Integrand size = 113, antiderivative size = 32 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {6+\frac {1}{3} e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{2 x} \]
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Time = 0.92 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {12, 6820, 6874, 2326} \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {3}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{6 x} \]
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Rule 12
Rule 2326
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx \\ & = \frac {1}{6} \int \frac {e^x+\log \left (\frac {x}{2}\right ) \left (e^x+\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \left (-18+e^x (-1+x) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx \\ & = \frac {1}{6} \int \left (-\frac {18}{x^2}+\frac {e^x \left (1+\log \left (\frac {x}{2}\right )-\log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )+x \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )}\right ) \, dx \\ & = \frac {3}{x}+\frac {1}{6} \int \frac {e^x \left (1+\log \left (\frac {x}{2}\right )-\log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )+x \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx \\ & = \frac {3}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{6 x} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {18+e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x} \]
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Time = 1.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\frac {18+\ln \left (\ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right ) {\mathrm e}^{x}}{6 x}\) | \(25\) |
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (\frac {x \log \left (\frac {1}{2} \, x\right )}{\log \left (2\right ) - 4}\right )\right ) + 18}{6 \, x} \]
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Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log {\left (\log {\left (\frac {x \log {\left (\frac {x}{2} \right )}}{-4 + \log {\left (2 \right )}} \right )} \right )}}{6 x} + \frac {3}{x} \]
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Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right )}{6 \, x} + \frac {3}{x} \]
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Time = 0.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right ) + 18}{6 \, x} \]
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Timed out. \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\int \frac {\frac {{\mathrm {e}}^x}{6}-3\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )+\frac {\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x}{6}+\frac {\ln \left (\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right )\,\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\,\left (x-1\right )}{6}}{x^2\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )} \,d x \]
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