Integrand size = 98, antiderivative size = 22 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x+x^2+\log \left (1+\left (2-e^{e^x}\right ) (-2+x)\right )\right ) \]
Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x+x^2+\log \left (-3-e^{e^x} (-2+x)+2 x\right )\right ) \]
Integrate[(1 + 4*x - 4*x^2 + E^E^x*(-1 + E^x*(-2 + x) - 3*x + 2*x^2))/(3*x + x^2 - 2*x^3 + E^E^x*(-2*x - x^2 + x^3) + (3 + E^E^x*(-2 + x) - 2*x)*Log [-3 + E^E^x*(2 - x) + 2*x]),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-4 x^2+e^{e^x} \left (2 x^2-3 x+e^x (x-2)-1\right )+4 x+1}{-2 x^3+x^2+e^{e^x} \left (x^3-x^2-2 x\right )+3 x+\left (e^{e^x} (x-2)-2 x+3\right ) \log \left (e^{e^x} (2-x)+2 x-3\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-4 x^2+e^{e^x} \left (2 x^2-3 x+e^x (x-2)-1\right )+4 x+1}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{e^x} x^2}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}-\frac {4 x^2}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}-\frac {3 e^{e^x} x}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}+\frac {4 x}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}-\frac {e^{e^x}}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}+\frac {e^{x+e^x} (x-2)}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}+\frac {1}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx-\int \frac {e^{e^x}}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx-2 \int \frac {e^{x+e^x}}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx+4 \int \frac {x}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx-3 \int \frac {e^{e^x} x}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx+\int \frac {e^{x+e^x} x}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx-4 \int \frac {x^2}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx+2 \int \frac {e^{e^x} x^2}{\left (e^{e^x} x-2 x-2 e^{e^x}+3\right ) \left (x^2+x+\log \left (-e^{e^x} (x-2)+2 x-3\right )\right )}dx\) |
Int[(1 + 4*x - 4*x^2 + E^E^x*(-1 + E^x*(-2 + x) - 3*x + 2*x^2))/(3*x + x^2 - 2*x^3 + E^E^x*(-2*x - x^2 + x^3) + (3 + E^E^x*(-2 + x) - 2*x)*Log[-3 + E^E^x*(2 - x) + 2*x]),x]
3.21.32.3.1 Defintions of rubi rules used
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 | \(\ln \left (x^{2}+x +\ln \left (\left (2-x \right ) {\mathrm e}^{{\mathrm e}^{x}}+2 x -3\right )\right )\) | \(22\) |
int(((exp(x)*(-2+x)+2*x^2-3*x-1)*exp(exp(x))-4*x^2+4*x+1)/(((-2+x)*exp(exp (x))+3-2*x)*ln((2-x)*exp(exp(x))+2*x-3)+(x^3-x^2-2*x)*exp(exp(x))-2*x^3+x^ 2+3*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x - 2\right )} e^{\left (e^{x}\right )} + 2 \, x - 3\right )\right ) \]
integrate(((exp(x)*(-2+x)+2*x^2-3*x-1)*exp(exp(x))-4*x^2+4*x+1)/(((-2+x)*e xp(exp(x))+3-2*x)*log((2-x)*exp(exp(x))+2*x-3)+(x^3-x^2-2*x)*exp(exp(x))-2 *x^3+x^2+3*x),x, algorithm=\
Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log {\left (x^{2} + x + \log {\left (2 x + \left (2 - x\right ) e^{e^{x}} - 3 \right )} \right )} \]
integrate(((exp(x)*(-2+x)+2*x**2-3*x-1)*exp(exp(x))-4*x**2+4*x+1)/(((-2+x) *exp(exp(x))+3-2*x)*ln((2-x)*exp(exp(x))+2*x-3)+(x**3-x**2-2*x)*exp(exp(x) )-2*x**3+x**2+3*x),x)
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x - 2\right )} e^{\left (e^{x}\right )} + 2 \, x - 3\right )\right ) \]
integrate(((exp(x)*(-2+x)+2*x^2-3*x-1)*exp(exp(x))-4*x^2+4*x+1)/(((-2+x)*e xp(exp(x))+3-2*x)*log((2-x)*exp(exp(x))+2*x-3)+(x^3-x^2-2*x)*exp(exp(x))-2 *x^3+x^2+3*x),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\log \left (x^{2} + x + \log \left (-{\left (x e^{\left (x + e^{x}\right )} - 2 \, x e^{x} - 2 \, e^{\left (x + e^{x}\right )} + 3 \, e^{x}\right )} e^{\left (-x\right )}\right )\right ) \]
integrate(((exp(x)*(-2+x)+2*x^2-3*x-1)*exp(exp(x))-4*x^2+4*x+1)/(((-2+x)*e xp(exp(x))+3-2*x)*log((2-x)*exp(exp(x))+2*x-3)+(x^3-x^2-2*x)*exp(exp(x))-2 *x^3+x^2+3*x),x, algorithm=\
Timed out. \[ \int \frac {1+4 x-4 x^2+e^{e^x} \left (-1+e^x (-2+x)-3 x+2 x^2\right )}{3 x+x^2-2 x^3+e^{e^x} \left (-2 x-x^2+x^3\right )+\left (3+e^{e^x} (-2+x)-2 x\right ) \log \left (-3+e^{e^x} (2-x)+2 x\right )} \, dx=\int \frac {4\,x-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x-{\mathrm {e}}^x\,\left (x-2\right )-2\,x^2+1\right )-4\,x^2+1}{3\,x+\ln \left (2\,x-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-2\right )-3\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-2\right )-2\,x+3\right )+x^2-2\,x^3-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (-x^3+x^2+2\,x\right )} \,d x \]
int((4*x - exp(exp(x))*(3*x - exp(x)*(x - 2) - 2*x^2 + 1) - 4*x^2 + 1)/(3* x + log(2*x - exp(exp(x))*(x - 2) - 3)*(exp(exp(x))*(x - 2) - 2*x + 3) + x ^2 - 2*x^3 - exp(exp(x))*(2*x + x^2 - x^3)),x)