Optimal. Leaf size=81 \[ \frac{e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.0515906, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2282, 390, 1158, 12, 288, 207} \[ \frac{e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 390
Rule 1158
Rule 12
Rule 288
Rule 207
Rubi steps
\begin{align*} \int e^{a+b x} \coth ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{2 \left (1+3 x^4\right )}{\left (-1+x^2\right )^3}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{e^{a+b x}}{b}+\frac{2 \operatorname{Subst}\left (\int \frac{1+3 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{e^{a+b x}}{b}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{12 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac{e^{a+b x}}{b}-\frac{2 e^{a+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{e^{a+b x}}{b}-\frac{2 e^{a+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{e^{a+b x}}{b}-\frac{2 e^{a+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 [C] time = 4.42973, size = 286, normalized size = 3.53 \[ -\frac{e^{-5 (a+b x)} \left (256 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^3 \, _6F_5\left (\frac{3}{2},2,2,2,2,2;1,1,1,1,\frac{11}{2};e^{2 (a+b x)}\right )+384 e^{8 (a+b x)} \left (5 e^{2 (a+b x)}+7\right ) \left (e^{2 (a+b x)}+1\right )^2 \, _5F_4\left (\frac{3}{2},2,2,2,2;1,1,1,\frac{11}{2};e^{2 (a+b x)}\right )-21 \left (507305 e^{2 (a+b x)}+173916 e^{4 (a+b x)}-154296 e^{6 (a+b x)}-73885 e^{8 (a+b x)}+4887 e^{10 (a+b x)}+252105\right )-\frac{315 \left (-28218 e^{2 (a+b x)}+1173 e^{4 (a+b x)}+17748 e^{6 (a+b x)}+4299 e^{8 (a+b x)}-1434 e^{10 (a+b x)}+7 e^{12 (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}\right )}{60480 b} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.025, size = 88, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{\sinh \left ( bx+a \right ) }}+2\,\sinh \left ( bx+a \right ) +{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}-3\,{\frac{\cosh \left ( bx+a \right ) }{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3\,{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}-3\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00041, size = 119, normalized size = 1.47 \begin{align*} \frac{e^{\left (b x + a\right )}}{b} - \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{3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b{\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, 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 = 1.86649, size = 1291, normalized size = 15.94 \begin{align*} \frac{2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, \sinh \left (b x + a\right )^{5} + 10 \,{\left (2 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 10 \, \cosh \left (b x + a\right )^{3} + 10 \,{\left (2 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 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 ) + 2 \,{\left (5 \, \cosh \left (b x + a\right )^{4} - 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 4 \, \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 = 1.13996, size = 105, normalized size = 1.3 \begin{align*} \frac{e^{\left (b x + a\right )}}{b} - \frac{3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac{3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} - \frac{3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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