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 = 0.10, antiderivative size = 129, normalized size of antiderivative = 1.00, 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 12
Rule 204
Rule 211
Rule 288
Rule 457
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2282
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*}
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Mathematica [A] time = 0.13, 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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fricas [B] time = 0.51, size = 210, normalized size = 1.63 \[ -\frac {4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 40 \, e^{\left (5 \, x\right )} + 8 \, e^{x}}{32 \, {\left (e^{\left (8 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 95, normalized size = 0.74 \[ \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \, {\left (e^{\left (4 \, x\right )} + 1\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.35, size = 44, normalized size = 0.34 \[ -\frac {{\mathrm e}^{x} \left (5 \,{\mathrm e}^{4 x}+1\right )}{4 \left (1+{\mathrm e}^{4 x}\right )^{2}}+4 \left (\munderset {\textit {\_R} =\RootOf \left (16777216 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+64 \textit {\_R} \right )\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 101, normalized size = 0.78 \[ \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \, {\left (e^{\left (8 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.03, size = 122, normalized size = 0.95 \[ -\frac {\frac {{\mathrm {e}}^{5\,x}}{2}-\frac {{\mathrm {e}}^x}{2}}{2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{8\,x}+1}-\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{4\,x}+1\right )}+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (\frac {1}{32}+\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (\frac {1}{32}-\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{32}+\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{32}-\frac {1}{32}{}\mathrm {i}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \tanh {\left (2 x \right )} \operatorname {sech}^{2}{\left (2 x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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