Optimal. Leaf size=34 \[ 25 e^{-5+e^{2 e^{-x}}} \left (x+x \left (1-\frac {x}{e^3-\log (x)}\right )\right ) \]
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Rubi [B] time = 0.36, antiderivative size = 79, normalized size of antiderivative = 2.32, number of steps used = 1, number of rules used = 1, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2288} \begin {gather*} \frac {25 e^{x+e^{2 e^{-x}}-5} \left (-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 e^6 x+2 x \log ^2(x)\right )}{e^{x+6}+e^x \log ^2(x)-2 e^{x+3} \log (x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {25 e^{-5+e^{2 e^{-x}}+x} \left (2 e^6 x-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 x \log ^2(x)\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 30, normalized size = 0.88 \begin {gather*} 25 e^{-5+e^{2 e^{-x}}} x \left (2+\frac {x}{-e^3+\log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 35, normalized size = 1.03 \begin {gather*} -\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \relax (x)\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{3} - \log \relax (x)} \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 {25 \, {\left (2 \, {\left (x - 2 \, e^{3}\right )} e^{x} \log \relax (x) + 2 \, e^{x} \log \relax (x)^{2} - {\left (2 \, x e^{3} + x - 2 \, e^{6}\right )} e^{x} + 2 \, {\left (x^{2} e^{3} - 2 \, x \log \relax (x)^{2} - 2 \, x e^{6} - {\left (x^{2} - 4 \, x e^{3}\right )} \log \relax (x)\right )} e^{\left (2 \, e^{\left (-x\right )}\right )}\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{x} \log \relax (x)^{2} - 2 \, e^{\left (x + 3\right )} \log \relax (x) + e^{\left (x + 6\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 35, normalized size = 1.03
method | result | size |
risch | \(\frac {25 \left (2 \,{\mathrm e}^{3}-x -2 \ln \relax (x )\right ) x \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \relax (x )}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 36, normalized size = 1.06 \begin {gather*} \frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \relax (x)\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )}\right )}}{e^{5} \log \relax (x) - e^{8}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.44, size = 38, normalized size = 1.12 \begin {gather*} 50\,x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}-\frac {25\,x^2\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}}{{\mathrm {e}}^3-\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.56, size = 34, normalized size = 1.00 \begin {gather*} \frac {\left (- 25 x^{2} - 50 x \log {\relax (x )} + 50 x e^{3}\right ) e^{e^{2 e^{- x}} - 5}}{- \log {\relax (x )} + e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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