Optimal. Leaf size=83 \[ \frac {\text {Li}_3\left (-e^{-a-b x}\right )}{2 b^2}-\frac {\text {Li}_3\left (e^{-a-b x}\right )}{2 b^2}+\frac {x \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x \text {Li}_2\left (e^{-a-b x}\right )}{2 b} \]
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Rubi [A] time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6214, 2531, 2282, 6589} \[ \frac {\text {PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac {\text {PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2}+\frac {x \text {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \text {PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 6214
Rule 6589
Rubi steps
\begin {align*} \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac {1}{2} \int x \log \left (1-e^{-a-b x}\right ) \, dx\right )+\frac {1}{2} \int x \log \left (1+e^{-a-b x}\right ) \, dx\\ &=\frac {x \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x \text {Li}_2\left (e^{-a-b x}\right )}{2 b}-\frac {\int \text {Li}_2\left (-e^{-a-b x}\right ) \, dx}{2 b}+\frac {\int \text {Li}_2\left (e^{-a-b x}\right ) \, dx}{2 b}\\ &=\frac {x \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x \text {Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}\\ &=\frac {x \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x \text {Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac {\text {Li}_3\left (-e^{-a-b x}\right )}{2 b^2}-\frac {\text {Li}_3\left (e^{-a-b x}\right )}{2 b^2}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 113, normalized size = 1.36 \[ \frac {b^2 x^2 \log \left (1-e^{a+b x}\right )-b^2 x^2 \log \left (e^{a+b x}+1\right )+2 b^2 x^2 \coth ^{-1}\left (e^{a+b x}\right )-2 b x \text {Li}_2\left (-e^{a+b x}\right )+2 b x \text {Li}_2\left (e^{a+b x}\right )+2 \text {Li}_3\left (-e^{a+b x}\right )-2 \text {Li}_3\left (e^{a+b x}\right )}{4 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.64, size = 198, normalized size = 2.39 \[ \frac {b^{2} x^{2} \log \left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, b x {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, b x {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{2}} \]
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 \[ \int x \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 153, normalized size = 1.84 \[ \frac {x^{2} \mathrm {arccoth}\left ({\mathrm e}^{b x +a}\right )}{2}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{2}}{4}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) a^{2}}{4 b^{2}}-\frac {x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{2 b}+\frac {\polylog \left (3, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{4}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{4 b^{2}}+\frac {x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b}-\frac {\polylog \left (3, {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{2 b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 108, normalized size = 1.30 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right ) - \frac {1}{4} \, b {\left (\frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} - \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}}\right )} \]
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 \[ \int x\,\mathrm {acoth}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]
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 \[ \int x \operatorname {acoth}{\left (e^{a} e^{b x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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