Optimal. Leaf size=104 \[ \frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}-\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x \sinh (a+b x)}{b^2}+\frac{2 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}+\frac{x^2 \text{sech}(a+b x)}{b} \]
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Rubi [A] time = 0.117554, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {5449, 3296, 2638, 5418, 4180, 2279, 2391} \[ \frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}-\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x \sinh (a+b x)}{b^2}+\frac{2 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}+\frac{x^2 \text{sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5449
Rule 3296
Rule 2638
Rule 5418
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^2 \sinh (a+b x) \tanh ^2(a+b x) \, dx &=\int x^2 \sinh (a+b x) \, dx-\int x^2 \text{sech}(a+b x) \tanh (a+b x) \, dx\\ &=\frac{x^2 \cosh (a+b x)}{b}+\frac{x^2 \text{sech}(a+b x)}{b}-\frac{2 \int x \cosh (a+b x) \, dx}{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 \cosh (a+b x)}{b}+\frac{x^2 \text{sech}(a+b x)}{b}-\frac{2 x \sinh (a+b x)}{b^2}+\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{2 \int \sinh (a+b x) \, dx}{b^2}\\ &=-\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{2 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}+\frac{x^2 \text{sech}(a+b x)}{b}-\frac{2 x \sinh (a+b x)}{b^2}+\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 \cosh (a+b x)}{b^3}+\frac{x^2 \cosh (a+b x)}{b}+\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}-\frac{2 x \sinh (a+b x)}{b^2}\\ \end{align*}
Mathematica [A] time = 0.492211, size = 172, normalized size = 1.65 \[ \frac{2 i \left (\text{PolyLog}\left (2,-i e^{a+b x}\right )-\text{PolyLog}\left (2,i e^{a+b x}\right )\right )+b^2 x^2 \text{sech}(a+b x)+\cosh (b x) \left (\cosh (a) \left (b^2 x^2+2\right )-2 b x \sinh (a)\right )+\sinh (b x) \left (\sinh (a) \left (b^2 x^2+2\right )-2 b x \cosh (a)\right )+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac{1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.074, size = 205, normalized size = 2. \begin{align*}{\frac{ \left ({x}^{2}{b}^{2}-2\,bx+2 \right ){{\rm e}^{bx+a}}}{2\,{b}^{3}}}+{\frac{ \left ({x}^{2}{b}^{2}+2\,bx+2 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{3}}}+2\,{\frac{{{\rm e}^{bx+a}}{x}^{2}}{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{{\left (b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + 2 \, e^{\left (4 \, a\right )}\right )} e^{\left (3 \, b x\right )} + 2 \,{\left (3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 2 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} +{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x\right )}}{2 \,{\left (b^{3} e^{\left (2 \, b x + 3 \, a\right )} + b^{3} e^{a}\right )}} - 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 = 2.47479, size = 2500, normalized size = 24.04 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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