Optimal. Leaf size=366 \[ e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]
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Rubi [A] time = 0.244167, antiderivative size = 366, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {2282, 388, 213, 1169, 634, 618, 204, 628} \[ e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 388
Rule 213
Rule 1169
Rule 634
Rule 618
Rule 204
Rule 628
Rubi steps
\begin{align*} \int e^x \tanh (4 x) \, dx &=\operatorname{Subst}\left (\int \frac{-1+x^8}{1+x^8} \, dx,x,e^x\right )\\ &=e^x-2 \operatorname{Subst}\left (\int \frac{1}{1+x^8} \, dx,x,e^x\right )\\ &=e^x-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}\\ &=e^x-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2-\sqrt{2}\right )}-\left (-1+\sqrt{2}\right ) x}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2-\sqrt{2}\right )}+\left (-1+\sqrt{2}\right ) x}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2+\sqrt{2}\right )}-\left (1+\sqrt{2}\right ) x}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2+\sqrt{2}\right )}+\left (1+\sqrt{2}\right ) x}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ &=e^x-\frac{1}{4} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )+\frac{1}{8} \sqrt{2-\sqrt{2}} \operatorname{Subst}\left (\int \frac{-\sqrt{2-\sqrt{2}}+2 x}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \operatorname{Subst}\left (\int \frac{\sqrt{2-\sqrt{2}}+2 x}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \operatorname{Subst}\left (\int \frac{-\sqrt{2+\sqrt{2}}+2 x}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \operatorname{Subst}\left (\int \frac{\sqrt{2+\sqrt{2}}+2 x}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )\\ &=e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,-\sqrt{2+\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,\sqrt{2+\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,-\sqrt{2-\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,\sqrt{2-\sqrt{2}}+2 e^x\right )\\ &=e^x+\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )+\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}+2 e^x}{\sqrt{2+\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}+2 e^x}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )\\ \end{align*}
Mathematica [C] time = 0.0117819, size = 24, normalized size = 0.07 \[ e^x-2 e^x \, _2F_1\left (\frac{1}{8},1;\frac{9}{8};-e^{8 x}\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.059, size = 24, normalized size = 0.1 \begin{align*}{{\rm e}^{x}}+\sum _{{\it \_R}={\it RootOf} \left ( 65536\,{{\it \_Z}}^{8}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}-4\,{\it \_R} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} e^{x} - 2 \, \int \frac{e^{x}}{e^{\left (8 \, x\right )} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.39636, size = 3330, normalized size = 9.1 \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} \tanh{\left (4 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.44803, size = 339, normalized size = 0.93 \begin{align*} -\frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (-\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (-\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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