Optimal. Leaf size=128 \[ -\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {6 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}+\frac {6 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^2} \]
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Rubi [A] time = 0.18, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5994, 5952, 4180, 2531, 2282, 6589} \[ -\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {6 i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}+\frac {6 \tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a^2} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 5952
Rule 5994
Rule 6589
Rubi steps
\begin {align*} \int \frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}+\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{a}\\ &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}+\frac {3 \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}-\frac {(6 i) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {(6 i) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}-\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^2}\\ &=\frac {6 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a^2}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}+\frac {6 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^2}-\frac {6 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^2}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 157, normalized size = 1.23 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3+6 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-6 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+6 i \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-6 i \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )+3 i \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-3 i \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{a^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \]
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 \[ \int \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {x \arctanh \left (a x \right )^{3}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]
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 \[ \int \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
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 \[ \int \frac {x\,{\mathrm {atanh}\left (a\,x\right )}^3}{\sqrt {1-a^2\,x^2}} \,d x \]
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 \[ \int \frac {x \operatorname {atanh}^{3}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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