Optimal. Leaf size=25 \[ e^{\frac {e^{-(-5+x)^2}}{5 x}} (1-\log (3)) \]
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Rubi [F] time = 1.89, 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^{-25+\frac {e^{-25+10 x-x^2}}{5 x}+10 x-x^2} \left (-1+10 x-2 x^2+\left (1-10 x+2 x^2\right ) \log (3)\right )}{5 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{-25+\frac {e^{-25+10 x-x^2}}{5 x}+10 x-x^2} \left (-1+10 x-2 x^2+\left (1-10 x+2 x^2\right ) \log (3)\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \left (1-10 x+2 x^2\right ) (-1+\log (3))}{x^2} \, dx\\ &=\frac {1}{5} (-1+\log (3)) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \left (1-10 x+2 x^2\right )}{x^2} \, dx\\ &=\frac {1}{5} (-1+\log (3)) \int \left (2 e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}+\frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x^2}-\frac {10 e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x}\right ) \, dx\\ &=-\left (\frac {1}{5} (2 (1-\log (3))) \int e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \, dx\right )+(2 (1-\log (3))) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x} \, dx+\frac {1}{5} (-1+\log (3)) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.32, size = 24, normalized size = 0.96 \begin {gather*} -e^{\frac {e^{-(-5+x)^2}}{5 x}} (-1+\log (3)) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.91, size = 47, normalized size = 1.88 \begin {gather*} -{\left (\log \relax (3) - 1\right )} e^{\left (x^{2} - 10 \, x - \frac {5 \, x^{3} - 50 \, x^{2} + 125 \, x - e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x} + 25\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 (2 \, x^{2} - {\left (2 \, x^{2} - 10 \, x + 1\right )} \log \relax (3) - 10 \, x + 1\right )} e^{\left (-x^{2} + 10 \, x + \frac {e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x} - 25\right )}}{5 \, x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 25, normalized size = 1.00
method | result | size |
norman | \(\left (1-\ln \relax (3)\right ) {\mathrm e}^{\frac {{\mathrm e}^{-x^{2}+10 x -25}}{5 x}}\) | \(25\) |
risch | \(-{\mathrm e}^{\frac {{\mathrm e}^{-\left (x -5\right )^{2}}}{5 x}} \ln \relax (3)+{\mathrm e}^{\frac {{\mathrm e}^{-\left (x -5\right )^{2}}}{5 x}}\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 23, normalized size = 0.92 \begin {gather*} -{\left (\log \relax (3) - 1\right )} e^{\left (\frac {e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.52, size = 24, normalized size = 0.96 \begin {gather*} -{\mathrm {e}}^{\frac {{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-25}\,{\mathrm {e}}^{-x^2}}{5\,x}}\,\left (\ln \relax (3)-1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.66, size = 19, normalized size = 0.76 \begin {gather*} \left (1 - \log {\relax (3 )}\right ) e^{\frac {e^{- x^{2} + 10 x - 25}}{5 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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