Optimal. Leaf size=30 \[ 5 e^{-x} \left (\frac {1+x}{5}+\log ^2(3)\right ) \left (2+x+\frac {x^2}{\log (x)}\right ) \]
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Rubi [F] time = 0.97, 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 {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-x^2+x \left (-1-5 \log ^2(3)\right )+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\int \left (-e^{-x} x^2+e^{-x} \left (1-5 \log ^2(3)\right )-e^{-x} x \left (1+5 \log ^2(3)\right )-\frac {e^{-x} x \left (1+x+5 \log ^2(3)\right )}{\log ^2(x)}+\frac {e^{-x} x \left (-x^2+x \left (2-5 \log ^2(3)\right )+2 \left (1+5 \log ^2(3)\right )\right )}{\log (x)}\right ) \, dx\\ &=\left (-1-5 \log ^2(3)\right ) \int e^{-x} x \, dx+\left (1-5 \log ^2(3)\right ) \int e^{-x} \, dx-\int e^{-x} x^2 \, dx-\int \frac {e^{-x} x \left (1+x+5 \log ^2(3)\right )}{\log ^2(x)} \, dx+\int \frac {e^{-x} x \left (-x^2+x \left (2-5 \log ^2(3)\right )+2 \left (1+5 \log ^2(3)\right )\right )}{\log (x)} \, dx\\ &=e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )-2 \int e^{-x} x \, dx+\left (-1-5 \log ^2(3)\right ) \int e^{-x} \, dx-\int \left (\frac {e^{-x} x^2}{\log ^2(x)}+\frac {e^{-x} x \left (1+5 \log ^2(3)\right )}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {e^{-x} x^3}{\log (x)}+\frac {e^{-x} x^2 \left (2-5 \log ^2(3)\right )}{\log (x)}+\frac {2 e^{-x} x \left (1+5 \log ^2(3)\right )}{\log (x)}\right ) \, dx\\ &=2 e^{-x} x+e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} \left (1+5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )-2 \int e^{-x} \, dx+\left (2-5 \log ^2(3)\right ) \int \frac {e^{-x} x^2}{\log (x)} \, dx-\left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log ^2(x)} \, dx+\left (2 \left (1+5 \log ^2(3)\right )\right ) \int \frac {e^{-x} x}{\log (x)} \, dx-\int \frac {e^{-x} x^2}{\log ^2(x)} \, dx-\int \frac {e^{-x} x^3}{\log (x)} \, dx\\ &=2 e^{-x}+2 e^{-x} x+e^{-x} x^2-e^{-x} \left (1-5 \log ^2(3)\right )+e^{-x} \left (1+5 \log ^2(3)\right )+e^{-x} x \left (1+5 \log ^2(3)\right )+\left (2-5 \log ^2(3)\right ) \int \frac {e^{-x} x^2}{\log (x)} \, dx-\left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log ^2(x)} \, dx+\left (2 \left (1+5 \log ^2(3)\right )\right ) \int \frac {e^{-x} x}{\log (x)} \, dx-\int \frac {e^{-x} x^2}{\log ^2(x)} \, dx-\int \frac {e^{-x} x^3}{\log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 29, normalized size = 0.97 \begin {gather*} \frac {e^{-x} \left (1+x+5 \log ^2(3)\right ) \left (x^2+(2+x) \log (x)\right )}{\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.90, size = 51, normalized size = 1.70 \begin {gather*} \frac {{\left (5 \, {\left (x + 2\right )} \log \relax (3)^{2} + x^{2} + 3 \, x + 2\right )} e^{\left (-x\right )} \log \relax (x) + {\left (5 \, x^{2} \log \relax (3)^{2} + x^{3} + x^{2}\right )} e^{\left (-x\right )}}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (5 \, x \log \relax (3)^{2} + {\left (5 \, {\left (x + 1\right )} \log \relax (3)^{2} + x^{2} + x - 1\right )} \log \relax (x)^{2} + x^{2} + {\left (x^{3} + 5 \, {\left (x^{2} - 2 \, x\right )} \log \relax (3)^{2} - 2 \, x^{2} - 2 \, x\right )} \log \relax (x) + x\right )} e^{\left (-x\right )}}{\log \relax (x)^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 49, normalized size = 1.63
method | result | size |
risch | \(\left (5 x \ln \relax (3)^{2}+10 \ln \relax (3)^{2}+x^{2}+3 x +2\right ) {\mathrm e}^{-x}+\frac {x^{2} {\mathrm e}^{-x} \left (5 \ln \relax (3)^{2}+x +1\right )}{\ln \relax (x )}\) | \(49\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.78, size = 76, normalized size = 2.53 \begin {gather*} 5 \, {\left (x + 1\right )} e^{\left (-x\right )} \log \relax (3)^{2} + 5 \, e^{\left (-x\right )} \log \relax (3)^{2} + {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + {\left (x + 1\right )} e^{\left (-x\right )} + \frac {{\left ({\left (5 \, \log \relax (3)^{2} + 1\right )} x^{2} + x^{3}\right )} e^{\left (-x\right )}}{\log \relax (x)} - e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.51, size = 39, normalized size = 1.30 \begin {gather*} {\mathrm {e}}^{-x}\,\left (x+2\right )\,\left (x+5\,{\ln \relax (3)}^2+1\right )+\frac {x^2\,{\mathrm {e}}^{-x}\,\left (x+5\,{\ln \relax (3)}^2+1\right )}{\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.33, size = 63, normalized size = 2.10 \begin {gather*} \frac {\left (x^{3} + x^{2} \log {\relax (x )} + x^{2} + 5 x^{2} \log {\relax (3 )}^{2} + 3 x \log {\relax (x )} + 5 x \log {\relax (3 )}^{2} \log {\relax (x )} + 2 \log {\relax (x )} + 10 \log {\relax (3 )}^{2} \log {\relax (x )}\right ) e^{- x}}{\log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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