Optimal. Leaf size=25 \[ \frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \]
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Rubi [F] time = 4.94, 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^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{\frac {1}{x}}} \left (-\frac {x^2}{(25+x) (-1+\log (25+x))}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (25+x))\right )}{2 \log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} \left (-\frac {x^2}{(25+x) (-1+\log (25+x))}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (25+x))\right )}{\log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \left (-\frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))}+\frac {e^{e^{\frac {1}{x}}} x (-x-50 \log (1-\log (25+x))-2 x \log (1-\log (25+x))+50 \log (25+x) \log (1-\log (25+x))+2 x \log (25+x) \log (1-\log (25+x)))}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\right )+\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x (-x-50 \log (1-\log (25+x))-2 x \log (1-\log (25+x))+50 \log (25+x) \log (1-\log (25+x))+2 x \log (25+x) \log (1-\log (25+x)))}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\right )+\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x (x-2 (25+x) (-1+\log (25+x)) \log (1-\log (25+x)))}{(25+x) (1-\log (25+x)) \log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \left (-\frac {e^{e^{\frac {1}{x}}} x^2}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {2 e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))}\right ) \, dx-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x^2}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \left (-\frac {25 e^{e^{\frac {1}{x}}}}{(-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {e^{e^{\frac {1}{x}}} x}{(-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {625 e^{e^{\frac {1}{x}}}}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}\right ) \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x}{(-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\frac {25}{2} \int \frac {e^{e^{\frac {1}{x}}}}{(-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx-\frac {625}{2} \int \frac {e^{e^{\frac {1}{x}}}}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.33, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.91, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 22, normalized size = 0.88
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{\frac {1}{x}}}}{2 \ln \left (-\ln \left (x +25\right )+1\right )}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.99, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}}}{2\,\ln \left (1-\ln \left (x+25\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.47, size = 19, normalized size = 0.76 \begin {gather*} \frac {x^{2} e^{e^{\frac {1}{x}}}}{2 \log {\left (1 - \log {\left (x + 25 \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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