Integrand size = 70, antiderivative size = 24 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=-\frac {1}{3}+x+\left (-x+\log \left (2-e^{e^x}-x\right )\right )^2 \] Output:
(ln(-exp(exp(x))+2-x)-x)^2+x-1/3
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=x+x^2-2 x \log \left (2-e^{e^x}-x\right )+\log ^2\left (2-e^{e^x}-x\right ) \] Input:
Integrate[(-2 - 5*x + 2*x^2 + E^E^x*(1 + 2*x - 2*E^x*x) + (6 + E^E^x*(-2 + 2*E^x) - 2*x)*Log[2 - E^E^x - x])/(-2 + E^E^x + x),x]
Output:
x + x^2 - 2*x*Log[2 - E^E^x - x] + Log[2 - E^E^x - x]^2
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.
\(\displaystyle \int \frac {2 x^2-5 x+e^{e^x} \left (-2 e^x x+2 x+1\right )+\left (e^{e^x} \left (2 e^x-2\right )-2 x+6\right ) \log \left (-x-e^{e^x}+2\right )-2}{x+e^{e^x}-2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2}{x+e^{e^x}-2}+\frac {2 e^{e^x} x}{x+e^{e^x}-2}-\frac {5 x}{x+e^{e^x}-2}+\frac {e^{e^x}}{x+e^{e^x}-2}-\frac {2}{x+e^{e^x}-2}-\frac {2 x \log \left (-x-e^{e^x}+2\right )}{x+e^{e^x}-2}-\frac {2 e^{x+e^x} \left (x-\log \left (-x-e^{e^x}+2\right )\right )}{x+e^{e^x}-2}-\frac {2 e^{e^x} \log \left (-x-e^{e^x}+2\right )}{x+e^{e^x}-2}+\frac {6 \log \left (-x-e^{e^x}+2\right )}{x+e^{e^x}-2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {x^2}{x+e^{e^x}-2}dx-2 \int \frac {1}{x+e^{e^x}-2}dx+\int \frac {e^{e^x}}{x+e^{e^x}-2}dx-5 \int \frac {x}{x+e^{e^x}-2}dx+2 \int \frac {e^{e^x} x}{x+e^{e^x}-2}dx-2 \int \frac {e^{x+e^x} x}{x+e^{e^x}-2}dx-6 \int \frac {\int \frac {1}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx-6 \int \frac {e^{x+e^x} \int \frac {1}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx+2 \int \frac {\int \frac {e^{e^x}}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx+2 \int \frac {e^{x+e^x} \int \frac {e^{e^x}}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx-2 \int \frac {\int \frac {e^{x+e^x}}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx-2 \int \frac {e^{x+e^x} \int \frac {e^{x+e^x}}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx+2 \int \frac {\int \frac {x}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx+2 \int \frac {e^{x+e^x} \int \frac {x}{x+e^{e^x}-2}dx}{x+e^{e^x}-2}dx+6 \log \left (-x-e^{e^x}+2\right ) \int \frac {1}{x+e^{e^x}-2}dx-2 \log \left (-x-e^{e^x}+2\right ) \int \frac {e^{e^x}}{x+e^{e^x}-2}dx+2 \log \left (-x-e^{e^x}+2\right ) \int \frac {e^{x+e^x}}{x+e^{e^x}-2}dx-2 \log \left (-x-e^{e^x}+2\right ) \int \frac {x}{x+e^{e^x}-2}dx\) |
Input:
Int[(-2 - 5*x + 2*x^2 + E^E^x*(1 + 2*x - 2*E^x*x) + (6 + E^E^x*(-2 + 2*E^x ) - 2*x)*Log[2 - E^E^x - x])/(-2 + E^E^x + x),x]
Output:
$Aborted
Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38
method | result | size |
risch | \(x^{2}-2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2-x \right ) x +\ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2-x \right )^{2}+x\) | \(33\) |
parallelrisch | \(x^{2}-2 \ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2-x \right ) x +\ln \left (-{\mathrm e}^{{\mathrm e}^{x}}+2-x \right )^{2}+x\) | \(33\) |
Input:
int((((2*exp(x)-2)*exp(exp(x))+6-2*x)*ln(-exp(exp(x))+2-x)+(-2*exp(x)*x+2* x+1)*exp(exp(x))+2*x^2-5*x-2)/(exp(exp(x))+x-2),x,method=_RETURNVERBOSE)
Output:
x^2-2*ln(-exp(exp(x))+2-x)*x+ln(-exp(exp(x))+2-x)^2+x
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=x^{2} - 2 \, x \log \left (-x - e^{\left (e^{x}\right )} + 2\right ) + \log \left (-x - e^{\left (e^{x}\right )} + 2\right )^{2} + x \] Input:
integrate((((2*exp(x)-2)*exp(exp(x))+6-2*x)*log(-exp(exp(x))+2-x)+(-2*exp( x)*x+2*x+1)*exp(exp(x))+2*x^2-5*x-2)/(exp(exp(x))+x-2),x, algorithm="frica s")
Output:
x^2 - 2*x*log(-x - e^(e^x) + 2) + log(-x - e^(e^x) + 2)^2 + x
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=x^{2} - 2 x \log {\left (- x - e^{e^{x}} + 2 \right )} + x + \log {\left (- x - e^{e^{x}} + 2 \right )}^{2} \] Input:
integrate((((2*exp(x)-2)*exp(exp(x))+6-2*x)*ln(-exp(exp(x))+2-x)+(-2*exp(x )*x+2*x+1)*exp(exp(x))+2*x**2-5*x-2)/(exp(exp(x))+x-2),x)
Output:
x**2 - 2*x*log(-x - exp(exp(x)) + 2) + x + log(-x - exp(exp(x)) + 2)**2
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=x^{2} - 2 \, x \log \left (-x - e^{\left (e^{x}\right )} + 2\right ) + \log \left (-x - e^{\left (e^{x}\right )} + 2\right )^{2} + x \] Input:
integrate((((2*exp(x)-2)*exp(exp(x))+6-2*x)*log(-exp(exp(x))+2-x)+(-2*exp( x)*x+2*x+1)*exp(exp(x))+2*x^2-5*x-2)/(exp(exp(x))+x-2),x, algorithm="maxim a")
Output:
x^2 - 2*x*log(-x - e^(e^x) + 2) + log(-x - e^(e^x) + 2)^2 + x
\[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=\int { \frac {2 \, x^{2} - {\left (2 \, x e^{x} - 2 \, x - 1\right )} e^{\left (e^{x}\right )} + 2 \, {\left ({\left (e^{x} - 1\right )} e^{\left (e^{x}\right )} - x + 3\right )} \log \left (-x - e^{\left (e^{x}\right )} + 2\right ) - 5 \, x - 2}{x + e^{\left (e^{x}\right )} - 2} \,d x } \] Input:
integrate((((2*exp(x)-2)*exp(exp(x))+6-2*x)*log(-exp(exp(x))+2-x)+(-2*exp( x)*x+2*x+1)*exp(exp(x))+2*x^2-5*x-2)/(exp(exp(x))+x-2),x, algorithm="giac" )
Output:
integrate((2*x^2 - (2*x*e^x - 2*x - 1)*e^(e^x) + 2*((e^x - 1)*e^(e^x) - x + 3)*log(-x - e^(e^x) + 2) - 5*x - 2)/(x + e^(e^x) - 2), x)
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=x^2-2\,x\,\ln \left (2-{\mathrm {e}}^{{\mathrm {e}}^x}-x\right )+x+{\ln \left (2-{\mathrm {e}}^{{\mathrm {e}}^x}-x\right )}^2 \] Input:
int((log(2 - exp(exp(x)) - x)*(exp(exp(x))*(2*exp(x) - 2) - 2*x + 6) - 5*x + exp(exp(x))*(2*x - 2*x*exp(x) + 1) + 2*x^2 - 2)/(x + exp(exp(x)) - 2),x )
Output:
x - 2*x*log(2 - exp(exp(x)) - x) + log(2 - exp(exp(x)) - x)^2 + x^2
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-2-5 x+2 x^2+e^{e^x} \left (1+2 x-2 e^x x\right )+\left (6+e^{e^x} \left (-2+2 e^x\right )-2 x\right ) \log \left (2-e^{e^x}-x\right )}{-2+e^{e^x}+x} \, dx=\mathrm {log}\left (-e^{e^{x}}-x +2\right )^{2}-2 \,\mathrm {log}\left (-e^{e^{x}}-x +2\right ) x +x^{2}+x \] Input:
int((((2*exp(x)-2)*exp(exp(x))+6-2*x)*log(-exp(exp(x))+2-x)+(-2*exp(x)*x+2 *x+1)*exp(exp(x))+2*x^2-5*x-2)/(exp(exp(x))+x-2),x)
Output:
log( - e**(e**x) - x + 2)**2 - 2*log( - e**(e**x) - x + 2)*x + x**2 + x