Integrand size = 77, antiderivative size = 26 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=e^3+\log (x)+\log \left (3+e+e^x+x-\frac {x}{5-e^2}\right ) \]
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\[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{(-15-5 e) x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx \\ & = \int \frac {15 \left (1+\frac {e}{3}\right )+8 x-e^2 (3+e+2 x)-e^x \left (-5-5 x+e^2 (1+x)\right )}{-((-15-5 e) x)+4 x^2-e^x \left (-5 x+e^2 x\right )-e^2 \left (3 x+e x+x^2\right )} \, dx \\ & = \int \left (\frac {1+x}{x}+\frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx \\ & = \int \frac {1+x}{x} \, dx+\int \frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx \\ & = \int \left (1+\frac {1}{x}\right ) \, dx+\int \left (\frac {11 \left (1-\frac {1}{11} e \left (-5+2 e+e^2\right )\right )}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x}+\frac {(-2-e) (2-e) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx \\ & = x+\log (x)+((-2-e) (2-e)) \int \frac {x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx+\left (11+e \left (5-2 e-e^2\right )\right ) \int \frac {1}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x} \, dx \\ \end{align*}
Time = 2.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log (x)+\log \left (15+5 e-3 e^2-e^3+5 e^x-e^{2+x}+4 x-e^2 x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\ln \left (x \right )+\ln \left ({\mathrm e}^{x}+\frac {{\mathrm e}^{3}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-4 x -15}{{\mathrm e}^{2}-5}\right )\) | \(34\) |
norman | \(\ln \left (x \right )+\ln \left ({\mathrm e} \,{\mathrm e}^{2}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15\right )\) | \(36\) |
parallelrisch | \(\ln \left (x \right )+\ln \left (\frac {{\mathrm e} \,{\mathrm e}^{2}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15}{{\mathrm e}^{2}-4}\right )\) | \(43\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left ({\left (x + 3\right )} e^{2} + {\left (e^{2} - 5\right )} e^{x} - 4 \, x + e^{3} - 5 \, e - 15\right ) + \log \left (x\right ) \]
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Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log {\left (x \right )} + \log {\left (\frac {- 4 x + x e^{2} - 15 - 5 e + e^{3} + 3 e^{2}}{-5 + e^{2}} + e^{x} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x {\left (e^{2} - 4\right )} + {\left (e^{2} - 5\right )} e^{x} + e^{3} + 3 \, e^{2} - 5 \, e - 15}{e^{2} - 5}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\log \left (-x e^{2} + 4 \, x - e^{3} - 3 \, e^{2} + 5 \, e - e^{\left (x + 2\right )} + 5 \, e^{x} + 15\right ) + \log \left (x\right ) \]
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Time = 0.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx=\ln \left (4\,x-{\mathrm {e}}^{x+2}+5\,\mathrm {e}-3\,{\mathrm {e}}^2-{\mathrm {e}}^3+5\,{\mathrm {e}}^x-x\,{\mathrm {e}}^2+15\right )+\ln \left (x\right ) \]
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