Optimal. Leaf size=26 \[ \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \]
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Rubi [F] time = 2.88, 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 {1}{50} \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{50} \int \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) \left (10-\frac {50 e^{e^2+x}}{x^2}-5 x\right ) x \, dx\\ &=\frac {1}{50} \int \left (-\frac {50 \exp \left (\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right )}{x}-5 \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) (-2+x) x\right ) \, dx\\ &=-\left (\frac {1}{10} \int \exp \left (-e^2-x+\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right ) (-2+x) x \, dx\right )-\int \frac {\exp \left (\frac {1}{50} e^{-e^2-x} x^2 \left (5+\frac {50 e^{e^2+x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right )\right )}{x} \, dx\\ &=-\left (\frac {1}{10} \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} (-2+x) \log (4) \, dx\right )-\int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x^2} \, dx\\ &=-\left (\frac {1}{10} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} (-2+x) \, dx\right )-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx\\ &=-\left (\frac {1}{10} \log (4) \int \left (-2 e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2}+e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} x\right ) \, dx\right )-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx\\ &=-\left (\frac {1}{10} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} x \, dx\right )+\frac {1}{5} \log (4) \int e^{-e^2-x+\frac {1}{10} e^{-e^2-x} x^2} \, dx-\log (4) \int \frac {e^{\frac {1}{10} e^{-e^2-x} x^2}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.22, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{10} e^{-e^2-x} x^2} \log (4)}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.38, size = 67, normalized size = 2.58 \begin {gather*} e^{\left (-\frac {1}{2} \, {\left ({\left (2 \, x + 2 \, e^{2} - \log \left (\frac {2}{25} \, \log \relax (2)^{2}\right ) + \log \left (\frac {50}{x^{2}}\right )\right )} e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} + x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \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*} \int -\frac {{\left (5 \, x + e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} - 10\right )} e^{\left ({\left (e^{\left (x + e^{2} + \log \left (\frac {50}{x^{2}}\right )\right )} \log \left (\frac {2 \, \log \relax (2)}{x}\right ) + 5\right )} e^{\left (-x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )} - x - e^{2} - \log \left (\frac {50}{x^{2}}\right )\right )}}{x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.40, size = 38, normalized size = 1.46
method | result | size |
default | \({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \relax (2)}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\) | \(38\) |
risch | \(\frac {2 \ln \relax (2) {\mathrm e}^{-\frac {x^{2} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}-x -{\mathrm e}^{2}}}{10}}}{x}\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{50} \, \int 5 \, {\left (x + \frac {10 \, e^{\left (x + e^{2}\right )}}{x^{2}} - 2\right )} x e^{\left (\frac {1}{10} \, x^{2} {\left (\frac {10 \, e^{\left (x + e^{2}\right )} \log \left (\frac {2 \, \log \relax (2)}{x}\right )}{x^{2}} + 1\right )} e^{\left (-x - e^{2}\right )} - x - e^{2}\right )}\,{d x} \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 {x\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \relax (2)}{5}-\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \relax (2)}{5}+\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}}\,\ln \relax (2)}{x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.42, size = 36, normalized size = 1.38 \begin {gather*} e^{\frac {x^{2} \left (5 + \frac {50 e^{x + e^{2}} \log {\left (\frac {2 \log {\relax (2 )}}{x} \right )}}{x^{2}}\right ) e^{- x - e^{2}}}{50}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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