Optimal. Leaf size=95 \[ -\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac{2 \sinh (a+b x)}{b^3}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{x^2 \sinh (a+b x)}{b}-\frac{x^2 \text{csch}(a+b x)}{b} \]
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Rubi [A] time = 0.117816, antiderivative size = 95, 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 = {5450, 3296, 2637, 5419, 4182, 2279, 2391} \[ -\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac{2 \sinh (a+b x)}{b^3}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{x^2 \sinh (a+b x)}{b}-\frac{x^2 \text{csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5450
Rule 3296
Rule 2637
Rule 5419
Rule 4182
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^2 \cosh (a+b x) \coth ^2(a+b x) \, dx &=\int x^2 \cosh (a+b x) \, dx+\int x^2 \coth (a+b x) \text{csch}(a+b x) \, dx\\ &=-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{x^2 \sinh (a+b x)}{b}+\frac{2 \int x \text{csch}(a+b x) \, dx}{b}-\frac{2 \int x \sinh (a+b x) \, dx}{b}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{x^2 \sinh (a+b x)}{b}+\frac{2 \int \cosh (a+b x) \, dx}{b^2}-\frac{2 \int \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{2 \int \log \left (1+e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{2 \sinh (a+b x)}{b^3}+\frac{x^2 \sinh (a+b x)}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{2 \sinh (a+b x)}{b^3}+\frac{x^2 \sinh (a+b x)}{b}\\ \end{align*}
Mathematica [B] time = 0.438756, size = 230, normalized size = 2.42 \[ \frac{\text{csch}\left (\frac{1}{2} (a+b x)\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right ) \left (4 \sinh (a+b x) \text{PolyLog}\left (2,-e^{-a-b x}\right )-4 \sinh (a+b x) \text{PolyLog}\left (2,e^{-a-b x}\right )+b^2 x^2 \cosh (2 (a+b x))-2 b x \sinh (2 (a+b x))+2 \cosh (2 (a+b x))+4 b x \log \left (1-e^{-a-b x}\right ) \sinh (a+b x)-4 b x \log \left (e^{-a-b x}+1\right ) \sinh (a+b x)+4 a \log \left (1-e^{-a-b x}\right ) \sinh (a+b x)-4 a \log \left (e^{-a-b x}+1\right ) \sinh (a+b x)-4 a \sinh (a+b x) \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )-3 b^2 x^2-2\right )}{4 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.075, size = 185, 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 ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-2\,{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{a{\it Artanh} \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 [A] time = 1.44217, size = 212, normalized size = 2.23 \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 )}} - \frac{2 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac{2 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.41477, size = 1952, normalized size = 20.55 \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} \cosh \left (b x + a\right )^{3} \operatorname{csch}\left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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