Optimal. Leaf size=59 \[ \frac{e^{2 a+2 b x}}{2 b}+\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
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Rubi [A] time = 0.0496781, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2282, 444, 43} \[ \frac{e^{2 a+2 b x}}{2 b}+\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 444
Rule 43
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \coth ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x \left (1+x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^2}{(1-x)^2} \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{4}{(-1+x)^2}+\frac{4}{-1+x}\right ) \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac{e^{2 a+2 b x}}{2 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 = 0.0620764, size = 48, normalized size = 0.81 \[ \frac{e^{2 (a+b x)}-\frac{4}{e^{2 (a+b x)}-1}+4 \log \left (1-e^{2 (a+b x)}\right )}{2 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 57, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{2\,bx+2\,a}}}{2\,b}}-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 = 1.01709, size = 116, normalized size = 1.97 \begin{align*} \frac{4 \,{\left (b x + a\right )}}{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{5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1}{2 \, b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.87767, size = 540, normalized size = 9.15 \begin{align*} \frac{\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} +{\left (6 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - \cosh \left (b x + a\right )^{2} + 4 \,{\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 ) + 2 \,{\left (2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 4}{2 \,{\left (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\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 = 1.20936, size = 76, normalized size = 1.29 \begin{align*} -\frac{\frac{4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - e^{\left (2 \, b x + 2 \, a\right )} - 4 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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