Integrand size = 98, antiderivative size = 22 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x+x^2+\log \left (1+\left (2-e^{e^x}\right ) (-2+x)\right )\right ) \]
[Out]
\[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx \\ & = \int \left (\frac {1}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}-\frac {e^{e^x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}+\frac {e^{e^x+x} (-2+x)}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}+\frac {4 x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}-\frac {3 e^{e^x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}-\frac {4 x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}+\frac {2 e^{e^x} x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}\right ) \, dx \\ & = 2 \int \frac {e^{e^x} x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-3 \int \frac {e^{e^x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+4 \int \frac {x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-4 \int \frac {x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \frac {1}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-\int \frac {e^{e^x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \frac {e^{e^x+x} (-2+x)}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx \\ & = 2 \int \frac {e^{e^x} x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-3 \int \frac {e^{e^x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+4 \int \frac {x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-4 \int \frac {x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \frac {1}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-\int \frac {e^{e^x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \left (-\frac {2 e^{e^x+x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}+\frac {e^{e^x+x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )}\right ) \, dx \\ & = -\left (2 \int \frac {e^{e^x+x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx\right )+2 \int \frac {e^{e^x} x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-3 \int \frac {e^{e^x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+4 \int \frac {x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-4 \int \frac {x^2}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \frac {1}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx-\int \frac {e^{e^x}}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx+\int \frac {e^{e^x+x} x}{\left (3-2 e^{e^x}-2 x+e^{e^x} x\right ) \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right )} \, dx \\ \end{align*}
Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right ) \]
[In]
[Out]
Time = 1.81 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\ln \left (x^{2}+x +\ln \left (\left (2-x \right ) {\mathrm e}^{{\mathrm e}^{x}}+2 x -3\right )\right )\) | \(22\) |
parallelrisch | \(\ln \left (x^{2}+x +\ln \left (\left (2-x \right ) {\mathrm e}^{{\mathrm e}^{x}}+2 x -3\right )\right )\) | \(22\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x - 2\right )} e^{\left (e^{x}\right )} + 2 \, x - 3\right )\right ) \]
[In]
[Out]
Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log {\left (x^{2} + x + \log {\left (2 x + \left (2 - x\right ) e^{e^{x}} - 3 \right )} \right )} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x - 2\right )} e^{\left (e^{x}\right )} + 2 \, x - 3\right )\right ) \]
[In]
[Out]
none
Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x e^{\left (x + e^{x}\right )} - 2 \, x e^{x} - 2 \, e^{\left (x + e^{x}\right )} + 3 \, e^{x}\right )} e^{\left (-x\right )}\right )\right ) \]
[In]
[Out]
Timed out. \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\int \frac {4\,x-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x-{\mathrm {e}}^x\,\left (x-2\right )-2\,x^2+1\right )-4\,x^2+1}{3\,x+\ln \left (2\,x-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-2\right )-3\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-2\right )-2\,x+3\right )+x^2-2\,x^3-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (-x^3+x^2+2\,x\right )} \,d x \]
[In]
[Out]