Optimal. Leaf size=69 \[ -\frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b} \]
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Rubi [A] time = 0.0614134, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5418, 4180, 2279, 2391} \[ -\frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5418
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^2 \text{sech}(a+b x) \tanh (a+b x) \, dx &=-\frac{x^2 \text{sech}(a+b x)}{b}+\frac{2 \int x \text{sech}(a+b x) \, dx}{b}\\ &=\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b}-\frac{(2 i) \int \log \left (1-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(2 i) \int \log \left (1+i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 i \text{Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac{x^2 \text{sech}(a+b x)}{b}\\ \end{align*}
Mathematica [A] time = 0.435018, size = 125, normalized size = 1.81 \[ -\frac{2 i \left (\text{PolyLog}\left (2,-i e^{a+b x}\right )-\text{PolyLog}\left (2,i e^{a+b x}\right )\right )+b^2 x^2 \text{sech}(a+b x)+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac{1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.054, size = 154, normalized size = 2.2 \begin{align*} -2\,{\frac{{x}^{2}{{\rm e}^{bx+a}}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-{\frac{2\,i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-{\frac{2\,i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+{\frac{2\,i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+{\frac{2\,i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}-{\frac{2\,i{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{2\,i{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-4\,{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + 4 \, \int \frac{x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.06587, size = 1395, normalized size = 20.22 \begin{align*} -\frac{2 \, b^{2} x^{2} \cosh \left (b x + a\right ) + 2 \, b^{2} x^{2} \sinh \left (b x + a\right ) -{\left (2 i \, \cosh \left (b x + a\right )^{2} + 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )^{2} + 2 i\right )}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) -{\left (-2 i \, \cosh \left (b x + a\right )^{2} - 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )^{2} - 2 i\right )}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) -{\left (-2 i \, a \cosh \left (b x + a\right )^{2} - 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )^{2} - 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) -{\left (2 i \, a \cosh \left (b x + a\right )^{2} + 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )^{2} + 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) -{\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right )^{2} +{\left (-4 i \, b x - 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )^{2} - 2 i \, b x - 2 i \, a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) -{\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right )^{2} +{\left (4 i \, b x + 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )^{2} + 2 i \, b x + 2 i \, a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sinh{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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