Integrand size = 58, antiderivative size = 23 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{2} e^{e^x-x} \left (-\frac {3}{2}+x\right ) x^3 \log (x) \]
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\[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} \int e^{e^x-x} x^2 \left (-3+2 x+\left (-9+\left (11-3 e^x\right ) x+2 \left (-1+e^x\right ) x^2\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} \int \left (e^{e^x} x^3 (-3+2 x) \log (x)-e^{e^x-x} x^2 \left (3-2 x+9 \log (x)-11 x \log (x)+2 x^2 \log (x)\right )\right ) \, dx \\ & = \frac {1}{4} \int e^{e^x} x^3 (-3+2 x) \log (x) \, dx-\frac {1}{4} \int e^{e^x-x} x^2 \left (3-2 x+9 \log (x)-11 x \log (x)+2 x^2 \log (x)\right ) \, dx \\ & = -\left (\frac {1}{4} \int \left (-e^{e^x-x} x^2 (-3+2 x)+e^{e^x-x} x^2 \left (9-11 x+2 x^2\right ) \log (x)\right ) \, dx\right )-\frac {1}{4} \int \frac {-3 \int e^{e^x} x^3 \, dx+2 \int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx \\ & = \frac {1}{4} \int e^{e^x-x} x^2 (-3+2 x) \, dx-\frac {1}{4} \int e^{e^x-x} x^2 \left (9-11 x+2 x^2\right ) \log (x) \, dx-\frac {1}{4} \int \left (-\frac {3 \int e^{e^x} x^3 \, dx}{x}+\frac {2 \int e^{e^x} x^4 \, dx}{x}\right ) \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx \\ & = \frac {1}{4} \int \left (-3 e^{e^x-x} x^2+2 e^{e^x-x} x^3\right ) \, dx+\frac {1}{4} \int \frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx+2 \int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \left (\frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx}{x}+\frac {2 \int e^{e^x-x} x^4 \, dx}{x}\right ) \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \frac {9 \int e^{e^x-x} x^2 \, dx-11 \int e^{e^x-x} x^3 \, dx}{x} \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{4} \int \left (\frac {9 \int e^{e^x-x} x^2 \, dx}{x}-\frac {11 \int e^{e^x-x} x^3 \, dx}{x}\right ) \, dx+\frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ & = \frac {1}{2} \int e^{e^x-x} x^3 \, dx-\frac {1}{2} \int \frac {\int e^{e^x} x^4 \, dx}{x} \, dx+\frac {1}{2} \int \frac {\int e^{e^x-x} x^4 \, dx}{x} \, dx-\frac {3}{4} \int e^{e^x-x} x^2 \, dx+\frac {3}{4} \int \frac {\int e^{e^x} x^3 \, dx}{x} \, dx+\frac {9}{4} \int \frac {\int e^{e^x-x} x^2 \, dx}{x} \, dx-\frac {11}{4} \int \frac {\int e^{e^x-x} x^3 \, dx}{x} \, dx+\frac {1}{2} \log (x) \int e^{e^x} x^4 \, dx-\frac {1}{2} \log (x) \int e^{e^x-x} x^4 \, dx-\frac {1}{4} (3 \log (x)) \int e^{e^x} x^3 \, dx-\frac {1}{4} (9 \log (x)) \int e^{e^x-x} x^2 \, dx+\frac {1}{4} (11 \log (x)) \int e^{e^x-x} x^3 \, dx \\ \end{align*}
Time = 2.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} e^{e^x-x} x^3 (-3+2 x) \log (x) \]
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {\ln \left (x \right ) x^{3} \left (-3+2 x \right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{4}\) | \(20\) |
parallelrisch | \(\frac {\left (2 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (x \right ) x^{4}-3 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (x \right ) x^{3}\right ) {\mathrm e}^{-x}}{4}\) | \(28\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {\left (2 x^{4} \log {\left (x \right )} - 3 x^{3} \log {\left (x \right )}\right ) e^{- x} e^{e^{x}}}{4} \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\frac {1}{2} \, x^{4} e^{\left (-x + e^{x}\right )} \log \left (x\right ) - \frac {3}{4} \, x^{3} e^{\left (-x + e^{x}\right )} \log \left (x\right ) \]
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Timed out. \[ \int \frac {1}{4} e^{e^x-x} \left (-3 x^2+2 x^3+\left (-9 x^2+11 x^3-2 x^4+e^x \left (-3 x^3+2 x^4\right )\right ) \log (x)\right ) \, dx=\int -\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (3\,x^3-2\,x^4\right )+9\,x^2-11\,x^3+2\,x^4\right )+3\,x^2-2\,x^3\right )}{4} \,d x \]
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