Optimal. Leaf size=161 \[ \frac {\text {ArcSin}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^3}-\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^3} \]
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Rubi [A]
time = 0.25, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 = {6163, 6141,
222, 6099, 4265, 2611, 2320, 6724} \begin {gather*} \frac {\text {ArcSin}(a x)}{a^3}+\frac {\tanh ^{-1}(a x)^2 \text {ArcTan}\left (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 {i \tanh ^{-1}(a x) \text {Li}_2\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 {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 222
Rule 2320
Rule 2611
Rule 4265
Rule 6099
Rule 6141
Rule 6163
Rule 6724
Rubi steps
\begin {align*} \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 {\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 {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^3}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=\frac {\sin ^{-1}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^3}-\frac {i \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac {i \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=\frac {\sin ^{-1}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{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 {i \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {i \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=\frac {\sin ^{-1}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{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 {i \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=\frac {\sin ^{-1}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^2}+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{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 {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}\\ \end {align*}
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Mathematica [A]
time = 0.53, size = 188, normalized size = 1.17 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-2 \tanh ^{-1}(a x)-a x \tanh ^{-1}(a x)^2-\frac {i \left (4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )+\tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+2 \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{2 a^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.97, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \arctanh \left (a x \right )^{2}}{\sqrt {-a^{2} x^{2}+1}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\mathrm {atanh}\left (a\,x\right )}^2}{\sqrt {1-a^2\,x^2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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