Integrand size = 77, antiderivative size = 31 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\log \left (\frac {x^2}{\frac {4-x}{-1+x}+\left (4-e^x-x\right ) x}\right ) \]
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\[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{(1-x) x \left (4-5 x+e^x x+5 x^2-e^x x^2-x^3\right )} \, dx \\ & = \int \left (\frac {1-x}{x}+\frac {4-4 x-9 x^2+12 x^3-7 x^4+x^5}{(-1+x) x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}\right ) \, dx \\ & = \int \frac {1-x}{x} \, dx+\int \frac {4-4 x-9 x^2+12 x^3-7 x^4+x^5}{(-1+x) x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx \\ & = \int \left (-1+\frac {1}{x}\right ) \, dx+\int \left (-\frac {3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}-\frac {3}{(-1+x) \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}-\frac {4}{x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}+\frac {6 x}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}-\frac {6 x^2}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}+\frac {x^3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}\right ) \, dx \\ & = -x+\log (x)-3 \int \frac {1}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx-3 \int \frac {1}{(-1+x) \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx-4 \int \frac {1}{x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx+6 \int \frac {x}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx-6 \int \frac {x^2}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx+\int \frac {x^3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx \\ \end{align*}
Time = 8.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\log (x)+\log ((1-x) x)-\log \left (4-5 x+e^x x+5 x^2-e^x x^2-x^3\right ) \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\ln \left (x \right )-\ln \left ({\mathrm e}^{x}+\frac {x^{3}-5 x^{2}+5 x -4}{x \left (-1+x \right )}\right )\) | \(32\) |
norman | \(2 \ln \left (x \right )-\ln \left (x^{3}+{\mathrm e}^{x} x^{2}-5 x^{2}-{\mathrm e}^{x} x +5 x -4\right )+\ln \left (-1+x \right )\) | \(37\) |
parallelrisch | \(2 \ln \left (x \right )-\ln \left (x^{3}+{\mathrm e}^{x} x^{2}-5 x^{2}-{\mathrm e}^{x} x +5 x -4\right )+\ln \left (-1+x \right )\) | \(37\) |
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\log \left (x\right ) - \log \left (\frac {x^{3} - 5 \, x^{2} + {\left (x^{2} - x\right )} e^{x} + 5 \, x - 4}{x^{2} - x}\right ) \]
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Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\log {\left (x \right )} - \log {\left (e^{x} + \frac {x^{3} - 5 x^{2} + 5 x - 4}{x^{2} - x} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\log \left (x\right ) - \log \left (\frac {x^{3} - 5 \, x^{2} + {\left (x^{2} - x\right )} e^{x} + 5 \, x - 4}{x^{2} - x}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=-\log \left (x^{3} + x^{2} e^{x} - 5 \, x^{2} - x e^{x} + 5 \, x - 4\right ) + \log \left (x - 1\right ) + 2 \, \log \left (x\right ) \]
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Time = 8.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx=\ln \left (x-1\right )-\ln \left (5\,x+x^2\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x-5\,x^2+x^3-4\right )+2\,\ln \left (x\right ) \]
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