Optimal. Leaf size=105 \[ \frac{2 e^x}{3 \left (1-e^{6 x}\right )}+\frac{1}{18} \log \left (-e^x+e^{2 x}+1\right )-\frac{1}{18} \log \left (e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{1-2 e^x}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 e^x+1}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{2}{9} \tanh ^{-1}\left (e^x\right ) \]
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Rubi [A] time = 0.139335, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {2282, 12, 288, 210, 634, 618, 204, 628, 206} \[ \frac{2 e^x}{3 \left (1-e^{6 x}\right )}+\frac{1}{18} \log \left (-e^x+e^{2 x}+1\right )-\frac{1}{18} \log \left (e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{1-2 e^x}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 e^x+1}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{2}{9} \tanh ^{-1}\left (e^x\right ) \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 288
Rule 210
Rule 634
Rule 618
Rule 204
Rule 628
Rule 206
Rubi steps
\begin{align*} \int e^x \text{csch}^2(3 x) \, dx &=\operatorname{Subst}\left (\int \frac{4 x^6}{\left (1-x^6\right )^2} \, dx,x,e^x\right )\\ &=4 \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^6\right )^2} \, dx,x,e^x\right )\\ &=\frac{2 e^x}{3 \left (1-e^{6 x}\right )}-\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{1-x^6} \, dx,x,e^x\right )\\ &=\frac{2 e^x}{3 \left (1-e^{6 x}\right )}-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,e^x\right )-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1-\frac{x}{2}}{1-x+x^2} \, dx,x,e^x\right )-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1+\frac{x}{2}}{1+x+x^2} \, dx,x,e^x\right )\\ &=\frac{2 e^x}{3 \left (1-e^{6 x}\right )}-\frac{2}{9} \tanh ^{-1}\left (e^x\right )+\frac{1}{18} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,e^x\right )-\frac{1}{18} \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,e^x\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,e^x\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,e^x\right )\\ &=\frac{2 e^x}{3 \left (1-e^{6 x}\right )}-\frac{2}{9} \tanh ^{-1}\left (e^x\right )+\frac{1}{18} \log \left (1-e^x+e^{2 x}\right )-\frac{1}{18} \log \left (1+e^x+e^{2 x}\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 e^x\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 e^x\right )\\ &=\frac{2 e^x}{3 \left (1-e^{6 x}\right )}+\frac{\tan ^{-1}\left (\frac{1-2 e^x}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{1+2 e^x}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{2}{9} \tanh ^{-1}\left (e^x\right )+\frac{1}{18} \log \left (1-e^x+e^{2 x}\right )-\frac{1}{18} \log \left (1+e^x+e^{2 x}\right )\\ \end{align*}
Mathematica [C] time = 0.0262982, size = 34, normalized size = 0.32 \[ \frac{2}{3} e^x \left (\frac{1}{1-e^{6 x}}-\, _2F_1\left (\frac{1}{6},1;\frac{7}{6};e^{6 x}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.055, size = 148, normalized size = 1.4 \begin{align*} -{\frac{2\,{{\rm e}^{x}}}{3\,{{\rm e}^{6\,x}}-3}}-{\frac{1}{18}\ln \left ({{\rm e}^{x}}+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }+{\frac{i}{18}}\ln \left ({{\rm e}^{x}}+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}-{\frac{1}{18}\ln \left ({{\rm e}^{x}}+{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) }-{\frac{i}{18}}\ln \left ({{\rm e}^{x}}+{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}+{\frac{1}{18}\ln \left ({{\rm e}^{x}}-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }+{\frac{i}{18}}\ln \left ({{\rm e}^{x}}-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}+{\frac{1}{18}\ln \left ({{\rm e}^{x}}-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) }-{\frac{i}{18}}\ln \left ({{\rm e}^{x}}-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}-{\frac{\ln \left ({{\rm e}^{x}}+1 \right ) }{9}}+{\frac{\ln \left ({{\rm e}^{x}}-1 \right ) }{9}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.6852, size = 115, normalized size = 1.1 \begin{align*} -\frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, e^{x} + 1\right )}\right ) - \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, e^{x} - 1\right )}\right ) - \frac{2 \, e^{x}}{3 \,{\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac{1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac{1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac{1}{9} \, \log \left (e^{x} + 1\right ) + \frac{1}{9} \, \log \left (e^{x} - 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.14551, size = 1985, normalized size = 18.9 \begin{align*} \text{result too large to display} \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*} \int e^{x} \operatorname{csch}^{2}{\left (3 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.10899, size = 116, normalized size = 1.1 \begin{align*} -\frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, e^{x} + 1\right )}\right ) - \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, e^{x} - 1\right )}\right ) - \frac{2 \, e^{x}}{3 \,{\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac{1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac{1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac{1}{9} \, \log \left (e^{x} + 1\right ) + \frac{1}{9} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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