\(\int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} (-4 x+4 x^2)}{-4+5 x}} (e^x (16-40 x+25 x^2)+e^{2 e^{-x} x} (32 x^2-104 x^3+112 x^4-40 x^5+e^x (16 x-32 x^2+20 x^3)))}{16-40 x+25 x^2} \, dx\) [2966]

Optimal result

Integrand size = 117, antiderivative size = 34 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=e^{5+\frac {4 e^{2 e^{-x} x} x^2}{5 x+\frac {x}{-1+x}}} x \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=e^{5+\frac {4 e^{2 e^{-x} x} (-1+x) x}{-4+5 x}} x \] Input:

Integrate[(E^(-x + (-20 + 25*x + E^((2*x)/E^x)*(-4*x + 4*x^2))/(-4 + 5*x)) 
*(E^x*(16 - 40*x + 25*x^2) + E^((2*x)/E^x)*(32*x^2 - 104*x^3 + 112*x^4 - 4 
0*x^5 + E^x*(16*x - 32*x^2 + 20*x^3))))/(16 - 40*x + 25*x^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (e^x \left (25 x^2-40 x+16\right )+e^{2 e^{-x} x} \left (-40 x^5+112 x^4-104 x^3+32 x^2+e^x \left (20 x^3-32 x^2+16 x\right )\right )\right ) \exp \left (\frac {e^{2 e^{-x} x} \left (4 x^2-4 x\right )+25 x-20}{5 x-4}-x\right )}{25 x^2-40 x+16} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 100 \int \frac {\exp \left (\frac {-25 x+4 e^{2 e^{-x} x} \left (x-x^2\right )+20}{4-5 x}-x\right ) \left (e^x \left (25 x^2-40 x+16\right )+4 e^{2 e^{-x} x} \left (-10 x^5+28 x^4-26 x^3+8 x^2+e^x \left (5 x^3-8 x^2+4 x\right )\right )\right )}{100 (4-5 x)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\left (e^x \left (25 x^2-40 x+16\right )+4 e^{2 e^{-x} x} \left (-10 x^5+28 x^4-26 x^3+8 x^2+e^x \left (5 x^3-8 x^2+4 x\right )\right )\right ) \exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}-x\right )}{(4-5 x)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {4 e^{2 e^{-x} x} \left (x-x^2\right )-25 x+20}{4-5 x}+2 e^{-x} x-x\right )}{(5 x-4)^2}\right )dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} x^2-4 e^{2 e^{-x} x} x+25 x-20}{5 x-4}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {5 x^2}{4-5 x}+2 e^{-x} x-\frac {4 e^{2 e^{-x} x} (x-1) x}{4-5 x}-\frac {29 x}{4-5 x}+\frac {20}{4-5 x}\right )}{(5 x-4)^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\exp \left (\frac {4 e^{2 e^{-x} x} x^2-4 e^{2 e^{-x} x} x+25 x-20}{5 x-4}\right )-\frac {4 x \left (10 x^4-28 x^3-5 e^x x^2+26 x^2+8 e^x x-8 x-4 e^x\right ) \exp \left (\frac {5 x^2}{4-5 x}+2 e^{-x} x-\frac {4 e^{2 e^{-x} x} (x-1) x}{4-5 x}-\frac {29 x}{4-5 x}+\frac {20}{4-5 x}\right )}{(5 x-4)^2}\right )dx\)

Input:

Int[(E^(-x + (-20 + 25*x + E^((2*x)/E^x)*(-4*x + 4*x^2))/(-4 + 5*x))*(E^x* 
(16 - 40*x + 25*x^2) + E^((2*x)/E^x)*(32*x^2 - 104*x^3 + 112*x^4 - 40*x^5 
+ E^x*(16*x - 32*x^2 + 20*x^3))))/(16 - 40*x + 25*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 20.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(x \,{\mathrm e}^{\frac {\left (4 x^{2}-4 x \right ) {\mathrm e}^{2 x \,{\mathrm e}^{-x}}+25 x -20}{5 x -4}}\)	\(36\)
risch	\(x \,{\mathrm e}^{\frac {4 \,{\mathrm e}^{2 x \,{\mathrm e}^{-x}} x^{2}-4 \,{\mathrm e}^{2 x \,{\mathrm e}^{-x}} x +25 x -20}{5 x -4}}\)	\(41\)

Input:

int((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp(x/exp 
(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25*x-20)/ 
(5*x-4))/(25*x^2-40*x+16)/exp(x),x,method=_RETURNVERBOSE)
 

Output:

x*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25*x-20)/(5*x-4))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\left (x - \frac {5 \, x^{2} - 4 \, {\left (x^{2} - x\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} - 29 \, x + 20}{5 \, x - 4}\right )} \] Input:

integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp 
(x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* 
x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm="fricas")
 

Output:

x*e^(x - (5*x^2 - 4*(x^2 - x)*e^(2*x*e^(-x)) - 29*x + 20)/(5*x - 4))
 

Sympy [A] (verification not implemented)

Time = 59.96 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\frac {25 x + \left (4 x^{2} - 4 x\right ) e^{2 x e^{- x}} - 20}{5 x - 4}} \] Input:

integrate((((20*x**3-32*x**2+16*x)*exp(x)-40*x**5+112*x**4-104*x**3+32*x** 
2)*exp(x/exp(x))**2+(25*x**2-40*x+16)*exp(x))*exp(((4*x**2-4*x)*exp(x/exp( 
x))**2+25*x-20)/(5*x-4))/(25*x**2-40*x+16)/exp(x),x)
 

Output:

x*exp((25*x + (4*x**2 - 4*x)*exp(2*x*exp(-x)) - 20)/(5*x - 4))
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x e^{\left (\frac {4}{5} \, x e^{\left (2 \, x e^{\left (-x\right )}\right )} - \frac {16 \, e^{\left (2 \, x e^{\left (-x\right )}\right )}}{25 \, {\left (5 \, x - 4\right )}} - \frac {4}{25} \, e^{\left (2 \, x e^{\left (-x\right )}\right )} + 5\right )} \] Input:

integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp 
(x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* 
x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm="maxima")
 

Output:

x*e^(4/5*x*e^(2*x*e^(-x)) - 16/25*e^(2*x*e^(-x))/(5*x - 4) - 4/25*e^(2*x*e 
^(-x)) + 5)
 

Giac [F]

\[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=\int { -\frac {{\left (4 \, {\left (10 \, x^{5} - 28 \, x^{4} + 26 \, x^{3} - 8 \, x^{2} - {\left (5 \, x^{3} - 8 \, x^{2} + 4 \, x\right )} e^{x}\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} - {\left (25 \, x^{2} - 40 \, x + 16\right )} e^{x}\right )} e^{\left (-x + \frac {4 \, {\left (x^{2} - x\right )} e^{\left (2 \, x e^{\left (-x\right )}\right )} + 25 \, x - 20}{5 \, x - 4}\right )}}{25 \, x^{2} - 40 \, x + 16} \,d x } \] Input:

integrate((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp 
(x/exp(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25* 
x-20)/(5*x-4))/(25*x^2-40*x+16)/exp(x),x, algorithm="giac")
 

Output:

integrate(-(4*(10*x^5 - 28*x^4 + 26*x^3 - 8*x^2 - (5*x^3 - 8*x^2 + 4*x)*e^ 
x)*e^(2*x*e^(-x)) - (25*x^2 - 40*x + 16)*e^x)*e^(-x + (4*(x^2 - x)*e^(2*x* 
e^(-x)) + 25*x - 20)/(5*x - 4))/(25*x^2 - 40*x + 16), x)
 

Mupad [B] (verification not implemented)

Time = 2.51 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=x\,{\mathrm {e}}^{-\frac {4\,x\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{-x}}}{5\,x-4}}\,{\mathrm {e}}^{-\frac {20}{5\,x-4}}\,{\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{-x}}}{5\,x-4}}\,{\mathrm {e}}^{\frac {25\,x}{5\,x-4}} \] Input:

int((exp(-x)*exp(-(exp(2*x*exp(-x))*(4*x - 4*x^2) - 25*x + 20)/(5*x - 4))* 
(exp(x)*(25*x^2 - 40*x + 16) + exp(2*x*exp(-x))*(32*x^2 - 104*x^3 + 112*x^ 
4 - 40*x^5 + exp(x)*(16*x - 32*x^2 + 20*x^3))))/(25*x^2 - 40*x + 16),x)
 

Output:

x*exp(-(4*x*exp(2*x*exp(-x)))/(5*x - 4))*exp(-20/(5*x - 4))*exp((4*x^2*exp 
(2*x*exp(-x)))/(5*x - 4))*exp((25*x)/(5*x - 4))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-x+\frac {-20+25 x+e^{2 e^{-x} x} \left (-4 x+4 x^2\right )}{-4+5 x}} \left (e^x \left (16-40 x+25 x^2\right )+e^{2 e^{-x} x} \left (32 x^2-104 x^3+112 x^4-40 x^5+e^x \left (16 x-32 x^2+20 x^3\right )\right )\right )}{16-40 x+25 x^2} \, dx=\frac {e^{\frac {4 e^{\frac {2 x}{e^{x}}} x^{2}}{5 x -4}} e^{5} x}{e^{\frac {4 e^{\frac {2 x}{e^{x}}} x}{5 x -4}}} \] Input:

int((((20*x^3-32*x^2+16*x)*exp(x)-40*x^5+112*x^4-104*x^3+32*x^2)*exp(x/exp 
(x))^2+(25*x^2-40*x+16)*exp(x))*exp(((4*x^2-4*x)*exp(x/exp(x))^2+25*x-20)/ 
(5*x-4))/(25*x^2-40*x+16)/exp(x),x)
 

Output:

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