Integrand size = 56, antiderivative size = 24 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=e^{e^{e^{-1+x}}} \left (4-4 e^{-x}\right )+\log \left (x^4\right ) \]
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\[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=\int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (4 e^{e^{e^{-1+x}}-x}+4 e^{-1+e^{e^{-1+x}}+e^{-1+x}+x}+\frac {4 \left (e-e^{e^{e^{-1+x}}+e^{-1+x}} x\right )}{e x}\right ) \, dx \\ & = 4 \int e^{e^{e^{-1+x}}-x} \, dx+4 \int e^{-1+e^{e^{-1+x}}+e^{-1+x}+x} \, dx+\frac {4 \int \frac {e-e^{e^{e^{-1+x}}+e^{-1+x}} x}{x} \, dx}{e} \\ & = 4 \text {Subst}\left (\int e^{-1+e^{\frac {x}{e}}+\frac {x}{e}} \, dx,x,e^x\right )+4 \text {Subst}\left (\int \frac {e^{e^{\frac {x}{e}}}}{x^2} \, dx,x,e^x\right )+\frac {4 \int \left (-e^{e^{e^{-1+x}}+e^{-1+x}}+\frac {e}{x}\right ) \, dx}{e} \\ & = 4 \log (x)+4 \text {Subst}\left (\int \frac {e^{e^{\frac {x}{e}}}}{x^2} \, dx,x,e^x\right )-\frac {4 \int e^{e^{e^{-1+x}}+e^{-1+x}} \, dx}{e}+(4 e) \text {Subst}\left (\int e^{-1+x} \, dx,x,e^{e^{-1+x}}\right ) \\ & = 4 e^{e^{e^{-1+x}}}+4 \log (x)+4 \text {Subst}\left (\int \frac {e^{e^{\frac {x}{e}}}}{x^2} \, dx,x,e^x\right )-\frac {4 \text {Subst}\left (\int \frac {e^{e^x+x}}{x} \, dx,x,e^{-1+x}\right )}{e} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=4 \left (e^{e^{e^{-1+x}}-x} \left (-1+e^x\right )+\log (x)\right ) \]
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Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(4 \ln \left (x \right )+4 \left ({\mathrm e}^{x}-1\right ) {\mathrm e}^{-x +{\mathrm e}^{{\mathrm e}^{-1+x}}}\) | \(22\) |
parts | \(-4 \,{\mathrm e}^{-x} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1} {\mathrm e}^{x}}}+4 \ln \left (x \right )+4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1+x}}}\) | \(27\) |
parallelrisch | \(\left (4 \,{\mathrm e}^{x} \ln \left (x \right )+4 \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1+x}}}-4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1+x}}}\right ) {\mathrm e}^{-x}\) | \(31\) |
norman | \(\left (4 \,{\mathrm e}^{-1} {\mathrm e} \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1} {\mathrm e}^{x}}}-4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-1} {\mathrm e}^{x}}}\right ) {\mathrm e}^{-x}+4 \ln \left (x \right )\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=4 \, {\left ({\left (e^{x} - 1\right )} e^{\left (e^{\left (e^{\left (x - 1\right )}\right )}\right )} + e^{x} \log \left (x\right )\right )} e^{\left (-x\right )} \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=\left (4 e^{x} - 4\right ) e^{- x} e^{e^{\frac {e^{x}}{e}}} + 4 \log {\left (x \right )} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=4 \, {\left (e^{x} - 1\right )} e^{\left (-x + e^{\left (e^{\left (x - 1\right )}\right )}\right )} + 4 \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=4 \, {\left (e^{\left (x + e^{\left (x - 1\right )} + e^{\left (e^{\left (x - 1\right )}\right )}\right )} - e^{\left (e^{\left (x - 1\right )} + e^{\left (e^{\left (x - 1\right )}\right )}\right )}\right )} e^{\left (-x - e^{\left (x - 1\right )}\right )} + 4 \, \log \left (x\right ) \]
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Time = 9.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (4 e^x+e^{e^{e^{-1+x}}} \left (4 x+e^{e^{-1+x}} \left (-4 e^{-1+x} x+4 e^{-1+2 x} x\right )\right )\right )}{x} \, dx=4\,\ln \left (x\right )+{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}-x}\,\left (4\,{\mathrm {e}}^x-4\right ) \]
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