Optimal. Leaf size=28 \[ e^x+\frac {1}{4} x^2 \log (x) \left (10-x^2+e^{-x} \log (x)\right ) \]
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Rubi [F] time = 0.66, 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^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(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^{-x} \left (4 e^{2 x}+e^x \left (10 x-x^3\right )+\left (2 x+e^x \left (20 x-4 x^3\right )\right ) \log (x)+\left (2 x-x^2\right ) \log ^2(x)\right ) \, dx\\ &=\frac {1}{4} \int \left (4 e^x-x \left (-10+x^2\right )-2 e^{-x} x \left (-1-10 e^x+2 e^x x^2\right ) \log (x)-e^{-x} (-2+x) x \log ^2(x)\right ) \, dx\\ &=-\left (\frac {1}{4} \int x \left (-10+x^2\right ) \, dx\right )-\frac {1}{4} \int e^{-x} (-2+x) x \log ^2(x) \, dx-\frac {1}{2} \int e^{-x} x \left (-1-10 e^x+2 e^x x^2\right ) \log (x) \, dx+\int e^x \, dx\\ &=e^x-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int \left (-10 x+x^3\right ) \, dx-\frac {1}{4} \int \left (-2 e^{-x} x \log ^2(x)+e^{-x} x^2 \log ^2(x)\right ) \, dx+\frac {1}{2} \int \left (e^{-x} \left (1+\frac {1}{x}\right )+\frac {1}{2} x \left (-10+x^2\right )\right ) \, dx\\ &=e^x+\frac {5 x^2}{4}-\frac {x^4}{16}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)+\frac {1}{4} \int x \left (-10+x^2\right ) \, dx-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} \left (1+\frac {1}{x}\right ) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=e^x+\frac {5 x^2}{4}-\frac {x^4}{16}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)+\frac {1}{4} \int \left (-10 x+x^3\right ) \, dx-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int \left (e^{-x}+\frac {e^{-x}}{x}\right ) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=e^x-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} \, dx+\frac {1}{2} \int \frac {e^{-x}}{x} \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ &=-\frac {e^{-x}}{2}+e^x+\frac {\text {Ei}(-x)}{2}-\frac {1}{2} e^{-x} \log (x)-\frac {1}{2} e^{-x} x \log (x)+\frac {5}{2} x^2 \log (x)-\frac {1}{4} x^4 \log (x)-\frac {1}{4} \int e^{-x} x^2 \log ^2(x) \, dx+\frac {1}{2} \int e^{-x} x \log ^2(x) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 37, normalized size = 1.32 \begin {gather*} \frac {1}{4} \left (4 e^x+10 x^2 \log (x)-x^4 \log (x)+e^{-x} x^2 \log ^2(x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.00, size = 36, normalized size = 1.29 \begin {gather*} \frac {1}{4} \, {\left (x^{2} \log \relax (x)^{2} - {\left (x^{4} - 10 \, x^{2}\right )} e^{x} \log \relax (x) + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 30, normalized size = 1.07 \begin {gather*} -\frac {1}{4} \, x^{4} \log \relax (x) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \relax (x)^{2} + \frac {5}{2} \, x^{2} \log \relax (x) + e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 31, normalized size = 1.11
method | result | size |
default | \(\frac {\ln \relax (x )^{2} x^{2} {\mathrm e}^{-x}}{4}-\frac {x^{4} \ln \relax (x )}{4}+\frac {5 x^{2} \ln \relax (x )}{2}+{\mathrm e}^{x}\) | \(31\) |
risch | \(\frac {\ln \relax (x )^{2} x^{2} {\mathrm e}^{-x}}{4}+\frac {\left (-x^{4}+10 x^{2}\right ) \ln \relax (x )}{4}+{\mathrm e}^{x}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 30, normalized size = 1.07 \begin {gather*} -\frac {1}{4} \, x^{4} \log \relax (x) + \frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \relax (x)^{2} + \frac {5}{2} \, x^{2} \log \relax (x) + e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.04, size = 109, normalized size = 3.89 \begin {gather*} {\mathrm {e}}^x+\frac {2\,x\,{\mathrm {e}}^{-x}+2\,x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)+x^3\,{\mathrm {e}}^{-x}\,{\ln \relax (x)}^2+2\,x\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{4\,x}+\frac {x^2\,\left (40\,\ln \relax (x)-4\,x^2\,\ln \relax (x)+x^2-20\right )}{16}+\frac {5\,x^2}{4}-\frac {x^4}{16}-\frac {x\,{\mathrm {e}}^{-x}+x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)+x\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{2\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.40, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{2} e^{- x} \log {\relax (x )}^{2}}{4} + \left (- \frac {x^{4}}{4} + \frac {5 x^{2}}{2}\right ) \log {\relax (x )} + e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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