Optimal. Leaf size=19 \[ e^{9-\frac {\log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x}} \]
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Rubi [F] time = 6.05, 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-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}\right ) \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\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 {9^{\frac {1}{x}} e^9 (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}} \left (9 x-(-2+x) \log (-2+x) (-100+9 \log (\log (-2+x))) \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )\right )}{(2-x) x^2 \log (-2+x)} \, dx\\ &=e^9 \int \frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}} \left (9 x-(-2+x) \log (-2+x) (-100+9 \log (\log (-2+x))) \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )\right )}{(2-x) x^2 \log (-2+x)} \, dx\\ &=e^9 \int \left (-\frac {9^{1+\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)}+\frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2}\right ) \, dx\\ &=-\left (e^9 \int \frac {9^{1+\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)} \, dx\right )+e^9 \int \frac {9^{\frac {1}{x}} (-100+9 \log (\log (-2+x)))^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{(-2+x) x \log (-2+x)} \, dx\right )+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (e^9 \int \left (\frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{2 (-2+x) \log (-2+x)}-\frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{2 x \log (-2+x)}\right ) \, dx\right )+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ &=-\left (\frac {1}{2} e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{(-2+x) \log (-2+x)} \, dx\right )+\frac {1}{2} e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1-\frac {1}{x}}}{x \log (-2+x)} \, dx+e^9 \int \frac {\left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.68, size = 19, normalized size = 1.00 \begin {gather*} e^9 \left (-\frac {100}{9}+\log (\log (-2+x))\right )^{-1/x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 21, normalized size = 1.11 \begin {gather*} e^{\left (\frac {5 \, x - \log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 3.51, size = 16, normalized size = 0.84 \begin {gather*} e^{\left (-\frac {\log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 9\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 17, normalized size = 0.89
method | result | size |
risch | \(\left (\ln \left (\ln \left (x -2\right )\right )-\frac {100}{9}\right )^{-\frac {1}{x}} {\mathrm e}^{9}\) | \(17\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.22, size = 25, normalized size = 1.32 \begin {gather*} e^{\left (\frac {2 \, \log \relax (3)}{x} - \frac {\log \left (9 \, \log \left (\log \left (x - 2\right )\right ) - 100\right )}{x} + 9\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.15, size = 16, normalized size = 0.84 \begin {gather*} \frac {{\mathrm {e}}^9}{{\left (\ln \left (\ln \left (x-2\right )\right )-\frac {100}{9}\right )}^{1/x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.37, size = 20, normalized size = 1.05 \begin {gather*} e^{4} e^{\frac {5 x - \log {\left (\log {\left (\log {\left (x - 2 \right )} \right )} - \frac {100}{9} \right )}}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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