Integrand size = 104, antiderivative size = 25 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=\log \left (\frac {e^{-e^x+x} \log (x)}{-4+\log \left (\frac {5+x}{x}\right )}\right ) \]
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\[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=\int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {1}{\log (x)}+\frac {5+20 \left (-1+e^x\right ) x+4 \left (-1+e^x\right ) x^2-\left (-1+e^x\right ) x (5+x) \log \left (\frac {5+x}{x}\right )}{(5+x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )}}{x} \, dx \\ & = \int \left (-e^x+\frac {-20-4 x+5 \log (x)-20 x \log (x)-4 x^2 \log (x)+5 \log \left (\frac {5+x}{x}\right )+x \log \left (\frac {5+x}{x}\right )+5 x \log (x) \log \left (\frac {5+x}{x}\right )+x^2 \log (x) \log \left (\frac {5+x}{x}\right )}{x (5+x) \log (x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )}\right ) \, dx \\ & = -\int e^x \, dx+\int \frac {-20-4 x+5 \log (x)-20 x \log (x)-4 x^2 \log (x)+5 \log \left (\frac {5+x}{x}\right )+x \log \left (\frac {5+x}{x}\right )+5 x \log (x) \log \left (\frac {5+x}{x}\right )+x^2 \log (x) \log \left (\frac {5+x}{x}\right )}{x (5+x) \log (x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )} \, dx \\ & = -e^x+\int \frac {\frac {1}{\log (x)}+\frac {5-20 x-4 x^2+x (5+x) \log \left (\frac {5+x}{x}\right )}{(5+x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )}}{x} \, dx \\ & = -e^x+\int \left (\frac {1+x \log (x)}{x \log (x)}+\frac {5}{x (5+x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )}\right ) \, dx \\ & = -e^x+5 \int \frac {1}{x (5+x) \left (-4+\log \left (\frac {5+x}{x}\right )\right )} \, dx+\int \frac {1+x \log (x)}{x \log (x)} \, dx \\ & = -e^x+5 \int \frac {1}{x (5+x) \left (-4+\log \left (1+\frac {5}{x}\right )\right )} \, dx+\int \left (1+\frac {1}{x \log (x)}\right ) \, dx \\ & = -e^x+x+5 \int \left (\frac {1}{5 x \left (-4+\log \left (1+\frac {5}{x}\right )\right )}-\frac {1}{5 (5+x) \left (-4+\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -e^x+x+\int \frac {1}{x \left (-4+\log \left (1+\frac {5}{x}\right )\right )} \, dx-\int \frac {1}{(5+x) \left (-4+\log \left (1+\frac {5}{x}\right )\right )} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -e^x+x+\log (\log (x))+\int \frac {1}{x \left (-4+\log \left (1+\frac {5}{x}\right )\right )} \, dx-\int \frac {1}{(5+x) \left (-4+\log \left (1+\frac {5}{x}\right )\right )} \, dx \\ \end{align*}
Time = 1.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=-e^x+x+\log (\log (x))-\log \left (4-\log \left (\frac {5+x}{x}\right )\right ) \]
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Time = 1.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
default | \(x -\ln \left (\ln \left (1+\frac {5}{x}\right )-4\right )+\ln \left (\ln \left (x \right )\right )-{\mathrm e}^{x}\) | \(23\) |
parts | \(x -\ln \left (\ln \left (1+\frac {5}{x}\right )-4\right )+\ln \left (\ln \left (x \right )\right )-{\mathrm e}^{x}\) | \(23\) |
parallelrisch | \(-\frac {5}{2}+\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (\frac {5+x}{x}\right )-4\right )+x -{\mathrm e}^{x}\) | \(24\) |
risch | \(-{\mathrm e}^{x}+x +\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (5+x \right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i \left (5+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (5+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (5+x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi \operatorname {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}+\pi \operatorname {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i \ln \left (x \right )+8 i\right )}{2}\right )\) | \(113\) |
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (\frac {x + 5}{x}\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log {\left (\log {\left (\frac {x + 5}{x} \right )} - 4 \right )} + \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \left (x\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \left (x\right ) - 4\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 13.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx=x-\ln \left (\ln \left (\frac {x+5}{x}\right )-4\right )+\ln \left (\ln \left (x\right )\right )-{\mathrm {e}}^x \]
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