Optimal. Leaf size=129 \[ -\frac{e^x}{4 \left (e^{4 x}+1\right )}-\frac{e^{5 x}}{\left (e^{4 x}+1\right )^2}-\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}+\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{8 \sqrt{2}} \]
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Rubi [A] time = 0.0960216, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {2282, 12, 457, 288, 211, 1165, 628, 1162, 617, 204} \[ -\frac{e^x}{4 \left (e^{4 x}+1\right )}-\frac{e^{5 x}}{\left (e^{4 x}+1\right )^2}-\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}+\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{16 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{8 \sqrt{2}} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
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 x) \, dx &=\operatorname{Subst}\left (\int \frac{4 x^4 \left (-1+x^4\right )}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=4 \operatorname{Subst}\left (\int \frac{x^4 \left (-1+x^4\right )}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=-\frac{e^{5 x}}{\left (1+e^{4 x}\right )^2}+\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=-\frac{e^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac{e^x}{4 \left (1+e^{4 x}\right )}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,e^x\right )\\ &=-\frac{e^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac{e^x}{4 \left (1+e^{4 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^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac{e^x}{4 \left (1+e^{4 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^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac{e^x}{4 \left (1+e^{4 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^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac{e^x}{4 \left (1+e^{4 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{\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 [A] time = 0.118189, size = 120, normalized size = 0.93 \[ \frac{1}{32} \left (-\frac{40 e^x}{e^{4 x}+1}+\frac{32 e^x}{\left (e^{4 x}+1\right )^2}-\sqrt{2} \log \left (-\sqrt{2} e^x+e^{2 x}+1\right )+\sqrt{2} \log \left (\sqrt{2} e^x+e^{2 x}+1\right )-2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} e^x\right )+2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} e^x+1\right )\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 0.05, size = 44, normalized size = 0.3 \begin{align*} -{\frac{{{\rm e}^{x}} \left ( 5\,{{\rm e}^{4\,x}}+1 \right ) }{4\, \left ( 1+{{\rm e}^{4\,x}} \right ) ^{2}}}+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 = 1.69146, size = 136, normalized size = 1.05 \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{5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \,{\left (e^{\left (8 \, x\right )} + 2 \, 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 = 1.93438, size = 637, normalized size = 4.94 \begin{align*} -\frac{4 \,{\left (\sqrt{2} e^{\left (8 \, x\right )} + 2 \, \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 ) + 4 \,{\left (\sqrt{2} e^{\left (8 \, x\right )} + 2 \, \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 ) -{\left (\sqrt{2} e^{\left (8 \, x\right )} + 2 \, \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 ) +{\left (\sqrt{2} e^{\left (8 \, x\right )} + 2 \, \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 ) + 40 \, e^{\left (5 \, x\right )} + 8 \, e^{x}}{32 \,{\left (e^{\left (8 \, x\right )} + 2 \, 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{\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 = 1.18154, size = 128, normalized size = 0.99 \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{5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \,{\left (e^{\left (4 \, x\right )} + 1\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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