \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )+(2 a) \int \frac{\tanh ^{-1}(a x)}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-2 a \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-4 a \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )+2 a \text{Li}_2\left (-\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-2 a \text{Li}_2\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )+(2 i a) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )-(2 i a) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-2 a \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-4 a \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )+2 i a \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+2 a \text{Li}_2\left (-\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-2 a \text{Li}_2\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-(2 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )+(2 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-2 a \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-4 a \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )+2 i a \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+2 a \text{Li}_2\left (-\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-2 a \text{Li}_2\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-(2 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )+(2 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )\\ &=-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-2 a \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-4 a \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )+2 i a \tanh ^{-1}(a x) \text{Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )-2 i a \tanh ^{-1}(a x) \text{Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+2 a \text{Li}_2\left (-\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-2 a \text{Li}_2\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )-2 i a \text{Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )+2 i a \text{Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )\\ \end{align*}
Mathematica [A] time = 0.726134, size = 223, normalized size = 1.13 \[ a \left (2 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-2 i \tanh ^{-1}(a x) \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+2 \text{PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-2 \text{PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )+2 i \text{PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-2 i \text{PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)^2}{a x}+i \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )\right ) \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.273, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{{x}^{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 \frac{\sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )^{2}}{x^{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} \operatorname{artanh}\left (a x\right )^{2}}{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{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname{atanh}^{2}{\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{\sqrt{-a^{2} x^{2} + 1} \operatorname{artanh}\left (a x\right )^{2}}{x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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