Optimal. Leaf size=22 \[ 4-\log (5+x) \left (-e^{2+e^x} x+\log (\log (3))\right ) \]
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Rubi [F] time = 1.39, 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^{2+e^x} x \left (x+\left (5+x+e^x \left (5 x+x^2\right )\right ) \log (5+x)\right )-x \log (\log (3))}{5 x+x^2} \, 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^{2+e^x} x \left (x+\left (5+x+e^x \left (5 x+x^2\right )\right ) \log (5+x)\right )-x \log (\log (3))}{x (5+x)} \, dx\\ &=\int \left (e^{2+e^x+x} x \log (5+x)+\frac {e^{2+e^x} x+5 e^{2+e^x} \log (5+x)+e^{2+e^x} x \log (5+x)-\log (\log (3))}{5+x}\right ) \, dx\\ &=\int e^{2+e^x+x} x \log (5+x) \, dx+\int \frac {e^{2+e^x} x+5 e^{2+e^x} \log (5+x)+e^{2+e^x} x \log (5+x)-\log (\log (3))}{5+x} \, dx\\ &=\log (5+x) \int e^{2+e^x+x} x \, dx+\int \frac {e^{2+e^x} x+e^{2+e^x} (5+x) \log (5+x)-\log (\log (3))}{5+x} \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=\log (5+x) \int e^{2+e^x+x} x \, dx+\int \left (\frac {e^{2+e^x} (x+5 \log (5+x)+x \log (5+x))}{5+x}-\frac {\log (\log (3))}{5+x}\right ) \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=-\log (5+x) \log (\log (3))+\log (5+x) \int e^{2+e^x+x} x \, dx+\int \frac {e^{2+e^x} (x+5 \log (5+x)+x \log (5+x))}{5+x} \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=-\log (5+x) \log (\log (3))+\log (5+x) \int e^{2+e^x+x} x \, dx+\int \frac {e^{2+e^x} (x+(5+x) \log (5+x))}{5+x} \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=-\log (5+x) \log (\log (3))+\log (5+x) \int e^{2+e^x+x} x \, dx+\int \left (\frac {e^{2+e^x} x}{5+x}+e^{2+e^x} \log (5+x)\right ) \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=-\log (5+x) \log (\log (3))+\log (5+x) \int e^{2+e^x+x} x \, dx+\int \frac {e^{2+e^x} x}{5+x} \, dx+\int e^{2+e^x} \log (5+x) \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=e^2 \text {Ei}\left (e^x\right ) \log (5+x)-\log (5+x) \log (\log (3))+\log (5+x) \int e^{2+e^x+x} x \, dx+\int \left (e^{2+e^x}-\frac {5 e^{2+e^x}}{5+x}\right ) \, dx-\int \frac {e^2 \text {Ei}\left (e^x\right )}{5+x} \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=e^2 \text {Ei}\left (e^x\right ) \log (5+x)-\log (5+x) \log (\log (3))-5 \int \frac {e^{2+e^x}}{5+x} \, dx-e^2 \int \frac {\text {Ei}\left (e^x\right )}{5+x} \, dx+\log (5+x) \int e^{2+e^x+x} x \, dx+\int e^{2+e^x} \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ &=e^2 \text {Ei}\left (e^x\right ) \log (5+x)-\log (5+x) \log (\log (3))-5 \int \frac {e^{2+e^x}}{5+x} \, dx-e^2 \int \frac {\text {Ei}\left (e^x\right )}{5+x} \, dx+\log (5+x) \int e^{2+e^x+x} x \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx+\operatorname {Subst}\left (\int \frac {e^{2+x}}{x} \, dx,x,e^x\right )\\ &=e^2 \text {Ei}\left (e^x\right )+e^2 \text {Ei}\left (e^x\right ) \log (5+x)-\log (5+x) \log (\log (3))-5 \int \frac {e^{2+e^x}}{5+x} \, dx-e^2 \int \frac {\text {Ei}\left (e^x\right )}{5+x} \, dx+\log (5+x) \int e^{2+e^x+x} x \, dx-\int \frac {\int e^{2+e^x+x} x \, dx}{5+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 20, normalized size = 0.91 \begin {gather*} \log (5+x) \left (e^{2+e^x} x-\log (\log (3))\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 22, normalized size = 1.00 \begin {gather*} e^{\left (e^{x} + \log \relax (x) + 2\right )} \log \left (x + 5\right ) - \log \left (x + 5\right ) \log \left (\log \relax (3)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 29, normalized size = 1.32 \begin {gather*} {\left (x e^{\left (x + e^{x} + 2\right )} \log \left (x + 5\right ) - e^{x} \log \left (x + 5\right ) \log \left (\log \relax (3)\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 = 0.29, size = 22, normalized size = 1.00
method | result | size |
risch | \(-\ln \left (\ln \relax (3)\right ) \ln \left (5+x \right )+\ln \left (5+x \right ) {\mathrm e}^{{\mathrm e}^{x}+2} x\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (e^{x} + 2\right )} \log \left (x + 5\right ) - \log \left (x + 5\right ) \log \left (\log \relax (3)\right ) \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 = -0.05 \begin {gather*} -\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^x+\ln \relax (x)+2}\,\left (x+\ln \left (x+5\right )\,\left (x+{\mathrm {e}}^x\,\left (x^2+5\,x\right )+5\right )\right )-x\,\ln \left (\ln \relax (3)\right )}{x^2+5\,x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 11.45, size = 22, normalized size = 1.00 \begin {gather*} x e^{e^{x} + 2} \log {\left (x + 5 \right )} - \log {\left (x + 5 \right )} \log {\left (\log {\relax (3 )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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