Integrand size = 8, antiderivative size = 95 \[ \int e^x \tanh (2 x) \, dx=e^x+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}} \]
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Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2320, 396, 217, 1179, 642, 1176, 631, 210} \[ \int e^x \tanh (2 x) \, dx=\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}+e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}} \]
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Rule 210
Rule 217
Rule 396
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-1+x^4}{1+x^4} \, dx,x,e^x\right ) \\ & = e^x-2 \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right ) \\ & = e^x-\text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right ) \\ & = e^x-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^x\right )}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^x\right )}{2 \sqrt {2}} \\ & = e^x+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{\sqrt {2}} \\ & = e^x+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int e^x \tanh (2 x) \, dx=e^x+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{2 \sqrt {2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.25
method | result | size |
risch | \({\mathrm e}^{x}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-2 \textit {\_R} \right )\right )\) | \(24\) |
default | \(-\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3+2 \sqrt {2}\right )}{4}+\frac {\left (-\sqrt {2}-2\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{2+2 \sqrt {2}}+\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3-2 \sqrt {2}\right )}{4}-\frac {\left (2-\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2 \sqrt {2}-2}\right )}{2 \sqrt {2}-2}-\frac {2}{\tanh \left (\frac {x}{2}\right )-1}\) | \(117\) |
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Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int e^x \tanh (2 x) \, dx=-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + \left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) - \left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + \left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + \cosh \left (x\right ) + \sinh \left (x\right ) \]
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\[ \int e^x \tanh (2 x) \, dx=\int e^{x} \tanh {\left (2 x \right )}\, dx \]
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none
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int e^x \tanh (2 x) \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \]
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Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int e^x \tanh (2 x) \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \]
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Time = 1.84 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int e^x \tanh (2 x) \, dx={\mathrm {e}}^x-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}{2}\right )}{2}+\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}^2+2\right )}{4}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}{2}\right )}{2}-\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}^2+2\right )}{4} \]
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