Optimal. Leaf size=130 \[ -\frac {3 e^{3 x}}{4 \left (e^{4 x}+1\right )}+\frac {e^{3 x}}{\left (e^{4 x}+1\right )^2}+\frac {5 \log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {5 \log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {5 \tan ^{-1}\left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}} \]
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Rubi [A] time = 0.11, antiderivative size = 130, 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, 463, 457, 297, 1162, 617, 204, 1165, 628} \[ -\frac {3 e^{3 x}}{4 \left (e^{4 x}+1\right )}+\frac {e^{3 x}}{\left (e^{4 x}+1\right )^2}+\frac {5 \log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {5 \log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {5 \tan ^{-1}\left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 297
Rule 457
Rule 463
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2282
Rubi steps
\begin {align*} \int e^x \text {sech}(2 x) \tanh ^2(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {2 x^2 \left (1-x^4\right )^2}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^2}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=\frac {e^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^2 \left (4-8 x^4\right )}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=\frac {e^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {3 e^{3 x}}{4 \left (1+e^{4 x}\right )}+\frac {5}{4} \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,e^x\right )\\ &=\frac {e^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {3 e^{3 x}}{4 \left (1+e^{4 x}\right )}-\frac {5}{8} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=\frac {e^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {3 e^{3 x}}{4 \left (1+e^{4 x}\right )}+\frac {5}{16} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {5}{16} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {5 \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt {2}}+\frac {5 \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^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {3 e^{3 x}}{4 \left (1+e^{4 x}\right )}+\frac {5 \log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {5 \log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{8 \sqrt {2}}\\ &=\frac {e^{3 x}}{\left (1+e^{4 x}\right )^2}-\frac {3 e^{3 x}}{4 \left (1+e^{4 x}\right )}-\frac {5 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {5 \tan ^{-1}\left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {5 \log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {5 \log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}\\ \end {align*}
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Mathematica [C] time = 0.06, size = 58, normalized size = 0.45 \[ \frac {e^{3 x}-3 e^{7 x}}{4 \left (e^{4 x}+1\right )^2}-\frac {5}{16} \text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}}\& \right ] \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 213, normalized size = 1.64 \[ -\frac {20 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 20 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) + 5 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) - 5 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} + 2 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 24 \, e^{\left (7 \, x\right )} - 8 \, e^{\left (3 \, x\right )}}{32 \, {\left (e^{\left (8 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 99, normalized size = 0.76 \[ \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {5}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3 \, e^{\left (7 \, x\right )} - e^{\left (3 \, x\right )}}{4 \, {\left (e^{\left (4 \, x\right )} + 1\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.35, size = 48, normalized size = 0.37 \[ -\frac {{\mathrm e}^{3 x} \left (3 \,{\mathrm e}^{4 x}-1\right )}{4 \left (1+{\mathrm e}^{4 x}\right )^{2}}+2 \left (\munderset {\textit {\_R} =\RootOf \left (1048576 \textit {\_Z}^{4}+625\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\frac {32768 \textit {\_R}^{3}}{125}\right )\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 105, normalized size = 0.81 \[ \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {5}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3 \, e^{\left (7 \, x\right )} - e^{\left (3 \, x\right )}}{4 \, {\left (e^{\left (8 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.09, size = 112, normalized size = 0.86 \[ \frac {{\mathrm {e}}^{3\,x}}{2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{8\,x}+1}-\frac {3\,{\mathrm {e}}^{3\,x}}{4\,\left ({\mathrm {e}}^{4\,x}+1\right )}+\sqrt {2}\,\ln \left (\frac {25}{16}+\sqrt {2}\,{\mathrm {e}}^x\,\left (-\frac {25}{32}-\frac {25}{32}{}\mathrm {i}\right )\right )\,\left (\frac {5}{32}+\frac {5}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {25}{16}+\sqrt {2}\,{\mathrm {e}}^x\,\left (-\frac {25}{32}+\frac {25}{32}{}\mathrm {i}\right )\right )\,\left (\frac {5}{32}-\frac {5}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {25}{16}+\sqrt {2}\,{\mathrm {e}}^x\,\left (\frac {25}{32}-\frac {25}{32}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{32}+\frac {5}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {25}{16}+\sqrt {2}\,{\mathrm {e}}^x\,\left (\frac {25}{32}+\frac {25}{32}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{32}-\frac {5}{32}{}\mathrm {i}\right ) \]
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 \[ \int e^{x} \tanh ^{2}{\left (2 x \right )} \operatorname {sech}{\left (2 x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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