Optimal. Leaf size=42 \[ \frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
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Rubi [A] time = 0.0427926, antiderivative size = 42, 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, 266, 43} \[ \frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 266
Rule 43
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \text{csch}^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{4 x^3}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{x^3}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x}{(1-x)^2} \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{1}{(-1+x)^2}+\frac{1}{-1+x}\right ) \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.0489636, size = 37, normalized size = 0.88 \[ \frac{2 \left (\frac{1}{1-e^{2 a+2 b x}}+\log \left (1-e^{2 a+2 b x}\right )\right )}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.032, size = 43, normalized size = 1. \begin{align*} -4\,{\frac{a}{b}}-2\,{\frac{1}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+2\,{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }{b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.45041, size = 84, normalized size = 2. \begin{align*} 4 \, x + \frac{4 \, a}{b} + \frac{2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac{2}{b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.77485, size = 286, normalized size = 6.81 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) - 1\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \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}^{2}{\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 = 1.15745, size = 65, normalized size = 1.55 \begin{align*} \frac{2 \,{\left (e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right ) - \frac{e^{\left (2 \, b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )} e^{\left (2 \, a\right )}}{b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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