Optimal. Leaf size=149 \[ -\frac{3 e^x}{8 \left (e^{4 x}+1\right )}-\frac{5 e^{5 x}}{6 \left (e^{4 x}+1\right )^2}+\frac{4 e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac{3 \log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{32 \sqrt{2}}+\frac{3 \log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{32 \sqrt{2}}-\frac{3 \tan ^{-1}\left (1-\sqrt{2} e^x\right )}{16 \sqrt{2}}+\frac{3 \tan ^{-1}\left (\sqrt{2} e^x+1\right )}{16 \sqrt{2}} \]
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Rubi [A] time = 0.127663, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.688, Rules used = {2282, 12, 463, 457, 288, 211, 1165, 628, 1162, 617, 204} \[ -\frac{3 e^x}{8 \left (e^{4 x}+1\right )}-\frac{5 e^{5 x}}{6 \left (e^{4 x}+1\right )^2}+\frac{4 e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac{3 \log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{32 \sqrt{2}}+\frac{3 \log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{32 \sqrt{2}}-\frac{3 \tan ^{-1}\left (1-\sqrt{2} e^x\right )}{16 \sqrt{2}}+\frac{3 \tan ^{-1}\left (\sqrt{2} e^x+1\right )}{16 \sqrt{2}} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 463
Rule 457
Rule 288
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rubi steps
\begin{align*} \int e^x \text{sech}^2(2 x) \tanh ^2(2 x) \, dx &=\operatorname{Subst}\left (\int \frac{4 x^4 \left (1-x^4\right )^2}{\left (1+x^4\right )^4} \, dx,x,e^x\right )\\ &=4 \operatorname{Subst}\left (\int \frac{x^4 \left (1-x^4\right )^2}{\left (1+x^4\right )^4} \, dx,x,e^x\right )\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4 \left (8-12 x^4\right )}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}+\frac{3}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac{3 e^x}{8 \left (1+e^{4 x}\right )}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,e^x\right )\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac{3 e^x}{8 \left (1+e^{4 x}\right )}+\frac{3}{16} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,e^x\right )+\frac{3}{16} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac{3 e^x}{8 \left (1+e^{4 x}\right )}+\frac{3}{32} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,e^x\right )+\frac{3}{32} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,e^x\right )-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,e^x\right )}{32 \sqrt{2}}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,e^x\right )}{32 \sqrt{2}}\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac{3 e^x}{8 \left (1+e^{4 x}\right )}-\frac{3 \log \left (1-\sqrt{2} e^x+e^{2 x}\right )}{32 \sqrt{2}}+\frac{3 \log \left (1+\sqrt{2} e^x+e^{2 x}\right )}{32 \sqrt{2}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} e^x\right )}{16 \sqrt{2}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} e^x\right )}{16 \sqrt{2}}\\ &=\frac{4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac{5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac{3 e^x}{8 \left (1+e^{4 x}\right )}-\frac{3 \tan ^{-1}\left (1-\sqrt{2} e^x\right )}{16 \sqrt{2}}+\frac{3 \tan ^{-1}\left (1+\sqrt{2} e^x\right )}{16 \sqrt{2}}-\frac{3 \log \left (1-\sqrt{2} e^x+e^{2 x}\right )}{32 \sqrt{2}}+\frac{3 \log \left (1+\sqrt{2} e^x+e^{2 x}\right )}{32 \sqrt{2}}\\ \end{align*}
Mathematica [C] time = 0.0621056, size = 64, normalized size = 0.43 \[ \frac{1}{96} \left (-9 \text{RootSum}\left [\text{$\#$1}^4+1\& ,\frac{x-\log \left (e^x-\text{$\#$1}\right )}{\text{$\#$1}^3}\& \right ]-\frac{4 e^x \left (6 e^{4 x}+29 e^{8 x}+9\right )}{\left (e^{4 x}+1\right )^3}\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.067, size = 50, normalized size = 0.3 \begin{align*} -{\frac{{{\rm e}^{x}} \left ( 29\,{{\rm e}^{8\,x}}+6\,{{\rm e}^{4\,x}}+9 \right ) }{24\, \left ( 1+{{\rm e}^{4\,x}} \right ) ^{3}}}+4\,\sum _{{\it \_R}={\it RootOf} \left ( 268435456\,{{\it \_Z}}^{4}+81 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}+{\frac{128\,{\it \_R}}{3}} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.6337, size = 155, normalized size = 1.04 \begin{align*} \frac{3}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{x}\right )}\right ) + \frac{3}{32} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{x}\right )}\right ) + \frac{3}{64} \, \sqrt{2} \log \left (\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{3}{64} \, \sqrt{2} \log \left (-\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \,{\left (e^{\left (12 \, x\right )} + 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.24074, size = 798, normalized size = 5.36 \begin{align*} -\frac{36 \,{\left (\sqrt{2} e^{\left (12 \, x\right )} + 3 \, \sqrt{2} e^{\left (8 \, x\right )} + 3 \, \sqrt{2} e^{\left (4 \, 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 ) + 36 \,{\left (\sqrt{2} e^{\left (12 \, x\right )} + 3 \, \sqrt{2} e^{\left (8 \, x\right )} + 3 \, \sqrt{2} e^{\left (4 \, 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 ) - 9 \,{\left (\sqrt{2} e^{\left (12 \, x\right )} + 3 \, \sqrt{2} e^{\left (8 \, x\right )} + 3 \, \sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \log \left (4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 9 \,{\left (\sqrt{2} e^{\left (12 \, x\right )} + 3 \, \sqrt{2} e^{\left (8 \, x\right )} + 3 \, \sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \log \left (-4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 232 \, e^{\left (9 \, x\right )} + 48 \, e^{\left (5 \, x\right )} + 72 \, e^{x}}{192 \,{\left (e^{\left (12 \, x\right )} + 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, 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} \tanh ^{2}{\left (2 x \right )} \operatorname{sech}^{2}{\left (2 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.13499, size = 139, normalized size = 0.93 \begin{align*} \frac{3}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{x}\right )}\right ) + \frac{3}{32} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{x}\right )}\right ) + \frac{3}{64} \, \sqrt{2} \log \left (\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{3}{64} \, \sqrt{2} \log \left (-\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \,{\left (e^{\left (4 \, x\right )} + 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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