Optimal. Leaf size=158 \[ \frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {\sin ^{-1}(a x)}{a}+\frac {\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]
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Rubi [A] time = 0.15, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5944, 5952, 4180, 2531, 2282, 6589, 216} \[ -\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}-\frac {\sin ^{-1}(a x)}{a}+\frac {\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]
Antiderivative was successfully verified.
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Rule 216
Rule 2282
Rule 2531
Rule 4180
Rule 5944
Rule 5952
Rule 6589
Rubi steps
\begin {align*} \int \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {1}{2} \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx-\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sin ^{-1}(a x)}{a}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a}\\ &=-\frac {\sin ^{-1}(a x)}{a}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \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}-\frac {i \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {i \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {\sin ^{-1}(a x)}{a}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \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}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}-\frac {i \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {\sin ^{-1}(a x)}{a}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \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}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}\\ &=-\frac {\sin ^{-1}(a x)}{a}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{a}+\frac {1}{2} x \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}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 187, normalized size = 1.18 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {i \left (2 \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )\right )}{\sqrt {1-a^2 x^2}}+a x \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)\right )}{2 a} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.80, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2}+1}\, \arctanh \left (a x \right )^{2}\, dx \]
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 \[ \int \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}\,{d x} \]
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 \[ \int {\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^2} \,d x \]
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 \[ \int \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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