Optimal. Leaf size=24 \[ \frac {\left (x+e^{1+x} x\right ) \log (x)}{\left (\frac {1}{2}+e^x\right ) x} \]
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Rubi [F] time = 0.85, 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 {2+4 e^x+e^{1+x} \left (2+4 e^x\right )+\left (-4 e^x x+2 e^{1+x} x\right ) \log (x)}{x+4 e^x x+4 e^{2 x} 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 {2 \left (\left (1+2 e^x\right ) \left (1+e^{1+x}\right )+(-2+e) e^x x \log (x)\right )}{\left (1+2 e^x\right )^2 x} \, dx\\ &=2 \int \frac {\left (1+2 e^x\right ) \left (1+e^{1+x}\right )+(-2+e) e^x x \log (x)}{\left (1+2 e^x\right )^2 x} \, dx\\ &=2 \int \left (\frac {e}{2 x}-\frac {(-2+e) \log (x)}{2 \left (1+2 e^x\right )^2}+\frac {(-2+e) (-1+x \log (x))}{2 \left (1+2 e^x\right ) x}\right ) \, dx\\ &=e \log (x)+(2-e) \int \frac {\log (x)}{\left (1+2 e^x\right )^2} \, dx+(-2+e) \int \frac {-1+x \log (x)}{\left (1+2 e^x\right ) x} \, dx\\ &=e \log (x)+(2-e) \int \frac {\log (x)}{\left (1+2 e^x\right )^2} \, dx+(-2+e) \int \left (-\frac {1}{\left (1+2 e^x\right ) x}+\frac {\log (x)}{1+2 e^x}\right ) \, dx\\ &=e \log (x)+(2-e) \int \frac {1}{\left (1+2 e^x\right ) x} \, dx+(2-e) \int \frac {\log (x)}{\left (1+2 e^x\right )^2} \, dx+(-2+e) \int \frac {\log (x)}{1+2 e^x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 \left (1+e^{1+x}\right ) \log (x)}{1+2 e^x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 22, normalized size = 0.92 \begin {gather*} \frac {2 \, {\left (e + e^{\left (x + 2\right )}\right )} \log \relax (x)}{e + 2 \, e^{\left (x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (e^{\left (x + 1\right )} \log \relax (x) + \log \relax (x)\right )}}{2 \, e^{x} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 23, normalized size = 0.96
method | result | size |
norman | \(\frac {2 \ln \relax (x )+2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \relax (x )}{2 \,{\mathrm e}^{x}+1}\) | \(23\) |
risch | \(-\frac {\left ({\mathrm e}-2\right ) \ln \relax (x )}{2 \,{\mathrm e}^{x}+1}+{\mathrm e} \ln \relax (x )\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 22, normalized size = 0.92 \begin {gather*} e \log \relax (x) - \frac {{\left (e - 2\right )} \log \relax (x)}{2 \, e^{x} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.19, size = 18, normalized size = 0.75 \begin {gather*} \frac {2\,\ln \relax (x)\,\left ({\mathrm {e}}^{x+1}+1\right )}{2\,{\mathrm {e}}^x+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.25, size = 24, normalized size = 1.00 \begin {gather*} \frac {- e \log {\relax (x )} + 2 \log {\relax (x )}}{2 e^{x} + 1} + e \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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