Optimal. Leaf size=382 \[ e^x+\frac{e^x}{2 \left (e^{8 x}+1\right )}+\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{8 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{8 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{8 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{8 \sqrt{2 \left (2+\sqrt{2}\right )}} \]
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Rubi [A] time = 0.40506, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {2282, 390, 288, 213, 1169, 634, 618, 204, 628} \[ e^x+\frac{e^x}{2 \left (e^{8 x}+1\right )}+\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{8 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{8 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{8 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{8 \sqrt{2 \left (2+\sqrt{2}\right )}} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 390
Rule 288
Rule 213
Rule 1169
Rule 634
Rule 618
Rule 204
Rule 628
Rubi steps
\begin{align*} \int e^x \tanh ^2(4 x) \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^8\right )^2}{\left (1+x^8\right )^2} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \left (1-\frac{4 x^8}{\left (1+x^8\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x-4 \operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^8\right )^2} \, dx,x,e^x\right )\\ &=e^x+\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 )\\ &=e^x+\frac{e^x}{2 \left (1+e^{8 x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{4 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{4 \sqrt{2}}\\ &=e^x+\frac{e^x}{2 \left (1+e^{8 x}\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 )}{8 \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 )}{8 \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 )}{8 \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 )}{8 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ &=e^x+\frac{e^x}{2 \left (1+e^{8 x}\right )}-\frac{1}{16} \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}{16} \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}{32} \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}{32} \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}{32} \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}{32} \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}{16} \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}{16} \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{e^x}{2 \left (1+e^{8 x}\right )}+\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{8} \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}{8} \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}{8} \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}{8} \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{e^x}{2 \left (1+e^{8 x}\right )}+\frac{1}{16} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )+\frac{1}{16} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{16} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}+2 e^x}{\sqrt{2+\sqrt{2}}}\right )-\frac{1}{16} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}+2 e^x}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{32} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{32} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )\\ \end{align*}
Mathematica [C] time = 0.0480806, size = 51, normalized size = 0.13 \[ \frac{1}{16} \text{RootSum}\left [\text{$\#$1}^8+1\& ,\frac{x-\log \left (e^x-\text{$\#$1}\right )}{\text{$\#$1}^7}\& \right ]+e^x+\frac{e^x}{2 \left (e^{8 x}+1\right )} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.095, size = 36, normalized size = 0.1 \begin{align*}{{\rm e}^{x}}+{\frac{{{\rm e}^{x}}}{2+2\,{{\rm e}^{8\,x}}}}+\sum _{{\it \_R}={\it RootOf} \left ( 4294967296\,{{\it \_Z}}^{8}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}-16\,{\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*} \frac{2 \, e^{\left (9 \, x\right )} + 3 \, e^{x}}{2 \,{\left (e^{\left (8 \, x\right )} + 1\right )}} - \int \frac{e^{x}}{2 \,{\left (e^{\left (8 \, x\right )} + 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.64269, size = 4162, normalized size = 10.9 \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} \tanh ^{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 = 1.33246, size = 355, normalized size = 0.93 \begin{align*} -\frac{1}{16} \, \sqrt{-\sqrt{2} + 2} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{16} \, \sqrt{-\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{16} \, \sqrt{\sqrt{2} + 2} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{16} \, \sqrt{\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{32} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{32} \, \sqrt{\sqrt{2} + 2} \log \left (-\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{32} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{32} \, \sqrt{-\sqrt{2} + 2} \log \left (-\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{e^{x}}{2 \,{\left (e^{\left (8 \, x\right )} + 1\right )}} + e^{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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