Integrand size = 70, antiderivative size = 26 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\log \left (x (1+2 x) \left (1-x+\frac {1}{2} e^{-5+e^4+x} x\right )\right ) \]
[Out]
\[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 e^5 \left (-1-x+x^2\right )}{x \left (-2 e^5+2 e^5 x-e^{e^4+x} x\right )}+\frac {2+7 x+2 x^2}{x (1+2 x)}\right ) \, dx \\ & = -\left (\left (2 e^5\right ) \int \frac {-1-x+x^2}{x \left (-2 e^5+2 e^5 x-e^{e^4+x} x\right )} \, dx\right )+\int \frac {2+7 x+2 x^2}{x (1+2 x)} \, dx \\ & = -\left (\left (2 e^5\right ) \int \left (-\frac {1}{-2 e^5+2 e^5 x-e^{e^4+x} x}-\frac {1}{x \left (-2 e^5+2 e^5 x-e^{e^4+x} x\right )}+\frac {x}{-2 e^5+2 e^5 x-e^{e^4+x} x}\right ) \, dx\right )+\int \left (1+\frac {2}{x}+\frac {2}{1+2 x}\right ) \, dx \\ & = x+2 \log (x)+\log (1+2 x)+\left (2 e^5\right ) \int \frac {1}{-2 e^5+2 e^5 x-e^{e^4+x} x} \, dx+\left (2 e^5\right ) \int \frac {1}{x \left (-2 e^5+2 e^5 x-e^{e^4+x} x\right )} \, dx-\left (2 e^5\right ) \int \frac {x}{-2 e^5+2 e^5 x-e^{e^4+x} x} \, dx \\ \end{align*}
Time = 3.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\log \left (x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} x^2 (1+2 x)\right ) \]
[In]
[Out]
Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\ln \left (x \right )+\ln \left (\frac {1}{2}+x \right )+\ln \left (x \,{\mathrm e}^{-\ln \left (2\right )+{\mathrm e}^{4}+x -5}-x +1\right )\) | \(26\) |
norman | \(\ln \left (x \right )+\ln \left (1+2 x \right )+\ln \left (x \,{\mathrm e}^{-\ln \left (2\right )+{\mathrm e}^{4}+x -5}-x +1\right )\) | \(28\) |
risch | \(\ln \left (1+2 x \right )+2 \ln \left (x \right )+\ln \left (2\right )-{\mathrm e}^{4}+5+\ln \left (\frac {{\mathrm e}^{x +{\mathrm e}^{4}-5}}{2}-\frac {-1+x}{x}\right )\) | \(37\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\log \left (2 \, x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (\frac {x e^{\left (x + e^{4} - \log \left (2\right ) - 5\right )} - x + 1}{x}\right ) \]
[In]
[Out]
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=2 \log {\left (x \right )} + \log {\left (x + \frac {1}{2} \right )} + \log {\left (e^{x - 5 + e^{4}} + \frac {2 - 2 x}{x} \right )} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\log \left (2 \, x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (-\frac {{\left (2 \, x e^{5} - x e^{\left (x + e^{4}\right )} - 2 \, e^{5}\right )} e^{\left (-e^{4}\right )}}{x}\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\log \left (x e^{\left (x + e^{4} - 5\right )} - 2 \, x + 2\right ) + \log \left (2 \, x + 1\right ) + \log \left (x\right ) \]
[In]
[Out]
Time = 13.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x-6 x^2+\frac {1}{2} e^{-5+e^4+x} \left (2 x+7 x^2+2 x^3\right )}{x+x^2-2 x^3+\frac {1}{2} e^{-5+e^4+x} \left (x^2+2 x^3\right )} \, dx=\ln \left (x+\frac {1}{2}\right )+\ln \left (\frac {x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^4}\,{\mathrm {e}}^x}{2}-x+1\right )+\ln \left (x\right ) \]
[In]
[Out]