Optimal. Leaf size=24 \[ \log \left (\frac {8 \left (\frac {x}{e^{(5-x) x}+x}+\log (5)\right )}{x}\right ) \]
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Rubi [F]
time = 5.12, 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 {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{e^{10 x-2 x^2} x \log (5)+x^3 (1+\log (5))+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {e^{2 x^2} \left (x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )\right )}{x \left (e^{5 x}+e^{x^2} x\right ) \left (e^{5 x} \log (5)+e^{x^2} x (1+\log (5))\right )} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {e^{2 x^2} x \left (1-5 x+2 x^2\right )}{e^{10 x}+e^{x (5+x)} x}+\frac {e^{-5 x+x^2} \left (1-5 x+2 x^2\right )}{\log (5)}+\frac {e^{-5 x+2 x^2} x \left (-1+5 x-2 x^2\right ) (1+\log (5))^2}{\log (5) \left (e^{5 x} \log (5)+e^{x^2} x (1+\log (5))\right )}\right ) \, dx\\ &=-\log (x)+\frac {\int e^{-5 x+x^2} \left (1-5 x+2 x^2\right ) \, dx}{\log (5)}+\frac {(1+\log (5))^2 \int \frac {e^{-5 x+2 x^2} x \left (-1+5 x-2 x^2\right )}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\int \frac {e^{2 x^2} x \left (1-5 x+2 x^2\right )}{e^{10 x}+e^{x (5+x)} x} \, dx\\ &=\frac {e^{-5 x+x^2} \left (5 x-2 x^2\right )}{(5-2 x) \log (5)}-\log (x)+\frac {(1+\log (5))^2 \int \left (\frac {e^{-5 x+2 x^2} x}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))}+\frac {2 e^{-5 x+2 x^2} x^3}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))}+\frac {5 e^{-5 x+2 x^2} x^2}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))}\right ) \, dx}{\log (5)}+\int \left (\frac {e^{2 x^2} x}{e^{10 x}+e^{x (5+x)} x}-\frac {5 e^{2 x^2} x^2}{e^{10 x}+e^{x (5+x)} x}+\frac {2 e^{2 x^2} x^3}{e^{10 x}+e^{x (5+x)} x}\right ) \, dx\\ &=\frac {e^{-5 x+x^2} \left (5 x-2 x^2\right )}{(5-2 x) \log (5)}-\log (x)+2 \int \frac {e^{2 x^2} x^3}{e^{10 x}+e^{x (5+x)} x} \, dx-5 \int \frac {e^{2 x^2} x^2}{e^{10 x}+e^{x (5+x)} x} \, dx+\frac {(1+\log (5))^2 \int \frac {e^{-5 x+2 x^2} x}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\frac {\left (2 (1+\log (5))^2\right ) \int \frac {e^{-5 x+2 x^2} x^3}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\frac {\left (5 (1+\log (5))^2\right ) \int \frac {e^{-5 x+2 x^2} x^2}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\int \frac {e^{2 x^2} x}{e^{10 x}+e^{x (5+x)} x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.12, size = 47, normalized size = 1.96 \begin {gather*} -\log (x)-\log \left (e^{5 x}+e^{x^2} x\right )+\log \left (e^{x^2} x+e^{5 x} \log (5)+e^{x^2} x \log (5)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.59, size = 37, normalized size = 1.54
method | result | size |
risch | \(-\ln \left (x \right )+\ln \left ({\mathrm e}^{-\left (x -5\right ) x}+\frac {\left (\ln \left (5\right )+1\right ) x}{\ln \left (5\right )}\right )-\ln \left (x +{\mathrm e}^{-\left (x -5\right ) x}\right )\) | \(37\) |
norman | \(-\ln \left (x \right )-\ln \left (x +{\mathrm e}^{-x^{2}+5 x}\right )+\ln \left (x \ln \left (5\right )+{\mathrm e}^{-x^{2}+5 x} \ln \left (5\right )+x \right )\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 52 vs.
\(2 (22) = 44\).
time = 0.52, size = 52, normalized size = 2.17 \begin {gather*} -\log \left (x\right ) - \log \left (\frac {x e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )}}{x}\right ) + \log \left (\frac {x {\left (\log \left (5\right ) + 1\right )} e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )} \log \left (5\right )}{x {\left (\log \left (5\right ) + 1\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 40, normalized size = 1.67 \begin {gather*} \log \left (x \log \left (5\right ) + e^{\left (-x^{2} + 5 \, x\right )} \log \left (5\right ) + x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 39 vs.
\(2 (19) = 38\).
time = 0.28, size = 39, normalized size = 1.62 \begin {gather*} - \log {\left (x \right )} - \log {\left (x + e^{- x^{2} + 5 x} \right )} + \log {\left (\frac {2 x + 2 x \log {\left (5 \right )}}{2 \log {\left (5 \right )}} + e^{- x^{2} + 5 x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.53, size = 44, normalized size = 1.83 \begin {gather*} \log \left (-x \log \left (5\right ) - e^{\left (-x^{2} + 5 \, x\right )} \log \left (5\right ) - x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.13, size = 34, normalized size = 1.42 \begin {gather*} \ln \left (x+x\,\ln \left (5\right )+{\mathrm {e}}^{-x\,\left (x-5\right )}\,\ln \left (5\right )\right )-\ln \left (x+{\mathrm {e}}^{-x\,\left (x-5\right )}\right )-\ln \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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