Optimal. Leaf size=23 \[ e^{\frac {e^{-x} x}{-1+\frac {x}{5-9 \log (x)}}} \]
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Rubi [F] time = 10.48, 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 {\exp \left (\frac {5 x-9 x \log (x)}{e^x (-5+x)+9 e^x \log (x)}\right ) \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{e^x \left (25-10 x+x^2\right )+e^x (-90+18 x) \log (x)+81 e^x \log ^2(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 {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) \left (-25+16 x-5 x^2+\left (90-90 x+9 x^2\right ) \log (x)+(-81+81 x) \log ^2(x)\right )}{(5-x-9 \log (x))^2} \, dx\\ &=\int \left (-\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right )+\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x-\frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x (9+x)}{(-5+x+9 \log (x))^2}-\frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) (-2+x) x}{-5+x+9 \log (x)}\right ) \, dx\\ &=-\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) \, dx+\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x \, dx-\int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x (9+x)}{(-5+x+9 \log (x))^2} \, dx-\int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) (-2+x) x}{-5+x+9 \log (x)} \, dx\\ &=-\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) \, dx+\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x \, dx-\int \left (\frac {9 \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x}{(-5+x+9 \log (x))^2}+\frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x^2}{(-5+x+9 \log (x))^2}\right ) \, dx-\int \left (-\frac {2 \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x}{-5+x+9 \log (x)}+\frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x^2}{-5+x+9 \log (x)}\right ) \, dx\\ &=2 \int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x}{-5+x+9 \log (x)} \, dx-9 \int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x}{(-5+x+9 \log (x))^2} \, dx-\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) \, dx+\int \exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x \, dx-\int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x^2}{(-5+x+9 \log (x))^2} \, dx-\int \frac {\exp \left (-x+\frac {e^{-x} (5 x-9 x \log (x))}{-5+x+9 \log (x)}\right ) x^2}{-5+x+9 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.24, size = 39, normalized size = 1.70 \begin {gather*} e^{\frac {5 e^{-x} x}{-5+x+9 \log (x)}} x^{-\frac {9 e^{-x} x}{-5+x+9 \log (x)}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.02, size = 27, normalized size = 1.17 \begin {gather*} e^{\left (-\frac {9 \, x \log \relax (x) - 5 \, x}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \relax (x)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 44, normalized size = 1.91 \begin {gather*} e^{\left (-\frac {9 \, x \log \relax (x)}{x e^{x} + 9 \, e^{x} \log \relax (x) - 5 \, e^{x}} + \frac {5 \, x}{x e^{x} + 9 \, e^{x} \log \relax (x) - 5 \, e^{x}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 24, normalized size = 1.04
method | result | size |
risch | \({\mathrm e}^{-\frac {x \left (9 \ln \relax (x )-5\right ) {\mathrm e}^{-x}}{9 \ln \relax (x )+x -5}}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.60, size = 73, normalized size = 3.17 \begin {gather*} e^{\left (-9 \, e^{\left (-x\right )} \log \relax (x) + \frac {81 \, \log \relax (x)^{2}}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \relax (x)} - \frac {90 \, \log \relax (x)}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \relax (x)} + \frac {25}{{\left (x - 5\right )} e^{x} + 9 \, e^{x} \log \relax (x)} + 5 \, e^{\left (-x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.54, size = 45, normalized size = 1.96 \begin {gather*} {\mathrm {e}}^{\frac {5\,x}{9\,{\mathrm {e}}^x\,\ln \relax (x)-5\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {9\,x\,\ln \relax (x)}{9\,{\mathrm {e}}^x\,\ln \relax (x)-5\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.39, size = 26, normalized size = 1.13 \begin {gather*} e^{\frac {- 9 x \log {\relax (x )} + 5 x}{\left (x - 5\right ) e^{x} + 9 e^{x} \log {\relax (x )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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