Integrand size = 39, antiderivative size = 17 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\log \left (5+\frac {2 e^{-7+e^5} \log (x)}{x^2}\right ) \]
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Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2641, 6844, 31} \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\log \left (\frac {2 e^{e^5} \log (x)}{x^2}+5 e^7\right ) \]
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Rule 31
Rule 2641
Rule 6844
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{x \left (5 e^7 x^2+2 e^{e^5} \log (x)\right )} \, dx \\ & = \left (2 e^{e^5}\right ) \text {Subst}\left (\int \frac {1}{5 e^7+2 e^{e^5} x} \, dx,x,\frac {\log (x)}{x^2}\right ) \\ & = \log \left (5 e^7+\frac {2 e^{e^5} \log (x)}{x^2}\right ) \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=-2 \log (x)+\log \left (5 e^7 x^2+2 e^{e^5} \log (x)\right ) \]
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Time = 1.91 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29
method | result | size |
risch | \(-2 \ln \left (x \right )+\ln \left (\ln \left (x \right )+\frac {5 x^{2} {\mathrm e}^{7-{\mathrm e}^{5}}}{2}\right )\) | \(22\) |
norman | \(-2 \ln \left (x \right )+\ln \left (5 x^{2} {\mathrm e} \,{\mathrm e}^{6}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \ln \left (x \right )\right )\) | \(26\) |
parallelrisch | \(\ln \left (\frac {\left (5 x^{2} {\mathrm e} \,{\mathrm e}^{6}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \ln \left (x \right )\right ) {\mathrm e}^{-1} {\mathrm e}^{-6}}{5}\right )-2 \ln \left (x \right )\) | \(36\) |
default | \(-2 \,{\mathrm e}^{{\mathrm e}^{5}} \left ({\mathrm e}^{-{\mathrm e}^{5}} \ln \left (x \right )-\frac {{\mathrm e}^{-{\mathrm e}^{5}} \ln \left (5 x^{2} {\mathrm e} \,{\mathrm e}^{6}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \ln \left (x \right )\right )}{2}\right )\) | \(42\) |
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\log \left (5 \, x^{2} e^{7} + 2 \, e^{\left (e^{5}\right )} \log \left (x\right )\right ) - 2 \, \log \left (x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=- 2 \log {\left (x \right )} + \log {\left (\frac {5 x^{2} e^{7}}{2 e^{e^{5}}} + \log {\left (x \right )} \right )} \]
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Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\log \left (\frac {1}{2} \, {\left (5 \, x^{2} e^{7} + 2 \, e^{\left (e^{5}\right )} \log \left (x\right )\right )} e^{\left (-e^{5}\right )}\right ) - 2 \, \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\log \left (-5 \, x^{2} e^{7} - 2 \, e^{\left (e^{5}\right )} \log \left (x\right )\right ) - 2 \, \log \left (x\right ) \]
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Time = 16.74 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx=\ln \left (\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^5-7}\,\ln \left (x\right )}{5}+x^2\right )-2\,\ln \left (x\right ) \]
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