Optimal. Leaf size=73 \[ \frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.044398, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2282, 12, 288, 207} \[ \frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 288
Rule 207
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \text{csch}^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{8 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{8 \operatorname{Subst}\left (\int \frac{x^4}{\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{6 \operatorname{Subst}\left (\int \frac{x^2}{\left (-1+x^2\right )^2} \, 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{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, 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{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0583474, size = 61, normalized size = 0.84 \[ \frac{3 e^{a+b x}-5 e^{3 (a+b x)}-3 \left (e^{2 (a+b x)}-1\right )^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b \left (e^{2 (a+b x)}-1\right )^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.037, size = 67, normalized size = 0.9 \begin{align*} -{\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{3\,\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,b}}+{\frac{3\,\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 = 1.05462, size = 119, normalized size = 1.63 \begin{align*} -\frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac{5 \, e^{\left (-b x - a\right )} - 3 \, e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.72627, size = 1098, normalized size = 15.04 \begin{align*} -\frac{10 \, \cosh \left (b x + a\right )^{3} + 30 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 10 \, \sinh \left (b x + a\right )^{3} + 3 \,{\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 ) - 3 \,{\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 ) + 6 \,{\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 6 \, \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] time = 0., size = 0, normalized size = 0. \begin{align*} e^{2 a} \int e^{2 b x} \operatorname{csch}^{3}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15136, size = 105, normalized size = 1.44 \begin{align*} -\frac{{\left (3 \, e^{\left (-2 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right ) + \frac{2 \,{\left (5 \, e^{\left (3 \, b x + 2 \, a\right )} - 3 \, e^{\left (b x\right )}\right )} e^{\left (-a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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