Optimal. Leaf size=30 \[ \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \]
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Rubi [F] time = 9.75, 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 {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (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 {x-2 x^3-e^x \left (-1+2 x^2\right )-\left (e^x+x\right ) \log (x)-\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{3 x^3 \left (e^x+x\right )^2 \left (2 x^2-\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \frac {x-2 x^3-e^x \left (-1+2 x^2\right )-\left (e^x+x\right ) \log (x)-\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{x^3 \left (e^x+x\right )^2 \left (2 x^2-\log (x)\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {(-1+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right )^2}-\frac {-1+2 x^2+\log (x)+4 x^2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )+2 x^3 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-x \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {(-1+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right )^2} \, dx-\frac {1}{3} \int \frac {-1+2 x^2+\log (x)+4 x^2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )+2 x^3 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-x \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx\\ &=-\left (\frac {1}{3} \int \frac {-1+2 x^2+\log (x)+2 x^2 (2+x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )-(2+x) \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx\right )+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \left (-\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\int \frac {1}{x \left (e^x+x\right )^2} \, dx\right )}{2 x^3-x \log (x)} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=-\left (\frac {1}{3} \int \left (-\frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {2}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {2 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )}+\frac {4 \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}-\frac {2 \log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}-\frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )}\right ) \, dx\right )+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \left (-\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\int \frac {1}{x \left (e^x+x\right )^2} \, dx\right )}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \left (-\frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\left (-1+2 x^2+\log (x)\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \left (-\frac {\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {2 x \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)}-\frac {\log (x) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )}\right ) \, dx+\frac {1}{3} \int \left (-\frac {\int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )}+\frac {2 x \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)}-\frac {\log (x) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x)}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^2 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log (x) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \int \frac {\log (x) \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{x \left (-2 x^2+\log (x)\right )} \, dx-\frac {2}{3} \int \frac {1}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{\left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx+\frac {2}{3} \int \frac {\log (x) \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x^3 \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {2}{3} \int \frac {x \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)} \, dx+\frac {2}{3} \int \frac {x \int \frac {1}{x \left (e^x+x\right )^2} \, dx}{2 x^2-\log (x)} \, dx-\frac {4}{3} \int \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{x \left (e^x+x\right ) \left (2 x^2-\log (x)\right )} \, dx-\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x^2 \left (e^x+x\right )^2} \, dx+\frac {1}{3} \log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right ) \int \frac {1}{x \left (e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 30, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.13, size = 30, normalized size = 1.00 \begin {gather*} \frac {\log \left (-\frac {x}{5 \, {\left (2 \, x^{2} - \log \relax (x)\right )}}\right )}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.47, size = 30, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) + \log \left (-2 \, x^{2} + \log \relax (x)\right ) - \log \relax (x)}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.22, size = 205, normalized size = 6.83
method | result | size |
risch | \(-\frac {\ln \left (x^{2}-\frac {\ln \relax (x )}{2}\right )}{3 x^{2} \left ({\mathrm e}^{x}+x \right )}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \relax (x )}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \relax (x )}{2}}\right ) \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \relax (x )}{2}}\right )^{3}-2 i \pi +2 \ln \relax (5)+2 \ln \relax (2)-2 \ln \relax (x )}{6 x^{2} \left ({\mathrm e}^{x}+x \right )}\) | \(205\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 30, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) + \log \left (-2 \, x^{2} + \log \relax (x)\right ) - \log \relax (x)}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.48, size = 26, normalized size = 0.87 \begin {gather*} \frac {\ln \left (\frac {x}{5\,\left (\ln \relax (x)-2\,x^2\right )}\right )}{3\,x^2\,\left (x+{\mathrm {e}}^x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.49, size = 26, normalized size = 0.87 \begin {gather*} \frac {\log {\left (\frac {x}{- 10 x^{2} + 5 \log {\relax (x )}} \right )}}{3 x^{3} + 3 x^{2} e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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