Optimal. Leaf size=131 \[ \frac{e^x}{2 \left (1-e^{8 x}\right )}+\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{1}{8} \tan ^{-1}\left (e^x\right )+\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{8 \sqrt{2}}-\frac{1}{8} \tanh ^{-1}\left (e^x\right ) \]
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Rubi [A] time = 0.0884536, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.3, Rules used = {2282, 12, 288, 214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \[ \frac{e^x}{2 \left (1-e^{8 x}\right )}+\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{1}{8} \tan ^{-1}\left (e^x\right )+\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{8 \sqrt{2}}-\frac{1}{8} \tanh ^{-1}\left (e^x\right ) \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 288
Rule 214
Rule 212
Rule 206
Rule 203
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rubi steps
\begin{align*} \int e^x \text{csch}^2(4 x) \, dx &=\operatorname{Subst}\left (\int \frac{4 x^8}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=4 \operatorname{Subst}\left (\int \frac{x^8}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^8} \, dx,x,e^x\right )\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-x^4} \, dx,x,e^x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,e^x\right )\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,e^x\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^x\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,e^x\right )-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{8} \tan ^{-1}\left (e^x\right )-\frac{1}{8} \tanh ^{-1}\left (e^x\right )-\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,e^x\right )-\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,e^x\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt{2}}\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{8} \tan ^{-1}\left (e^x\right )-\frac{1}{8} \tanh ^{-1}\left (e^x\right )+\frac{\log \left (1-\sqrt{2} e^x+e^{2 x}\right )}{16 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} e^x+e^{2 x}\right )}{16 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} e^x\right )}{8 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} e^x\right )}{8 \sqrt{2}}\\ &=\frac{e^x}{2 \left (1-e^{8 x}\right )}-\frac{1}{8} \tan ^{-1}\left (e^x\right )+\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (1+\sqrt{2} e^x\right )}{8 \sqrt{2}}-\frac{1}{8} \tanh ^{-1}\left (e^x\right )+\frac{\log \left (1-\sqrt{2} e^x+e^{2 x}\right )}{16 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} e^x+e^{2 x}\right )}{16 \sqrt{2}}\\ \end{align*}
Mathematica [C] time = 0.0248644, size = 34, normalized size = 0.26 \[ \frac{1}{2} e^x \left (\frac{1}{1-e^{8 x}}-\, _2F_1\left (\frac{1}{8},1;\frac{9}{8};e^{8 x}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.058, size = 68, normalized size = 0.5 \begin{align*} -{\frac{{{\rm e}^{x}}}{2\,{{\rm e}^{8\,x}}-2}}+{\frac{i}{16}}\ln \left ({{\rm e}^{x}}-i \right ) -{\frac{i}{16}}\ln \left ({{\rm e}^{x}}+i \right ) -{\frac{\ln \left ({{\rm e}^{x}}+1 \right ) }{16}}+{\frac{\ln \left ({{\rm e}^{x}}-1 \right ) }{16}}+4\,\sum _{{\it \_R}={\it RootOf} \left ( 16777216\,{{\it \_Z}}^{4}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}-64\,{\it \_R} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.59527, size = 144, normalized size = 1.1 \begin{align*} -\frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{x}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{x}\right )}\right ) - \frac{1}{32} \, \sqrt{2} \log \left (\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{32} \, \sqrt{2} \log \left (-\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{e^{x}}{2 \,{\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac{1}{8} \, \arctan \left (e^{x}\right ) - \frac{1}{16} \, \log \left (e^{x} + 1\right ) + \frac{1}{16} \, \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 = 2.79839, size = 620, normalized size = 4.73 \begin{align*} \frac{4 \,{\left (\sqrt{2} e^{\left (8 \, x\right )} - \sqrt{2}\right )} \arctan \left (-\sqrt{2} e^{x} + \sqrt{2} \sqrt{\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \,{\left (\sqrt{2} e^{\left (8 \, x\right )} - \sqrt{2}\right )} \arctan \left (-\sqrt{2} e^{x} + \frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - 4 \,{\left (e^{\left (8 \, x\right )} - 1\right )} \arctan \left (e^{x}\right ) -{\left (\sqrt{2} e^{\left (8 \, x\right )} - \sqrt{2}\right )} \log \left (4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) +{\left (\sqrt{2} e^{\left (8 \, x\right )} - \sqrt{2}\right )} \log \left (-4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) - 2 \,{\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) + 2 \,{\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) - 16 \, e^{x}}{32 \,{\left (e^{\left (8 \, x\right )} - 1\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*} \int e^{x} \operatorname{csch}^{2}{\left (4 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.11866, size = 146, normalized size = 1.11 \begin{align*} -\frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{x}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{x}\right )}\right ) - \frac{1}{32} \, \sqrt{2} \log \left (\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{32} \, \sqrt{2} \log \left (-\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{e^{x}}{2 \,{\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac{1}{8} \, \arctan \left (e^{x}\right ) - \frac{1}{16} \, \log \left (e^{x} + 1\right ) + \frac{1}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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