Integrand size = 95, antiderivative size = 23 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-5+\frac {4 e^x \log \left (3+e^x\right )}{\log \left (\frac {1}{3} (-5+x)\right )} \]
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\[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^x \left (\frac {e^x \log \left (\frac {1}{3} (-5+x)\right )}{3+e^x}+\frac {\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )}{-5+x}\right )}{\log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \frac {e^x \left (\frac {e^x \log \left (\frac {1}{3} (-5+x)\right )}{3+e^x}+\frac {\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )}{-5+x}\right )}{\log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \left (\frac {e^x \left (\log \left (3+e^x\right )+5 \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (\frac {1}{3} (-5+x)\right )+5 \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x (\log (27)-3 \log (-5+x))}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx \\ & = 4 \int \frac {e^x \left (\log \left (3+e^x\right )+5 \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (\frac {1}{3} (-5+x)\right )+5 \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x (\log (27)-3 \log (-5+x))}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \frac {e^x \left (-\left ((-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )-\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \left (\frac {e^x \log (27)}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {3 e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx \\ & = 4 \int \left (\frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x \left (1+\log \left (3+e^x\right )\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x \left (1+\log \left (3+e^x\right )\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \left (\frac {e^x}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x \log \left (3+e^x\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx+4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ & = 4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x \log \left (3+e^x\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx \\ \end{align*}
Time = 5.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4 e^x \log \left (3+e^x\right )}{\log \left (\frac {1}{3} (-5+x)\right )} \]
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Time = 0.96 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {4 \ln \left (3+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )}\) | \(18\) |
parallelrisch | \(\frac {4 \ln \left (3+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )}\) | \(18\) |
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Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (\frac {1}{3} \, x - \frac {5}{3}\right )} \]
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Timed out. \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\text {Timed out} \]
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Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (3\right ) - \log \left (x - 5\right )} \]
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=-\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (3\right ) - \log \left (x - 5\right )} \]
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Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx=\frac {4\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+3\right )}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )} \]
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