Integrand size = 117, antiderivative size = 28 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=x+\frac {\log \left (x \left (-5 e^{-e^x-x}+(-1+x) x\right )\right )}{x} \] Output:
ln(x*((-1+x)*x-exp(ln(5/exp(exp(x)))-x)))/x+x
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=1+x+\frac {\log \left (x \left (-5 e^{-e^x-x}+(-1+x) x\right )\right )}{x} \] Input:
Integrate[(2*x - 3*x^2 + x^3 - x^4 + 5*E^(-E^x - x)*(1 - x - E^x*x + x^2) + (-5*E^(-E^x - x) - x + x^2)*Log[-5*E^(-E^x - x)*x - x^2 + x^3])/(5*E^(-E ^x - x)*x^2 + x^3 - x^4),x]
Output:
1 + x + Log[x*(-5*E^(-E^x - x) + (-1 + x)*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.
\(\displaystyle \int \frac {-x^4+x^3-3 x^2+5 e^{-x-e^x} \left (x^2-e^x x-x+1\right )+\left (x^2-x-5 e^{-x-e^x}\right ) \log \left (x^3-x^2-5 e^{-x-e^x} x\right )+2 x}{-x^4+x^3+5 e^{-x-e^x} x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 e^{-e^x} \left (e^{e^x} x^2+e^{e^x} x-e^{e^x}+5\right )}{(x-1) x^2 \left (e^{x+e^x} x^2-e^{x+e^x} x-5\right )}+\frac {e^{-e^x} \left (e^{e^x} x^3-e^{e^x} x^2+3 e^{e^x} x-2 e^{e^x}-e^{e^x} x \log \left (x \left ((x-1) x-5 e^{-x-e^x}\right )\right )+e^{e^x} \log \left (x \left ((x-1) x-5 e^{-x-e^x}\right )\right )+5\right )}{(x-1) x^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {5 e^{-e^x} \left (e^{e^x} x^2+e^{e^x} x-e^{e^x}+5\right )}{(x-1) x^2 \left (e^{x+e^x} x^2-e^{x+e^x} x-5\right )}+\frac {e^{-e^x} \left (e^{e^x} x^3-e^{e^x} x^2+3 e^{e^x} x-2 e^{e^x}-e^{e^x} x \log \left (x \left ((x-1) x-5 e^{-x-e^x}\right )\right )+e^{e^x} \log \left (x \left ((x-1) x-5 e^{-x-e^x}\right )\right )+5\right )}{(x-1) x^2}\right )dx\) |
Input:
Int[(2*x - 3*x^2 + x^3 - x^4 + 5*E^(-E^x - x)*(1 - x - E^x*x + x^2) + (-5* E^(-E^x - x) - x + x^2)*Log[-5*E^(-E^x - x)*x - x^2 + x^3])/(5*E^(-E^x - x )*x^2 + x^3 - x^4),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(95\) vs. \(2(30)=60\).
Time = 15.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.43
method | result | size |
parallelrisch | \(-\frac {2 x \ln \left (x \right )+2 \ln \left (-{\mathrm e}^{\ln \left (5 \,{\mathrm e}^{-{\mathrm e}^{x}}\right )-x}+x^{2}-x \right ) x -2 x^{2}-2 \ln \left (-x \left (-x^{2}+{\mathrm e}^{\ln \left (5 \,{\mathrm e}^{-{\mathrm e}^{x}}\right )-x}+x \right )\right ) x -2 \ln \left (-x \left (-x^{2}+{\mathrm e}^{\ln \left (5 \,{\mathrm e}^{-{\mathrm e}^{x}}\right )-x}+x \right )\right )}{2 x}\) | \(96\) |
Input:
int(((-exp(ln(5/exp(exp(x)))-x)+x^2-x)*ln(-x*exp(ln(5/exp(exp(x)))-x)+x^3- x^2)+(-exp(x)*x+x^2-x+1)*exp(ln(5/exp(exp(x)))-x)-x^4+x^3-3*x^2+2*x)/(x^2* exp(ln(5/exp(exp(x)))-x)-x^4+x^3),x,method=_RETURNVERBOSE)
Output:
-1/2*(2*x*ln(x)+2*ln(-exp(ln(5/exp(exp(x)))-x)+x^2-x)*x-2*x^2-2*ln(-x*(-x^ 2+exp(ln(5/exp(exp(x)))-x)+x))*x-2*ln(-x*(-x^2+exp(ln(5/exp(exp(x)))-x)+x) ))/x
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=\frac {x^{2} + \log \left (x^{3} - x^{2} - x e^{\left (-x - e^{x} + \log \left (5\right )\right )}\right )}{x} \] Input:
integrate(((-exp(log(5/exp(exp(x)))-x)+x^2-x)*log(-x*exp(log(5/exp(exp(x)) )-x)+x^3-x^2)+(-exp(x)*x+x^2-x+1)*exp(log(5/exp(exp(x)))-x)-x^4+x^3-3*x^2+ 2*x)/(x^2*exp(log(5/exp(exp(x)))-x)-x^4+x^3),x, algorithm="fricas")
Output:
(x^2 + log(x^3 - x^2 - x*e^(-x - e^x + log(5))))/x
Time = 0.65 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=x + \frac {\log {\left (x^{3} - x^{2} - 5 x e^{- x} e^{- e^{x}} \right )}}{x} \] Input:
integrate(((-exp(ln(5/exp(exp(x)))-x)+x**2-x)*ln(-x*exp(ln(5/exp(exp(x)))- x)+x**3-x**2)+(-exp(x)*x+x**2-x+1)*exp(ln(5/exp(exp(x)))-x)-x**4+x**3-3*x* *2+2*x)/(x**2*exp(ln(5/exp(exp(x)))-x)-x**4+x**3),x)
Output:
x + log(x**3 - x**2 - 5*x*exp(-x)*exp(-exp(x)))/x
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=\frac {x^{2} - e^{x} + \log \left ({\left (x^{2} - x\right )} e^{\left (x + e^{x}\right )} - 5\right ) + \log \left (x\right )}{x} \] Input:
integrate(((-exp(log(5/exp(exp(x)))-x)+x^2-x)*log(-x*exp(log(5/exp(exp(x)) )-x)+x^3-x^2)+(-exp(x)*x+x^2-x+1)*exp(log(5/exp(exp(x)))-x)-x^4+x^3-3*x^2+ 2*x)/(x^2*exp(log(5/exp(exp(x)))-x)-x^4+x^3),x, algorithm="maxima")
Output:
(x^2 - e^x + log((x^2 - x)*e^(x + e^x) - 5) + log(x))/x
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=\frac {x^{2} - e^{x} + \log \left (x^{2} e^{\left (x + e^{x}\right )} - x e^{\left (x + e^{x}\right )} - 5\right ) + \log \left (x\right )}{x} \] Input:
integrate(((-exp(log(5/exp(exp(x)))-x)+x^2-x)*log(-x*exp(log(5/exp(exp(x)) )-x)+x^3-x^2)+(-exp(x)*x+x^2-x+1)*exp(log(5/exp(exp(x)))-x)-x^4+x^3-3*x^2+ 2*x)/(x^2*exp(log(5/exp(exp(x)))-x)-x^4+x^3),x, algorithm="giac")
Output:
(x^2 - e^x + log(x^2*e^(x + e^x) - x*e^(x + e^x) - 5) + log(x))/x
Time = 1.90 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=x+\frac {\ln \left (x^3-x^2-5\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\right )}{x} \] Input:
int(-(exp(log(5*exp(-exp(x))) - x)*(x + x*exp(x) - x^2 - 1) - 2*x + log(x^ 3 - x^2 - x*exp(log(5*exp(-exp(x))) - x))*(x + exp(log(5*exp(-exp(x))) - x ) - x^2) + 3*x^2 - x^3 + x^4)/(x^2*exp(log(5*exp(-exp(x))) - x) + x^3 - x^ 4),x)
Output:
x + log(x^3 - x^2 - 5*x*exp(-x)*exp(-exp(x)))/x
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.29 \[ \int \frac {2 x-3 x^2+x^3-x^4+5 e^{-e^x-x} \left (1-x-e^x x+x^2\right )+\left (-5 e^{-e^x-x}-x+x^2\right ) \log \left (-5 e^{-e^x-x} x-x^2+x^3\right )}{5 e^{-e^x-x} x^2+x^3-x^4} \, dx=\frac {-e^{x} x +\mathrm {log}\left (e^{e^{x}+x} x^{2}-e^{e^{x}+x} x -5\right ) x -\mathrm {log}\left (\frac {e^{e^{x}+x} x^{3}-e^{e^{x}+x} x^{2}-5 x}{e^{e^{x}+x}}\right ) x +\mathrm {log}\left (\frac {e^{e^{x}+x} x^{3}-e^{e^{x}+x} x^{2}-5 x}{e^{e^{x}+x}}\right )+\mathrm {log}\left (x \right ) x}{x} \] Input:
int(((-exp(log(5/exp(exp(x)))-x)+x^2-x)*log(-x*exp(log(5/exp(exp(x)))-x)+x ^3-x^2)+(-exp(x)*x+x^2-x+1)*exp(log(5/exp(exp(x)))-x)-x^4+x^3-3*x^2+2*x)/( x^2*exp(log(5/exp(exp(x)))-x)-x^4+x^3),x)
Output:
( - e**x*x + log(e**(e**x + x)*x**2 - e**(e**x + x)*x - 5)*x - log((e**(e* *x + x)*x**3 - e**(e**x + x)*x**2 - 5*x)/e**(e**x + x))*x + log((e**(e**x + x)*x**3 - e**(e**x + x)*x**2 - 5*x)/e**(e**x + x)) + log(x)*x)/x