\(\int \frac {e^{e^{e^{\frac {2+x^2-x^3+(-x+x^2) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+(-x+x^2) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+(-x+x^2) \log (4 x)}{-x+x^2}} (2-2 x-5 x^2+4 x^3-x^4+e^5 (2-2 x-5 x^2+4 x^3-x^4))}{4 x-8 x^2+4 x^3} \, dx\) [1031]

Optimal result

Integrand size = 170, antiderivative size = 35 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {1}{4} e^{e^{4 e^{-x-\frac {2}{x-x^2}} x^2}} \left (1+e^5\right ) \] Output:

1/4*(exp(5)+1)*exp(exp(x*exp(ln(4*x)-x-2/(-x^2+x))))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {1}{4} e^{e^{4 e^{\frac {2}{-1+x}-\frac {2}{x}-x} x^2}} \left (1+e^5\right ) \] Input:

Integrate[(E^(E^(E^((2 + x^2 - x^3 + (-x + x^2)*Log[4*x])/(-x + x^2))*x) + 
 E^((2 + x^2 - x^3 + (-x + x^2)*Log[4*x])/(-x + x^2))*x + (2 + x^2 - x^3 + 
 (-x + x^2)*Log[4*x])/(-x + x^2))*(2 - 2*x - 5*x^2 + 4*x^3 - x^4 + E^5*(2 
- 2*x - 5*x^2 + 4*x^3 - x^4)))/(4*x - 8*x^2 + 4*x^3),x]
 

Output:

(E^E^(4*E^(2/(-1 + x) - 2/x - x)*x^2)*(1 + E^5))/4
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (-x^4+4 x^3-5 x^2+e^5 \left (-x^4+4 x^3-5 x^2-2 x+2\right )-2 x+2\right ) \exp \left (\exp \left (x e^{\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}}\right )+x e^{\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}}+\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}\right )}{4 x^3-8 x^2+4 x} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (-x^4+4 x^3-5 x^2+e^5 \left (-x^4+4 x^3-5 x^2-2 x+2\right )-2 x+2\right ) \exp \left (\exp \left (x e^{\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}}\right )+x e^{\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}}+\frac {-x^3+x^2+\left (x^2-x\right ) \log (4 x)+2}{x^2-x}\right )}{x \left (4 x^2-8 x+4\right )}dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 16 \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (-x^4+4 x^3-5 x^2-2 x+e^5 \left (-x^4+4 x^3-5 x^2-2 x+2\right )+2\right )}{64 (1-x)^2 x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (-x^4+4 x^3-5 x^2-2 x+e^5 \left (-x^4+4 x^3-5 x^2-2 x+2\right )+2\right )}{(1-x)^2 x}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {1}{4} \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (-\left (\left (1+e^5\right ) x^4\right )+4 \left (1+e^5\right ) x^3-5 \left (1+e^5\right ) x^2-2 \left (1+e^5\right ) x+2 \left (1+e^5\right )\right )}{(1-x)^2 x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{4} \int \left (-\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (1+e^5\right ) x-\frac {2 \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (1+e^5\right )}{x-1}-\frac {2 \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (1+e^5\right )}{(x-1)^2}+2 \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (1+e^5\right )+\frac {2 \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) \left (1+e^5\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (2 \left (1+e^5\right ) \int \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right )dx-2 \left (1+e^5\right ) \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right )}{(x-1)^2}dx-2 \left (1+e^5\right ) \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right )}{x-1}dx+2 \left (1+e^5\right ) \int \frac {\exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right )}{x}dx-\left (1+e^5\right ) \int \exp \left (4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2+e^{4 e^{-\frac {-x^3+x^2+2}{x-x^2}} x^2}-\frac {-x^3+x^2-\left (x-x^2\right ) \log (4 x)+2}{x-x^2}\right ) xdx\right )\)

Input:

Int[(E^(E^(E^((2 + x^2 - x^3 + (-x + x^2)*Log[4*x])/(-x + x^2))*x) + E^((2 
 + x^2 - x^3 + (-x + x^2)*Log[4*x])/(-x + x^2))*x + (2 + x^2 - x^3 + (-x + 
 x^2)*Log[4*x])/(-x + x^2))*(2 - 2*x - 5*x^2 + 4*x^3 - x^4 + E^5*(2 - 2*x 
- 5*x^2 + 4*x^3 - x^4)))/(4*x - 8*x^2 + 4*x^3),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(85\) vs. \(2(30)=60\).

Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.46

\[\frac {{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{\frac {\ln \left (4 x \right ) x^{2}-x^{3}-x \ln \left (4 x \right )+x^{2}+2}{x \left (-1+x \right )}}}} {\mathrm e}^{5}}{4}+\frac {{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{\frac {\ln \left (4 x \right ) x^{2}-x^{3}-x \ln \left (4 x \right )+x^{2}+2}{x \left (-1+x \right )}}}}}{4}\]

Input:

int(((-x^4+4*x^3-5*x^2-2*x+2)*exp(5)-x^4+4*x^3-5*x^2-2*x+2)*exp(((x^2-x)*l 
n(4*x)-x^3+x^2+2)/(x^2-x))*exp(x*exp(((x^2-x)*ln(4*x)-x^3+x^2+2)/(x^2-x))) 
*exp(exp(x*exp(((x^2-x)*ln(4*x)-x^3+x^2+2)/(x^2-x))))/(4*x^3-8*x^2+4*x),x)
 

Output:

1/4*exp(exp(x*exp((ln(4*x)*x^2-x^3-x*ln(4*x)+x^2+2)/x/(-1+x))))*exp(5)+1/4 
*exp(exp(x*exp((ln(4*x)*x^2-x^3-x*ln(4*x)+x^2+2)/x/(-1+x))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (30) = 60\).

Time = 0.09 (sec) , antiderivative size = 206, normalized size of antiderivative = 5.89 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {1}{4} \, {\left (e^{5} + 1\right )} e^{\left (-x e^{\left (-\frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )} - \frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} e^{\left (x e^{\left (-\frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )}\right )} - {\left (x^{3} - x^{2}\right )} e^{\left (-\frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x} + \frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )} \] Input:

integrate(((-x^4+4*x^3-5*x^2-2*x+2)*exp(5)-x^4+4*x^3-5*x^2-2*x+2)*exp(((x^ 
2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))*exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/( 
x^2-x)))*exp(exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))))/(4*x^3-8*x^ 
2+4*x),x, algorithm="fricas")
 

Output:

1/4*(e^5 + 1)*e^(-x*e^(-(x^3 - x^2 - (x^2 - x)*log(4*x) - 2)/(x^2 - x)) - 
(x^3 - x^2 - (x^2 - x)*e^(x*e^(-(x^3 - x^2 - (x^2 - x)*log(4*x) - 2)/(x^2 
- x))) - (x^3 - x^2)*e^(-(x^3 - x^2 - (x^2 - x)*log(4*x) - 2)/(x^2 - x)) - 
 (x^2 - x)*log(4*x) - 2)/(x^2 - x) + (x^3 - x^2 - (x^2 - x)*log(4*x) - 2)/ 
(x^2 - x))
 

Sympy [A] (verification not implemented)

Time = 3.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {\left (1 + e^{5}\right ) e^{e^{x e^{\frac {- x^{3} + x^{2} + \left (x^{2} - x\right ) \log {\left (4 x \right )} + 2}{x^{2} - x}}}}}{4} \] Input:

integrate(((-x**4+4*x**3-5*x**2-2*x+2)*exp(5)-x**4+4*x**3-5*x**2-2*x+2)*ex 
p(((x**2-x)*ln(4*x)-x**3+x**2+2)/(x**2-x))*exp(x*exp(((x**2-x)*ln(4*x)-x** 
3+x**2+2)/(x**2-x)))*exp(exp(x*exp(((x**2-x)*ln(4*x)-x**3+x**2+2)/(x**2-x) 
)))/(4*x**3-8*x**2+4*x),x)
 

Output:

(1 + exp(5))*exp(exp(x*exp((-x**3 + x**2 + (x**2 - x)*log(4*x) + 2)/(x**2 
- x))))/4
 

Maxima [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {1}{4} \, {\left (e^{5} + 1\right )} e^{\left (e^{\left (4 \, x^{2} e^{\left (-x + \frac {2}{x - 1} - \frac {2}{x}\right )}\right )}\right )} \] Input:

integrate(((-x^4+4*x^3-5*x^2-2*x+2)*exp(5)-x^4+4*x^3-5*x^2-2*x+2)*exp(((x^ 
2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))*exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/( 
x^2-x)))*exp(exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))))/(4*x^3-8*x^ 
2+4*x),x, algorithm="maxima")
 

Output:

1/4*(e^5 + 1)*e^(e^(4*x^2*e^(-x + 2/(x - 1) - 2/x)))
 

Giac [F]

\[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\int { -\frac {{\left (x^{4} - 4 \, x^{3} + 5 \, x^{2} + {\left (x^{4} - 4 \, x^{3} + 5 \, x^{2} + 2 \, x - 2\right )} e^{5} + 2 \, x - 2\right )} e^{\left (x e^{\left (-\frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )} - \frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x} + e^{\left (x e^{\left (-\frac {x^{3} - x^{2} - {\left (x^{2} - x\right )} \log \left (4 \, x\right ) - 2}{x^{2} - x}\right )}\right )}\right )}}{4 \, {\left (x^{3} - 2 \, x^{2} + x\right )}} \,d x } \] Input:

integrate(((-x^4+4*x^3-5*x^2-2*x+2)*exp(5)-x^4+4*x^3-5*x^2-2*x+2)*exp(((x^ 
2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))*exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/( 
x^2-x)))*exp(exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))))/(4*x^3-8*x^ 
2+4*x),x, algorithm="giac")
 

Output:

integrate(-1/4*(x^4 - 4*x^3 + 5*x^2 + (x^4 - 4*x^3 + 5*x^2 + 2*x - 2)*e^5 
+ 2*x - 2)*e^(x*e^(-(x^3 - x^2 - (x^2 - x)*log(4*x) - 2)/(x^2 - x)) - (x^3 
 - x^2 - (x^2 - x)*log(4*x) - 2)/(x^2 - x) + e^(x*e^(-(x^3 - x^2 - (x^2 - 
x)*log(4*x) - 2)/(x^2 - x))))/(x^3 - 2*x^2 + x), x)
 

Mupad [B] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.37 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {4\,x\,x^{\frac {x}{x-x^2}}\,{\mathrm {e}}^{\frac {x^3}{x-x^2}}\,{\mathrm {e}}^{-\frac {x^2}{x-x^2}}\,{\mathrm {e}}^{-\frac {2}{x-x^2}}}{x^{\frac {x^2}{x-x^2}}}}}\,\left (\frac {{\mathrm {e}}^5}{4}+\frac {1}{4}\right ) \] Input:

int(-(exp((log(4*x)*(x - x^2) - x^2 + x^3 - 2)/(x - x^2))*exp(x*exp((log(4 
*x)*(x - x^2) - x^2 + x^3 - 2)/(x - x^2)))*exp(exp(x*exp((log(4*x)*(x - x^ 
2) - x^2 + x^3 - 2)/(x - x^2))))*(2*x + exp(5)*(2*x + 5*x^2 - 4*x^3 + x^4 
- 2) + 5*x^2 - 4*x^3 + x^4 - 2))/(4*x - 8*x^2 + 4*x^3),x)
 

Output:

exp(exp((4*x*x^(x/(x - x^2))*exp(x^3/(x - x^2))*exp(-x^2/(x - x^2))*exp(-2 
/(x - x^2)))/x^(x^2/(x - x^2))))*(exp(5)/4 + 1/4)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {e^{e^{e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x}+e^{\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} x+\frac {2+x^2-x^3+\left (-x+x^2\right ) \log (4 x)}{-x+x^2}} \left (2-2 x-5 x^2+4 x^3-x^4+e^5 \left (2-2 x-5 x^2+4 x^3-x^4\right )\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {e^{e^{\frac {4 e^{\frac {2}{x^{2}-x}} x^{2}}{e^{x}}}} \left (e^{5}+1\right )}{4} \] Input:

int(((-x^4+4*x^3-5*x^2-2*x+2)*exp(5)-x^4+4*x^3-5*x^2-2*x+2)*exp(((x^2-x)*l 
og(4*x)-x^3+x^2+2)/(x^2-x))*exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/(x^2-x) 
))*exp(exp(x*exp(((x^2-x)*log(4*x)-x^3+x^2+2)/(x^2-x))))/(4*x^3-8*x^2+4*x) 
,x)
 

Output:

(e**(e**((4*e**(2/(x**2 - x))*x**2)/e**x))*(e**5 + 1))/4