Optimal. Leaf size=65 \[ -\frac {\text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}+\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^3}{3} \]
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Rubi [A] time = 0.14, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3718, 2190, 2531, 2282, 6589} \[ \frac {x \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^3}{3} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 6589
Rubi steps
\begin {align*} \int x^2 \tanh (a+b x) \, dx &=-\frac {x^3}{3}+2 \int \frac {e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}} \, dx\\ &=-\frac {x^3}{3}+\frac {x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac {2 \int x \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x^3}{3}+\frac {x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac {\int \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^3}{3}+\frac {x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^3}\\ &=-\frac {x^3}{3}+\frac {x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac {\text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}\\ \end {align*}
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Mathematica [A] time = 2.15, size = 66, normalized size = 1.02 \[ \frac {2 b^2 x^2 \left (3 \log \left (e^{-2 (a+b x)}+1\right )+b x\right )-6 b x \text {Li}_2\left (-e^{-2 (a+b x)}\right )-3 \text {Li}_3\left (-e^{-2 (a+b x)}\right )}{6 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.81, size = 207, normalized size = 3.18 \[ -\frac {b^{3} x^{3} - 6 \, b x {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 \, b x {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{3 \, b^{3}} \]
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 \[ \int x^{2} \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 94, normalized size = 1.45 \[ -\frac {x^{3}}{3}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 a^{2} x}{b^{2}}+\frac {4 a^{3}}{3 b^{3}}+\frac {x^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}+\frac {x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}-\frac {\polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 63, normalized size = 0.97 \[ -\frac {1}{3} \, x^{3} + \frac {2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,\mathrm {sinh}\left (a+b\,x\right )}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
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 \[ \int x^{2} \sinh {\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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