Optimal. Leaf size=24 \[ \frac {2+x}{\log \left (12 e^{-x} \left (-5+x-x^2+\log (x)\right )\right )} \]
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Rubi [F] time = 2.37, 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 {-2-13 x+x^3-x^4+\left (2 x+x^2\right ) \log (x)+\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (-5 x+x^2-x^3+x \log (x)\right ) \log ^2\left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\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 {2+13 x-x^3+x^4-\left (2 x+x^2\right ) \log (x)-\left (-5 x+x^2-x^3+x \log (x)\right ) \log \left (e^{-x} \left (-60+12 x-12 x^2+12 \log (x)\right )\right )}{\left (5 x-x^2+x^3-x \log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=\int \left (\frac {13}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {2}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}-\frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}-\frac {(2+x) \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}\right ) \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {(2+x) \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \left (\frac {2 \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}+\frac {x \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )}\right ) \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ &=2 \int \frac {1}{x \left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-2 \int \frac {\log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+13 \int \frac {1}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x^2}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {x^3}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx-\int \frac {x \log (x)}{\left (5-x+x^2-\log (x)\right ) \log ^2\left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx+\int \frac {1}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 26, normalized size = 1.08 \begin {gather*} \frac {2+x}{\log \left (-12 e^{-x} \left (5-x+x^2-\log (x)\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.67, size = 30, normalized size = 1.25 \begin {gather*} \frac {x + 2}{\log \left (-12 \, {\left (x^{2} - x + 5\right )} e^{\left (-x\right )} + 12 \, e^{\left (-x\right )} \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 26, normalized size = 1.08 \begin {gather*} -\frac {x + 2}{x - \log \left (12\right ) - \log \left (-x^{2} + x + \log \relax (x) - 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.15, size = 201, normalized size = 8.38
method | result | size |
risch | \(-\frac {2 i \left (2+x \right )}{-2 \pi \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right )\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\ln \relax (x )+x -x^{2}-5\right ) {\mathrm e}^{-x}\right )^{3}+2 \pi -4 i \ln \relax (2)-2 i \ln \relax (3)-2 i \ln \left (-\ln \relax (x )-x +x^{2}+5\right )+2 i \ln \left ({\mathrm e}^{x}\right )}\) | \(201\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.75, size = 30, normalized size = 1.25 \begin {gather*} -\frac {x + 2}{x - \log \relax (3) - 2 \, \log \relax (2) - \log \left (-x^{2} + x + \log \relax (x) - 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.82, size = 26, normalized size = 1.08 \begin {gather*} \frac {x+2}{\ln \left ({\mathrm {e}}^{-x}\,\left (12\,x+12\,\ln \relax (x)-12\,x^2-60\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.56, size = 22, normalized size = 0.92 \begin {gather*} \frac {x + 2}{\log {\left (\left (- 12 x^{2} + 12 x + 12 \log {\relax (x )} - 60\right ) e^{- x} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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