Optimal. Leaf size=27 \[ \frac {2}{4 e^x x+\frac {1}{-1+\frac {1}{3} \left (2-\frac {\log (x)}{5}\right )}} \]
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Rubi [F] time = 2.70, 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 {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \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 {2 \left (-15-100 e^x x (1+x)-40 e^x x (1+x) \log (x)-4 e^x x (1+x) \log ^2(x)\right )}{x \left (15-20 e^x x-4 e^x x \log (x)\right )^2} \, dx\\ &=2 \int \frac {-15-100 e^x x (1+x)-40 e^x x (1+x) \log (x)-4 e^x x (1+x) \log ^2(x)}{x \left (15-20 e^x x-4 e^x x \log (x)\right )^2} \, dx\\ &=2 \int \left (-\frac {15 (6+5 x+\log (x)+x \log (x))}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2}-\frac {(1+x) (5+\log (x))}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {(1+x) (5+\log (x))}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )} \, dx\right )-30 \int \frac {6+5 x+\log (x)+x \log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {5+\log (x)}{-15+20 e^x x+4 e^x x \log (x)}+\frac {5+\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )}\right ) \, dx\right )-30 \int \left (\frac {5}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2}+\frac {6}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2}+\frac {\log (x)}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2}+\frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {5+\log (x)}{-15+20 e^x x+4 e^x x \log (x)} \, dx\right )-2 \int \frac {5+\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )} \, dx-30 \int \frac {\log (x)}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-30 \int \frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-150 \int \frac {1}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-180 \int \frac {1}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {5}{-15+20 e^x x+4 e^x x \log (x)}+\frac {\log (x)}{-15+20 e^x x+4 e^x x \log (x)}\right ) \, dx\right )-2 \int \left (\frac {5}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )}+\frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )}\right ) \, dx-30 \int \frac {\log (x)}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-30 \int \frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-150 \int \frac {1}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-180 \int \frac {1}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx\\ &=-\left (2 \int \frac {\log (x)}{-15+20 e^x x+4 e^x x \log (x)} \, dx\right )-2 \int \frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )} \, dx-10 \int \frac {1}{-15+20 e^x x+4 e^x x \log (x)} \, dx-10 \int \frac {1}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )} \, dx-30 \int \frac {\log (x)}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-30 \int \frac {\log (x)}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-150 \int \frac {1}{\left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx-180 \int \frac {1}{x \left (-15+20 e^x x+4 e^x x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.42, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 (5+\log (x))}{15-20 e^x x-4 e^x x \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 22, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (\log \relax (x) + 5\right )}}{4 \, x e^{x} \log \relax (x) + 20 \, x e^{x} - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.00, size = 22, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (\log \relax (x) + 5\right )}}{4 \, x e^{x} \log \relax (x) + 20 \, x e^{x} - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 36, normalized size = 1.33
method | result | size |
risch | \(\frac {{\mathrm e}^{-x}}{2 x}+\frac {15 \,{\mathrm e}^{-x}}{2 x \left (4 x \,{\mathrm e}^{x} \ln \relax (x )+20 \,{\mathrm e}^{x} x -15\right )}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 22, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (\log \relax (x) + 5\right )}}{4 \, {\left (x \log \relax (x) + 5 \, x\right )} e^{x} - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left (8\,x^2+8\,x\right )\,{\ln \relax (x)}^2+{\mathrm {e}}^x\,\left (80\,x^2+80\,x\right )\,\ln \relax (x)+{\mathrm {e}}^x\,\left (200\,x^2+200\,x\right )+30}{225\,x-600\,x^2\,{\mathrm {e}}^x-\ln \relax (x)\,\left (120\,x^2\,{\mathrm {e}}^x-160\,x^3\,{\mathrm {e}}^{2\,x}\right )+400\,x^3\,{\mathrm {e}}^{2\,x}+16\,x^3\,{\mathrm {e}}^{2\,x}\,{\ln \relax (x)}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 20, normalized size = 0.74 \begin {gather*} \frac {2 \log {\relax (x )} + 10}{\left (4 x \log {\relax (x )} + 20 x\right ) e^{x} - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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