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 = 1.66, 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 = 0.12, 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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Maple [A]
time = 0.34, size = 38, normalized size = 1.46
method | result | size |
default | \({\mathrm e}^{\frac {\left (\ln \left (\frac {2 \ln \left (2\right )}{x}\right ) {\mathrm e}^{\ln \left (\frac {50}{x^{2}}\right )+x +{\mathrm e}^{2}}+5\right ) x^{2} {\mathrm e}^{-x -{\mathrm e}^{2}}}{50}}\) | \(38\) |
risch | \({\mathrm e}^{\frac {\left (-1\right )^{-\frac {\mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}} \left (-{\mathrm e}^{-x -{\mathrm e}^{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}} x^{2}-10 \ln \left (x \right ) {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}}+10 \ln \left (2\right ) {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}}+10 \ln \left (\ln \left (2\right )\right ) {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}}\right )}{10}}\) | \(121\) |
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs.
\(2 (29) = 58\).
time = 0.38, 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 \left (2\right )^{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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Sympy [A]
time = 0.15, size = 36, normalized size = 1.38 \begin {gather*} e^{\frac {x^{2} \cdot \left (5 + \frac {50 e^{x + e^{2}} \log {\left (\frac {2 \log {\left (2 \right )}}{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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 \left (2\right )}{5}-\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}-{\mathrm {e}}^2-x}\,\ln \left (2\right )}{5}+\frac {2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}}{10}}\,\ln \left (2\right )}{x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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