Optimal. Leaf size=101 \[ -\frac {\text {Li}_4\left (-e^{a+b x}\right )}{b^3}+\frac {\text {Li}_4\left (e^{a+b x}\right )}{b^3}+\frac {x \text {Li}_3\left (-e^{a+b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b} \]
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Rubi [A] time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6213, 2531, 6609, 2282, 6589} \[ \frac {x \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^2}-\frac {x \text {PolyLog}\left (3,e^{a+b x}\right )}{b^2}-\frac {\text {PolyLog}\left (4,-e^{a+b x}\right )}{b^3}+\frac {\text {PolyLog}\left (4,e^{a+b x}\right )}{b^3}-\frac {x^2 \text {PolyLog}\left (2,-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {PolyLog}\left (2,e^{a+b x}\right )}{2 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 6213
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \tanh ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac {1}{2} \int x^2 \log \left (1-e^{a+b x}\right ) \, dx\right )+\frac {1}{2} \int x^2 \log \left (1+e^{a+b x}\right ) \, dx\\ &=-\frac {x^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {\int x \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b}-\frac {\int x \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {x^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{a+b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{a+b x}\right )}{b^2}-\frac {\int \text {Li}_3\left (-e^{a+b x}\right ) \, dx}{b^2}+\frac {\int \text {Li}_3\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {x^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{a+b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{a+b x}\right )}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac {x^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b}+\frac {x^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{a+b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{a+b x}\right )}{b^2}-\frac {\text {Li}_4\left (-e^{a+b x}\right )}{b^3}+\frac {\text {Li}_4\left (e^{a+b x}\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 149, normalized size = 1.48 \[ \frac {b^3 x^3 \log \left (1-e^{a+b x}\right )-b^3 x^3 \log \left (e^{a+b x}+1\right )+2 b^3 x^3 \tanh ^{-1}\left (e^{a+b x}\right )-3 b^2 x^2 \text {Li}_2\left (-e^{a+b x}\right )+3 b^2 x^2 \text {Li}_2\left (e^{a+b x}\right )+6 b x \text {Li}_3\left (-e^{a+b x}\right )-6 b x \text {Li}_3\left (e^{a+b x}\right )-6 \text {Li}_4\left (-e^{a+b x}\right )+6 \text {Li}_4\left (e^{a+b x}\right )}{6 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.60, size = 248, normalized size = 2.46 \[ \frac {b^{3} x^{3} \log \left (-\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{6 \, b^{3}} \]
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 \[ \int x^{2} \operatorname {artanh}\left (e^{\left (b x + a\right )}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 185, normalized size = 1.83 \[ \frac {x^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{3}-\frac {\polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{6}-\frac {x^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{2 b}+\frac {x \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{6}+\frac {x^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b}-\frac {x \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {a^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{3 b^{3}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) a^{3}}{6 b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{6 b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 142, normalized size = 1.41 \[ \frac {1}{3} \, x^{3} \operatorname {artanh}\left (e^{\left (b x + a\right )}\right ) - \frac {1}{6} \, b {\left (\frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} - \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}}\right )} \]
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 \[ \int x^2\,\mathrm {atanh}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]
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 \[ \int x^{2} \operatorname {atanh}{\left (e^{a} e^{b x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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