Optimal. Leaf size=21 \[ -5+\frac {5 \log (x)}{6 \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \]
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Rubi [F] time = 0.73, 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 {-5+5 x+(-5-10 x) \log (x)+5 \log \left (e^{-1+x} x\right )}{6 x-12 x^2+6 x^3+\left (-12 x+12 x^2\right ) \log \left (e^{-1+x} x\right )+6 x \log ^2\left (e^{-1+x} x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (-1+x-\log (x)-2 x \log (x)+\log \left (e^{-1+x} x\right )\right )}{6 x \left (1-x-\log \left (e^{-1+x} x\right )\right )^2} \, dx\\ &=\frac {5}{6} \int \frac {-1+x-\log (x)-2 x \log (x)+\log \left (e^{-1+x} x\right )}{x \left (1-x-\log \left (e^{-1+x} x\right )\right )^2} \, dx\\ &=\frac {5}{6} \int \left (-\frac {(1+2 x) \log (x)}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )^2}+\frac {1}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )}\right ) \, dx\\ &=-\left (\frac {5}{6} \int \frac {(1+2 x) \log (x)}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )^2} \, dx\right )+\frac {5}{6} \int \frac {1}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \, dx\\ &=\frac {5}{6} \int \frac {1}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \, dx-\frac {5}{6} \int \left (\frac {2 \log (x)}{\left (-1+x+\log \left (e^{-1+x} x\right )\right )^2}+\frac {\log (x)}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {5}{6} \int \frac {\log (x)}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )^2} \, dx\right )+\frac {5}{6} \int \frac {1}{x \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \, dx-\frac {5}{3} \int \frac {\log (x)}{\left (-1+x+\log \left (e^{-1+x} x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.29, size = 32, normalized size = 1.52 \begin {gather*} -\frac {5 \left (-1+x-\log (x)+\log \left (e^{-1+x} x\right )\right )}{6 \left (-1+x+\log \left (e^{-1+x} x\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.64, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.10, size = 107, normalized size = 5.10
method | result | size |
risch | \(\frac {5 \ln \relax (x )}{3 \left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{2}-i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{x -1}\right )^{3}+2 x +2 \ln \relax (x )+2 \ln \left ({\mathrm e}^{x -1}\right )-2\right )}\) | \(107\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x - 1\right )}}{3 \, {\left (2 \, x + \log \relax (x) - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.68, size = 15, normalized size = 0.71 \begin {gather*} \frac {5\,\ln \relax (x)}{6\,\left (2\,x+\ln \relax (x)-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.36, size = 14, normalized size = 0.67 \begin {gather*} \frac {5 - 5 x}{6 x + 3 \log {\relax (x )} - 6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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