Optimal. Leaf size=30 \[ 2+x+e^{-x} x+\log (5)+\left (e^{e^x}-x^2\right ) (5+\log (x))^2 \]
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Rubi [F] time = 2.24, 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+e^x \left (x-60 x^2\right )-22 e^x x^2 \log (x)-2 e^x x^2 \log ^2(x)+e^{e^x} \left (10 e^x+25 e^{2 x} x+\left (2 e^x+10 e^{2 x} x\right ) \log (x)+e^{2 x} x \log ^2(x)\right )\right )}{x} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-x}-e^{-x} x+e^{e^x+x} (5+\log (x))^2-\frac {-10 e^{e^x}-x+60 x^2-2 e^{e^x} \log (x)+22 x^2 \log (x)+2 x^2 \log ^2(x)}{x}\right ) \, dx\\ &=\int e^{-x} \, dx-\int e^{-x} x \, dx+\int e^{e^x+x} (5+\log (x))^2 \, dx-\int \frac {-10 e^{e^x}-x+60 x^2-2 e^{e^x} \log (x)+22 x^2 \log (x)+2 x^2 \log ^2(x)}{x} \, dx\\ &=-e^{-x}+e^{-x} x-\int e^{-x} \, dx+\int \left (25 e^{e^x+x}+10 e^{e^x+x} \log (x)+e^{e^x+x} \log ^2(x)\right ) \, dx-\int \left (-1+60 x+22 x \log (x)+2 x \log ^2(x)-\frac {2 e^{e^x} (5+\log (x))}{x}\right ) \, dx\\ &=x+e^{-x} x-30 x^2-2 \int x \log ^2(x) \, dx+2 \int \frac {e^{e^x} (5+\log (x))}{x} \, dx+10 \int e^{e^x+x} \log (x) \, dx-22 \int x \log (x) \, dx+25 \int e^{e^x+x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ &=x+e^{-x} x-\frac {49 x^2}{2}+10 e^{e^x} \log (x)-11 x^2 \log (x)-x^2 \log ^2(x)+2 \int x \log (x) \, dx+2 \int \left (\frac {5 e^{e^x}}{x}+\frac {e^{e^x} \log (x)}{x}\right ) \, dx-10 \int \frac {e^{e^x}}{x} \, dx+25 \operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )+\int e^{e^x+x} \log ^2(x) \, dx\\ &=25 e^{e^x}+x+e^{-x} x-25 x^2+10 e^{e^x} \log (x)-10 x^2 \log (x)-x^2 \log ^2(x)+2 \int \frac {e^{e^x} \log (x)}{x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ &=25 e^{e^x}+x+e^{-x} x-25 x^2+10 e^{e^x} \log (x)-10 x^2 \log (x)-x^2 \log ^2(x)-2 \int \frac {\int \frac {e^{e^x}}{x} \, dx}{x} \, dx+(2 \log (x)) \int \frac {e^{e^x}}{x} \, dx+\int e^{e^x+x} \log ^2(x) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 42, normalized size = 1.40 \begin {gather*} x+e^{-x} x-25 x^2-10 x^2 \log (x)-x^2 \log ^2(x)+e^{e^x} (5+\log (x))^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.02, size = 64, normalized size = 2.13 \begin {gather*} -{\left (x^{2} e^{x} \log \relax (x)^{2} + 10 \, x^{2} e^{x} \log \relax (x) + {\left (25 \, x^{2} - x\right )} e^{x} - {\left (e^{x} \log \relax (x)^{2} + 10 \, e^{x} \log \relax (x) + 25 \, e^{x}\right )} e^{\left (e^{x}\right )} - x\right )} e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 68, normalized size = 2.27 \begin {gather*} -{\left (x^{2} e^{x} \log \relax (x)^{2} + 10 \, x^{2} e^{x} \log \relax (x) + 25 \, x^{2} e^{x} - e^{\left (x + e^{x}\right )} \log \relax (x)^{2} - x e^{x} - 10 \, e^{\left (x + e^{x}\right )} \log \relax (x) - x - 25 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 50, normalized size = 1.67
method | result | size |
risch | \(-x^{2} \ln \relax (x )^{2}-10 x^{2} \ln \relax (x )-x \left (25 \,{\mathrm e}^{x} x -{\mathrm e}^{x}-1\right ) {\mathrm e}^{-x}+\left (25+\ln \relax (x )^{2}+10 \ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{x}}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 51, normalized size = 1.70 \begin {gather*} -x^{2} \log \relax (x)^{2} - 10 \, x^{2} \log \relax (x) - 25 \, x^{2} + {\left (x + 1\right )} e^{\left (-x\right )} + {\left (\log \relax (x)^{2} + 10 \, \log \relax (x) + 25\right )} e^{\left (e^{x}\right )} + x - e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.33, size = 43, normalized size = 1.43 \begin {gather*} x+x\,{\mathrm {e}}^{-x}-10\,x^2\,\ln \relax (x)-x^2\,{\ln \relax (x)}^2+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\ln \relax (x)}^2+10\,\ln \relax (x)+25\right )-25\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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