Optimal. Leaf size=85 \[ \frac{2 e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.0684383, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 12, 463, 455, 388, 207} \[ \frac{2 e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 463
Rule 455
Rule 388
Rule 207
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \coth ^2(a+b x) \text{csch}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{2 x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (8+4 x^2\right )}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{-12-8 x^2}{-1+x^2} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{2 e^{a+b x}}{b}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 e^{a+b x}}{b}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [C] time = 4.02851, size = 247, normalized size = 2.91 \[ -\frac{e^{-3 (a+b x)} \left (128 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^2 \, _5F_4\left (2,2,2,2,\frac{5}{2};1,1,1,\frac{11}{2};e^{2 (a+b x)}\right )+128 e^{8 (a+b x)} \left (16 e^{2 (a+b x)}+7 e^{4 (a+b x)}+9\right ) \, _4F_3\left (2,2,2,\frac{5}{2};1,1,\frac{11}{2};e^{2 (a+b x)}\right )-21 \left (62725 e^{2 (a+b x)}-12071 e^{4 (a+b x)}-19353 e^{6 (a+b x)}+768 e^{8 (a+b x)}+56595\right )+\frac{315 \left (2924 e^{2 (a+b x)}-2534 e^{4 (a+b x)}-1548 e^{6 (a+b x)}+297 e^{8 (a+b x)}+3773\right ) \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}\right )}{10080 b} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.08, size = 78, normalized size = 0.9 \begin{align*} 2\,{\frac{{{\rm e}^{bx+a}}}{b}}-{\frac{{{\rm e}^{bx+a}} \left ( 5\,{{\rm e}^{2\,bx+2\,a}}-3 \right ) }{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-{\frac{5\,\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,b}}+{\frac{5\,\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{2\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.23156, size = 130, normalized size = 1.53 \begin{align*} -\frac{5 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{5 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} - \frac{9 \, e^{\left (-2 \, b x - 2 \, a\right )} - 5 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2}{b{\left (e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.84851, size = 1295, normalized size = 15.24 \begin{align*} \frac{4 \, \cosh \left (b x + a\right )^{5} + 20 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 4 \, \sinh \left (b x + a\right )^{5} + 2 \,{\left (20 \, \cosh \left (b x + a\right )^{2} - 9\right )} \sinh \left (b x + a\right )^{3} - 18 \, \cosh \left (b x + a\right )^{3} + 2 \,{\left (20 \, \cosh \left (b x + a\right )^{3} - 27 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 5 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 5 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left (10 \, \cosh \left (b x + a\right )^{4} - 27 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 10 \, \cosh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \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 [A] time = 1.18629, size = 97, normalized size = 1.14 \begin{align*} -\frac{\frac{2 \,{\left (5 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 4 \, e^{\left (b x + a\right )} + 5 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 5 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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