Integrand size = 142, antiderivative size = 28 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=x \log (5)+\frac {e^x (-1+\log (5-\log (1-x)))}{6 x} \]
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\[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{6 (1-x) x^2 (5-\log (1-x))} \, dx \\ & = \frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{(1-x) x^2 (5-\log (1-x))} \, dx \\ & = \frac {1}{6} \int \left (6 \log (5)+\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))}\right ) \, dx \\ & = x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx \\ & = x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2+(-1+x)^2 \log (1-x) (-1+\log (5-\log (1-x)))-5 (-1+x)^2 \log (5-\log (1-x))\right )}{(1-x) x^2 (5-\log (1-x))} \, dx \\ & = x \log (5)+\frac {1}{6} \int \left (\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))}+\frac {e^x (-1+x) \log (5-\log (1-x))}{x^2}\right ) \, dx \\ & = x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx+\frac {1}{6} \int \frac {e^x (-1+x) \log (5-\log (1-x))}{x^2} \, dx \\ & = x \log (5)+\frac {1}{6} \int \left (\frac {e^x (1-x)}{x^2}+\frac {e^x}{(-1+x) x (-5+\log (1-x))}\right ) \, dx+\frac {1}{6} \int \left (-\frac {e^x \log (5-\log (1-x))}{x^2}+\frac {e^x \log (5-\log (1-x))}{x}\right ) \, dx \\ & = x \log (5)+\frac {1}{6} \int \frac {e^x (1-x)}{x^2} \, dx+\frac {1}{6} \int \frac {e^x}{(-1+x) x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx \\ & = -\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \left (\frac {e^x}{(-1+x) (-5+\log (1-x))}-\frac {e^x}{x (-5+\log (1-x))}\right ) \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx \\ & = -\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \frac {e^x}{(-1+x) (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x}{x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {1}{6} \left (-\frac {e^x}{x}+6 x \log (5)+\frac {e^x \log (5-\log (1-x))}{x}\right ) \]
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Time = 5.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )}{6 x}+\frac {6 x^{2} \ln \left (5\right )-{\mathrm e}^{x}}{6 x}\) | \(37\) |
parallelrisch | \(\frac {6 x^{2} \ln \left (5\right )+12 x \ln \left (5\right )+{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )-{\mathrm e}^{x}}{6 x}\) | \(37\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {6 \, x^{2} \log \left (5\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \]
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Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=x \log {\left (5 \right )} + \frac {\left (\log {\left (5 - \log {\left (1 - x \right )} \right )} - 1\right ) e^{x}}{6 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (25) = 50\).
Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.79 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=-\log \left (5\right ) \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + {\left ({\left (\log \left (-x + 1\right ) - 5\right )} \log \left (\log \left (-x + 1\right ) - 5\right ) - \log \left (-x + 1\right ) + 5\right )} \log \left (5\right ) + 5 \, \log \left (5\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + \frac {6 \, x^{2} \log \left (5\right ) + 6 \, x \log \left (5\right ) \log \left (-x + 1\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {6 \, x^{2} \log \left (5\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \]
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Timed out. \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\int \frac {\ln \left (5-\ln \left (1-x\right )\right )\,\left ({\mathrm {e}}^x\,\left (5\,x^2-10\,x+5\right )-{\mathrm {e}}^x\,\ln \left (1-x\right )\,\left (x^2-2\,x+1\right )\right )+\ln \left (1-x\right )\,\left ({\mathrm {e}}^x\,\left (x^2-2\,x+1\right )+\ln \left (5\right )\,\left (6\,x^2-6\,x^3\right )\right )-\ln \left (5\right )\,\left (30\,x^2-30\,x^3\right )-{\mathrm {e}}^x\,\left (5\,x^2-9\,x+5\right )}{\ln \left (1-x\right )\,\left (6\,x^2-6\,x^3\right )-30\,x^2+30\,x^3} \,d x \]
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