Optimal. Leaf size=23 \[ -1+25 \left (\frac {1}{x}+\log \left (-\frac {e^x}{x}+\log ^2(\log (x))\right )\right ) \]
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Rubi [F] time = 1.83, 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^x \left (25+25 x-25 x^2\right ) \log (x)+50 x^2 \log (\log (x))-25 x \log (x) \log ^2(\log (x))}{-e^x x^2 \log (x)+x^3 \log (x) \log ^2(\log (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^x \left (25+25 x-25 x^2\right ) \log (x)-50 x^2 \log (\log (x))+25 x \log (x) \log ^2(\log (x))}{x^2 \log (x) \left (e^x-x \log ^2(\log (x))\right )} \, dx\\ &=\int \left (\frac {25 \left (-1-x+x^2\right )}{x^2}-\frac {25 \log (\log (x)) (-2-\log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )}\right ) \, dx\\ &=25 \int \frac {-1-x+x^2}{x^2} \, dx-25 \int \frac {\log (\log (x)) (-2-\log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )} \, dx\\ &=25 \int \left (1-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx-25 \int \left (-\frac {2 \log (\log (x))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )}-\frac {\log ^2(\log (x))}{-e^x+x \log ^2(\log (x))}+\frac {x \log ^2(\log (x))}{-e^x+x \log ^2(\log (x))}\right ) \, dx\\ &=\frac {25}{x}+25 x-25 \log (x)+25 \int \frac {\log ^2(\log (x))}{-e^x+x \log ^2(\log (x))} \, dx-25 \int \frac {x \log ^2(\log (x))}{-e^x+x \log ^2(\log (x))} \, dx+50 \int \frac {\log (\log (x))}{\log (x) \left (-e^x+x \log ^2(\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.35, size = 25, normalized size = 1.09 \begin {gather*} \frac {25}{x}-25 \log (x)+25 \log \left (e^x-x \log ^2(\log (x))\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 26, normalized size = 1.13 \begin {gather*} \frac {25 \, {\left (x \log \left (\frac {x \log \left (\log \relax (x)\right )^{2} - e^{x}}{x}\right ) + 1\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.43, size = 27, normalized size = 1.17 \begin {gather*} \frac {25 \, {\left (x \log \left (x \log \left (\log \relax (x)\right )^{2} - e^{x}\right ) - x \log \relax (x) + 1\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 23, normalized size = 1.00
method | result | size |
risch | \(\frac {25}{x}+25 \ln \left (\ln \left (\ln \relax (x )\right )^{2}-\frac {{\mathrm e}^{x}}{x}\right )\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.90, size = 25, normalized size = 1.09 \begin {gather*} \frac {25}{x} + 25 \, \log \left (\frac {x \log \left (\log \relax (x)\right )^{2} - e^{x}}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.39, size = 24, normalized size = 1.04 \begin {gather*} 25\,\ln \left (\frac {{\mathrm {e}}^x-x\,{\ln \left (\ln \relax (x)\right )}^2}{x}\right )+\frac {25}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 22, normalized size = 0.96 \begin {gather*} - 25 \log {\relax (x )} + 25 \log {\left (- x \log {\left (\log {\relax (x )} \right )}^{2} + e^{x} \right )} + \frac {25}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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