Optimal. Leaf size=65 \[ -\frac {x^2}{b}+\frac {x^3}{3}+\frac {2 x \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {\text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac {x^2 \tanh (a+b x)}{b} \]
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Rubi [A]
time = 0.10, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3801, 3799,
2221, 2317, 2438, 30} \begin {gather*} \frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {2 x \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac {x^2 \tanh (a+b x)}{b}-\frac {x^2}{b}+\frac {x^3}{3} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2221
Rule 2317
Rule 2438
Rule 3799
Rule 3801
Rubi steps
\begin {align*} \int x^2 \tanh ^2(a+b x) \, dx &=-\frac {x^2 \tanh (a+b x)}{b}+\frac {2 \int x \tanh (a+b x) \, dx}{b}+\int x^2 \, dx\\ &=-\frac {x^2}{b}+\frac {x^3}{3}-\frac {x^2 \tanh (a+b x)}{b}+\frac {4 \int \frac {e^{2 (a+b x)} x}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac {x^2}{b}+\frac {x^3}{3}+\frac {2 x \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {x^2 \tanh (a+b x)}{b}-\frac {2 \int \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^2}{b}+\frac {x^3}{3}+\frac {2 x \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {x^2 \tanh (a+b x)}{b}-\frac {\text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{b^3}\\ &=-\frac {x^2}{b}+\frac {x^3}{3}+\frac {2 x \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {x^2 \tanh (a+b x)}{b}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 2.23, size = 168, normalized size = 2.58 \begin {gather*} \frac {3 i b \pi x+b^3 x^3-3 i \pi \log \left (1+e^{2 b x}\right )+6 b x \log \left (1-e^{-2 \left (b x+\tanh ^{-1}(\coth (a))\right )}\right )+3 i \pi \log (\cosh (b x))+6 \tanh ^{-1}(\coth (a)) \left (b x+\log \left (1-e^{-2 \left (b x+\tanh ^{-1}(\coth (a))\right )}\right )-\log \left (i \sinh \left (b x+\tanh ^{-1}(\coth (a))\right )\right )\right )-3 \text {PolyLog}\left (2,e^{-2 \left (b x+\tanh ^{-1}(\coth (a))\right )}\right )-3 b^2 x^2 \text {sech}(a) \text {sech}(a+b x) \sinh (b x)-3 b^2 e^{-\tanh ^{-1}(\coth (a))} x^2 \sqrt {-\text {csch}^2(a)} \tanh (a)}{3 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.11, size = 99, normalized size = 1.52
method | result | size |
risch | \(\frac {x^{3}}{3}+\frac {2 x^{2}}{\left ({\mathrm e}^{2 b x +2 a}+1\right ) b}-\frac {2 x^{2}}{b}-\frac {4 a x}{b^{2}}-\frac {2 a^{2}}{b^{3}}+\frac {2 x \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b^{2}}+\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}+\frac {4 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}\) | \(99\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.33, size = 84, normalized size = 1.29 \begin {gather*} -\frac {2 \, x^{2}}{b} + \frac {b x^{3} e^{\left (2 \, b x + 2 \, a\right )} + b x^{3} + 6 \, x^{2}}{3 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] Result contains complex when optimal does not.
time = 0.39, size = 515, normalized size = 7.92 \begin {gather*} \frac {b^{3} x^{3} + {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 6 \, a^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 6 \, {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} + a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - 6 \, {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} + a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 6 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} + b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} + b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{3 \, {\left (b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} + b^{3}\right )}} \end {gather*}
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 \begin {gather*} \int x^{2} \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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