Integrand size = 82, antiderivative size = 25 \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 e^{2 \left (-e^x+\log \left (-e^{e^x}+\log (4)\right )\right )^2} \]
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\[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\int \frac {\exp \left (2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )\right ) \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int 100 e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} \log (4) \left (-e^x+\log (4)\right )^{-1-4 x} \left (x-\log \left (-e^x+\log (4)\right )\right ) \, dx,x,e^x\right ) \\ & = (100 \log (4)) \text {Subst}\left (\int e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} \left (-e^x+\log (4)\right )^{-1-4 x} \left (x-\log \left (-e^x+\log (4)\right )\right ) \, dx,x,e^x\right ) \\ & = (100 \log (4)) \text {Subst}\left (\int \left (e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} x \left (-e^x+\log (4)\right )^{-1-4 x}-e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} \left (-e^x+\log (4)\right )^{-1-4 x} \log \left (-e^x+\log (4)\right )\right ) \, dx,x,e^x\right ) \\ & = (100 \log (4)) \text {Subst}\left (\int e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} x \left (-e^x+\log (4)\right )^{-1-4 x} \, dx,x,e^x\right )-(100 \log (4)) \text {Subst}\left (\int e^{2 \left (x^2+\log ^2\left (-e^x+\log (4)\right )\right )} \left (-e^x+\log (4)\right )^{-1-4 x} \log \left (-e^x+\log (4)\right ) \, dx,x,e^x\right ) \\ \end{align*}
Time = 1.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 e^{2 e^{2 x}+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-e^{e^x}+\log (4)\right )^{-4 e^x} \]
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Time = 4.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56
method | result | size |
parallelrisch | \(25 \,{\mathrm e}^{2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{2}-4 \,{\mathrm e}^{x} \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )+2 \,{\mathrm e}^{2 x}}\) | \(39\) |
risch | \(25 \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{-4 \,{\mathrm e}^{x}} {\mathrm e}^{2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (2\right )\right )^{2}+2 \,{\mathrm e}^{2 x}}\) | \(43\) |
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Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 \, e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )} \]
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Timed out. \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\text {Timed out} \]
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Time = 0.38 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=25 \, e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )} \]
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\[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\int { \frac {200 \, {\left (e^{x} \log \left (2\right ) \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) - e^{\left (2 \, x\right )} \log \left (2\right )\right )} e^{\left (-4 \, e^{x} \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right ) + 2 \, \log \left (-e^{\left (e^{x}\right )} + 2 \, \log \left (2\right )\right )^{2} + 2 \, e^{\left (2 \, x\right )}\right )}}{e^{\left (e^{x}\right )} - 2 \, \log \left (2\right )} \,d x } \]
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Time = 13.45 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {e^{2 e^{2 x}-4 e^x \log \left (-e^{e^x}+\log (4)\right )+2 \log ^2\left (-e^{e^x}+\log (4)\right )} \left (-100 e^{2 x} \log (4)+100 e^x \log (4) \log \left (-e^{e^x}+\log (4)\right )\right )}{e^{e^x}-\log (4)} \, dx=\frac {25\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,{\ln \left (\ln \left (4\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\right )}^2}}{{\left (2\,\ln \left (2\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\right )}^{4\,{\mathrm {e}}^x}} \]
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