Optimal. Leaf size=23 \[ \frac {1}{2} e^{e^x-x} \left (-\frac {3}{2}+x\right ) x^3 \log (x) \]
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Rubi [F] time = 2.02, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [A] time = 0.83, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{4} e^{e^x-x} x^3 (-3+2 x) \log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 29, normalized size = 1.26 \begin {gather*} \frac {1}{2} \, x^{4} e^{\left (-x + e^{x}\right )} \log \relax (x) - \frac {3}{4} \, x^{3} e^{\left (-x + e^{x}\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 20, normalized size = 0.87
method | result | size |
risch | \(\frac {\ln \relax (x ) x^{3} \left (2 x -3\right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{4}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (-x + e^{x}\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (\ln \relax (x)\,\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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.51, size = 29, normalized size = 1.26 \begin {gather*} \frac {\left (2 x^{4} e^{- x} \log {\relax (x )} - 3 x^{3} e^{- x} \log {\relax (x )}\right ) e^{e^{x}}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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