Optimal. Leaf size=23 \[ \frac {1}{-5 e^{2-x}+\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )} \]
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Rubi [F] time = 6.22, 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 {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\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 {e^x \left (-5 e^2-e^x-e^x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\\ &=\int \left (-\frac {5 e^{2+x} \left (1+\log (4)+\log \left (x \log \left (\frac {17}{7}\right )\right )+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}-\frac {e^x \left (1+\log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}\right ) \, dx\\ &=-\left (5 \int \frac {e^{2+x} \left (1+\log (4)+\log \left (x \log \left (\frac {17}{7}\right )\right )+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\right )-\int \frac {e^x \left (1+\log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx\\ &=-\left (5 \int \left (\frac {e^{2+x} (1+\log (4))}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}+\frac {e^{2+x} \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}+\frac {e^{2+x} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}\right ) \, dx\right )-\int \left (\frac {e^x}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}+\frac {e^x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}\right ) \, dx\\ &=-\left (5 \int \frac {e^{2+x} \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\right )-5 \int \frac {e^{2+x} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx-(5 (1+\log (4))) \int \frac {e^{2+x}}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx-\int \frac {e^x}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx-\int \frac {e^x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.35, size = 30, normalized size = 1.30 \begin {gather*} \frac {e^x}{-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{x \log \left (x \log \left (\frac {17}{7}\right )\right ) - 5 \, e^{\left (-x + 2\right )} + 2 \, \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 2.15, size = 808, normalized size = 35.13 result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 28, normalized size = 1.22
method | result | size |
risch | \(\frac {1}{x \ln \left (x \left (\ln \left (17\right )-\ln \relax (7)\right )\right )+2 \ln \relax (2)-5 \,{\mathrm e}^{2-x}}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.59, size = 32, normalized size = 1.39 \begin {gather*} \frac {e^{x}}{{\left (x \log \relax (x) + x \log \left (\log \left (17\right ) - \log \relax (7)\right ) + 2 \, \log \relax (2)\right )} e^{x} - 5 \, e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {5\,{\mathrm {e}}^{2-x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )+1}{25\,{\mathrm {e}}^{4-2\,x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )\,\left (4\,x\,\ln \relax (2)-10\,x\,{\mathrm {e}}^{2-x}\right )+x^2\,{\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )}^2-20\,{\mathrm {e}}^{2-x}\,\ln \relax (2)+4\,{\ln \relax (2)}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 24, normalized size = 1.04 \begin {gather*} - \frac {1}{- x \log {\left (x \log {\left (\frac {17}{7} \right )} \right )} + 5 e^{2 - x} - 2 \log {\relax (2 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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