Optimal. Leaf size=26 \[ \frac {3 e^{-x} \left (x+\frac {x}{3-\frac {3}{\log (x)}}\right )}{5 x^2} \]
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Rubi [F] time = 0.96, 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 {-4-3 x+(7+7 x) \log (x)+(-4-4 x) \log ^2(x)}{5 e^x x^2-10 e^x x^2 \log (x)+5 e^x x^2 \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 {e^{-x} \left (-4-3 x+7 (1+x) \log (x)-4 (1+x) \log ^2(x)\right )}{5 x^2 (1-\log (x))^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-x} \left (-4-3 x+7 (1+x) \log (x)-4 (1+x) \log ^2(x)\right )}{x^2 (1-\log (x))^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {4 e^{-x} (1+x)}{x^2}-\frac {e^{-x}}{x^2 (-1+\log (x))^2}+\frac {e^{-x} (-1-x)}{x^2 (-1+\log (x))}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx\right )+\frac {1}{5} \int \frac {e^{-x} (-1-x)}{x^2 (-1+\log (x))} \, dx-\frac {4}{5} \int \frac {e^{-x} (1+x)}{x^2} \, dx\\ &=\frac {4 e^{-x}}{5 x}+\frac {1}{5} \int \left (-\frac {e^{-x}}{x^2 (-1+\log (x))}-\frac {e^{-x}}{x (-1+\log (x))}\right ) \, dx-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx\\ &=\frac {4 e^{-x}}{5 x}-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))^2} \, dx-\frac {1}{5} \int \frac {e^{-x}}{x^2 (-1+\log (x))} \, dx-\frac {1}{5} \int \frac {e^{-x}}{x (-1+\log (x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 24, normalized size = 0.92 \begin {gather*} \frac {e^{-x} (-3+4 \log (x))}{5 x (-1+\log (x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 22, normalized size = 0.85 \begin {gather*} \frac {4 \, \log \relax (x) - 3}{5 \, {\left (x e^{x} \log \relax (x) - x e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 27, normalized size = 1.04 \begin {gather*} \frac {4 \, e^{\left (-x\right )} \log \relax (x) - 3 \, e^{\left (-x\right )}}{5 \, {\left (x \log \relax (x) - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 21, normalized size = 0.81
method | result | size |
norman | \(\frac {\left (-\frac {3}{5}+\frac {4 \ln \relax (x )}{5}\right ) {\mathrm e}^{-x}}{x \left (\ln \relax (x )-1\right )}\) | \(21\) |
risch | \(\frac {4 \,{\mathrm e}^{-x}}{5 x}+\frac {{\mathrm e}^{-x}}{5 x \left (\ln \relax (x )-1\right )}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.84, size = 22, normalized size = 0.85 \begin {gather*} \frac {{\left (4 \, \log \relax (x) - 3\right )} e^{\left (-x\right )}}{5 \, {\left (x \log \relax (x) - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.59, size = 28, normalized size = 1.08 \begin {gather*} -\frac {x\,\left (4\,\ln \relax (x)-3\right )}{5\,\left (x^2\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x\,\ln \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 19, normalized size = 0.73 \begin {gather*} \frac {\left (4 \log {\relax (x )} - 3\right ) e^{- x}}{5 x \log {\relax (x )} - 5 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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