Optimal. Leaf size=21 \[ \frac {e^{-\frac {x}{-2+x}}}{45 x \log (x)} \]
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Rubi [F] time = 2.20, 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 {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{\left (20 x^2-20 x^3+5 x^4\right ) \log (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^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{x^2 \left (20-20 x+5 x^2\right ) \log (x)} \, dx\\ &=\int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{5 (-2+x)^2 x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+4 x-x^2+\left (-4+6 x-x^2\right ) \log (x)\right )}{(-2+x)^2 x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+6 x-x^2\right )}{(-2+x)^2 x^2}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2 \log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}} \left (-4+6 x-x^2\right )}{(-2+x)^2 x^2} \, dx-\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{(-2+x)^2}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{2 (-2+x)}-\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2}+\frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{2 x}\right ) \, dx-\frac {1}{5} \int \frac {e^{-\frac {x}{-2+x}}}{9 x^2 \log ^2(x)} \, dx\\ &=-\left (\frac {1}{45} \int \frac {e^{-\frac {x}{-2+x}}}{x^2 \log ^2(x)} \, dx\right )-\frac {1}{10} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{-2+x} \, dx+\frac {1}{10} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x} \, dx+\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{(-2+x)^2} \, dx-\frac {1}{5} \int \frac {e^{-\frac {x+(-2+x) \log (9 \log (x))}{-2+x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 22, normalized size = 1.05 \begin {gather*} \frac {e^{-1-\frac {2}{-2+x}}}{45 x \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 24, normalized size = 1.14 \begin {gather*} \frac {e^{\left (-\frac {{\left (x - 2\right )} \log \left (9 \, \log \relax (x)\right ) + x}{x - 2}\right )}}{5 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 30, normalized size = 1.43
method | result | size |
risch | \(\frac {{\mathrm e}^{-\frac {\ln \left (9 \ln \relax (x )\right ) x -2 \ln \left (9 \ln \relax (x )\right )+x}{x -2}}}{5 x}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.56, size = 19, normalized size = 0.90 \begin {gather*} \frac {e^{\left (-\frac {2}{x - 2} - 1\right )}}{45 \, x \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.42, size = 18, normalized size = 0.86 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x}{x-2}}}{45\,x\,\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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