\(\int \frac {-25+4 x^2+e^2 (-625-1000 x-600 x^2-160 x^3-16 x^4)+e^2 (625+1000 x+600 x^2+160 x^3+16 x^4) \log ^2(4)+e^{2 e^x} (-e^2+e^2 \log ^2(4))+e^{e^x} (1-e^x x+e^2 (50+40 x+8 x^2)+e^2 (-50-40 x-8 x^2) \log ^2(4))}{-625-1000 x-600 x^2-160 x^3-16 x^4+(625+1000 x+600 x^2+160 x^3+16 x^4) \log ^2(4)+e^{2 e^x} (-1+\log ^2(4))+e^{e^x} (50+40 x+8 x^2+(-50-40 x-8 x^2) \log ^2(4))} \, dx\) [1553]

Optimal result

Integrand size = 220, antiderivative size = 36 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=e^2 x-\frac {x}{\left (e^{e^x}-(5+2 x)^2\right ) \left (1-\log ^2(4)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=\frac {\frac {x}{-25+e^{e^x}-20 x-4 x^2}+e^2 x \left (-1+\log ^2(4)\right )}{-1+\log ^2(4)} \] Input:

Integrate[(-25 + 4*x^2 + E^2*(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4) 
+ E^2*(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4)*Log[4]^2 + E^(2*E^x)*(-E 
^2 + E^2*Log[4]^2) + E^E^x*(1 - E^x*x + E^2*(50 + 40*x + 8*x^2) + E^2*(-50 
 - 40*x - 8*x^2)*Log[4]^2))/(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4 + 
(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4)*Log[4]^2 + E^(2*E^x)*(-1 + Log 
[4]^2) + E^E^x*(50 + 40*x + 8*x^2 + (-50 - 40*x - 8*x^2)*Log[4]^2)),x]
 

Output:

(x/(-25 + E^E^x - 20*x - 4*x^2) + E^2*x*(-1 + Log[4]^2))/(-1 + Log[4]^2)
 

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.

\(\displaystyle \int \frac {4 x^2+e^{e^x} \left (e^2 \left (8 x^2+40 x+50\right )+e^2 \left (-8 x^2-40 x-50\right ) \log ^2(4)-e^x x+1\right )+e^2 \left (-16 x^4-160 x^3-600 x^2-1000 x-625\right )+e^2 \left (16 x^4+160 x^3+600 x^2+1000 x+625\right ) \log ^2(4)+e^{2 e^x} \left (e^2 \log ^2(4)-e^2\right )-25}{-16 x^4-160 x^3-600 x^2+e^{e^x} \left (8 x^2+\left (-8 x^2-40 x-50\right ) \log ^2(4)+40 x+50\right )+\left (16 x^4+160 x^3+600 x^2+1000 x+625\right ) \log ^2(4)-1000 x+e^{2 e^x} \left (\log ^2(4)-1\right )-625} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {-4 x^2-e^{e^x}+e^{x+e^x} x-e^2 (2 x+5)^4 \left (\log ^2(4)-1\right )+2 e^{e^x+2} (2 x+5)^2 \left (\log ^2(4)-1\right )-e^{2 e^x+2} \left (\log ^2(4)-1\right )+25}{\left (e^{e^x}-(2 x+5)^2\right )^2 \left (1-\log ^2(4)\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {e^2 \left (1-\log ^2(4)\right ) (2 x+5)^4-2 e^{2+e^x} \left (1-\log ^2(4)\right ) (2 x+5)^2-e^{e^x}-4 x^2+e^{x+e^x} x+e^{2+2 e^x} \left (1-\log ^2(4)\right )+25}{\left (e^{e^x}-(2 x+5)^2\right )^2}dx}{1-\log ^2(4)}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {\int \left (-\frac {e^2 (-1+\log (4)) (1+\log (4)) (2 x+5)^4}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}+\frac {2 e^{2+e^x} (-1+\log (4)) (1+\log (4)) (2 x+5)^2}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}-\frac {e^{e^x}}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}+\frac {e^{x+e^x} x}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}+\frac {25}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}-\frac {4 x^2}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}+\frac {e^{2 \left (1+e^x\right )} (1-\log (4)) (1+\log (4))}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}\right )dx}{1-\log ^2(4)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {25 \int \frac {1}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}dx-\int \frac {e^{e^x}}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}dx+\int \frac {e^{x+e^x} x}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}dx-4 \int \frac {x^2}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx+\left (1-\log ^2(4)\right ) \int \frac {e^{2 \left (1+e^x\right )}}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx+625 e^2 (1-\log (4)) (1+\log (4)) \int \frac {1}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}dx-50 (1-\log (4)) (1+\log (4)) \int \frac {e^{2+e^x}}{\left (-4 x^2-20 x+e^{e^x}-25\right )^2}dx+1000 e^2 (1-\log (4)) (1+\log (4)) \int \frac {x}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx-40 (1-\log (4)) (1+\log (4)) \int \frac {e^{2+e^x} x}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx+600 e^2 (1-\log (4)) (1+\log (4)) \int \frac {x^2}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx-8 (1-\log (4)) (1+\log (4)) \int \frac {e^{2+e^x} x^2}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx+16 e^2 (1-\log (4)) (1+\log (4)) \int \frac {x^4}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx+160 e^2 (1-\log (4)) (1+\log (4)) \int \frac {x^3}{\left (4 x^2+20 x-e^{e^x}+25\right )^2}dx}{1-\log ^2(4)}\)

Input:

Int[(-25 + 4*x^2 + E^2*(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4) + E^2* 
(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4)*Log[4]^2 + E^(2*E^x)*(-E^2 + E 
^2*Log[4]^2) + E^E^x*(1 - E^x*x + E^2*(50 + 40*x + 8*x^2) + E^2*(-50 - 40* 
x - 8*x^2)*Log[4]^2))/(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4 + (625 + 
 1000*x + 600*x^2 + 160*x^3 + 16*x^4)*Log[4]^2 + E^(2*E^x)*(-1 + Log[4]^2) 
 + E^E^x*(50 + 40*x + 8*x^2 + (-50 - 40*x - 8*x^2)*Log[4]^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00

method	result	size
risch	\({\mathrm e}^{2} x -\frac {x}{\left (4 \ln \left (2\right )^{2}-1\right ) \left (4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25\right )}\)	\(36\)
norman	\(\frac {\frac {\left (100 \,{\mathrm e}^{2} \ln \left (2\right )^{2}-25 \,{\mathrm e}^{2}-1\right ) x}{4 \ln \left (2\right )^{2}-1}+20 x^{2} {\mathrm e}^{2}+4 x^{3} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{2} x}{4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25}\)	\(68\)
parallelrisch	\(\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{2} x^{3}+80 \ln \left (2\right )^{2} {\mathrm e}^{2} x^{2}-4 \ln \left (2\right )^{2} {\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{2} x +100 \,{\mathrm e}^{2} \ln \left (2\right )^{2} x -4 x^{3} {\mathrm e}^{2}-20 x^{2} {\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{2} x -25 \,{\mathrm e}^{2} x -x}{\left (4 \ln \left (2\right )^{2}-1\right ) \left (4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25\right )}\)	\(102\)

Input:

int(((4*exp(2)*ln(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50) 
*exp(2)*ln(2)^2+(8*x^2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+60 
0*x^2+1000*x+625)*exp(2)*ln(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x-625)*exp( 
2)+4*x^2-25)/((4*ln(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*ln(2)^2+8*x^ 
2+40*x+50)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*ln(2)^2-16*x^ 
4-160*x^3-600*x^2-1000*x-625),x,method=_RETURNVERBOSE)
 

Output:

exp(2)*x-x/(4*ln(2)^2-1)/(4*x^2+20*x-exp(exp(x))+25)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (33) = 66\).

Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.94 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=\frac {4 \, {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} \log \left (2\right )^{2} - {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} - {\left (4 \, x e^{2} \log \left (2\right )^{2} - x e^{2}\right )} e^{\left (e^{x}\right )} - x}{4 \, {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (2\right )^{2} - 4 \, x^{2} - {\left (4 \, \log \left (2\right )^{2} - 1\right )} e^{\left (e^{x}\right )} - 20 \, x - 25} \] Input:

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-4 
0*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+16 
0*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x- 
625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*lo 
g(2)^2+8*x^2+40*x+50)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*lo 
g(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="fricas")
 

Output:

(4*(4*x^3 + 20*x^2 + 25*x)*e^2*log(2)^2 - (4*x^3 + 20*x^2 + 25*x)*e^2 - (4 
*x*e^2*log(2)^2 - x*e^2)*e^(e^x) - x)/(4*(4*x^2 + 20*x + 25)*log(2)^2 - 4* 
x^2 - (4*log(2)^2 - 1)*e^(e^x) - 20*x - 25)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.50 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=x e^{2} + \frac {x}{- 16 x^{2} \log {\left (2 \right )}^{2} + 4 x^{2} - 80 x \log {\left (2 \right )}^{2} + 20 x + \left (-1 + 4 \log {\left (2 \right )}^{2}\right ) e^{e^{x}} - 100 \log {\left (2 \right )}^{2} + 25} \] Input:

integrate(((4*exp(2)*ln(2)**2-exp(2))*exp(exp(x))**2+(-exp(x)*x+4*(-8*x**2 
-40*x-50)*exp(2)*ln(2)**2+(8*x**2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x** 
4+160*x**3+600*x**2+1000*x+625)*exp(2)*ln(2)**2+(-16*x**4-160*x**3-600*x** 
2-1000*x-625)*exp(2)+4*x**2-25)/((4*ln(2)**2-1)*exp(exp(x))**2+(4*(-8*x**2 
-40*x-50)*ln(2)**2+8*x**2+40*x+50)*exp(exp(x))+4*(16*x**4+160*x**3+600*x** 
2+1000*x+625)*ln(2)**2-16*x**4-160*x**3-600*x**2-1000*x-625),x)
 

Output:

x*exp(2) + x/(-16*x**2*log(2)**2 + 4*x**2 - 80*x*log(2)**2 + 20*x + (-1 + 
4*log(2)**2)*exp(exp(x)) - 100*log(2)**2 + 25)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (33) = 66\).

Time = 0.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.08 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=\frac {4 \, {\left (4 \, \log \left (2\right )^{2} - 1\right )} x^{3} e^{2} + 20 \, {\left (4 \, \log \left (2\right )^{2} - 1\right )} x^{2} e^{2} - {\left (4 \, \log \left (2\right )^{2} - 1\right )} x e^{\left (e^{x} + 2\right )} + {\left (25 \, {\left (4 \, \log \left (2\right )^{2} - 1\right )} e^{2} - 1\right )} x}{4 \, {\left (4 \, \log \left (2\right )^{2} - 1\right )} x^{2} + 20 \, {\left (4 \, \log \left (2\right )^{2} - 1\right )} x - {\left (4 \, \log \left (2\right )^{2} - 1\right )} e^{\left (e^{x}\right )} + 100 \, \log \left (2\right )^{2} - 25} \] Input:

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-4 
0*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+16 
0*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x- 
625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*lo 
g(2)^2+8*x^2+40*x+50)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*lo 
g(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="maxima")
 

Output:

(4*(4*log(2)^2 - 1)*x^3*e^2 + 20*(4*log(2)^2 - 1)*x^2*e^2 - (4*log(2)^2 - 
1)*x*e^(e^x + 2) + (25*(4*log(2)^2 - 1)*e^2 - 1)*x)/(4*(4*log(2)^2 - 1)*x^ 
2 + 20*(4*log(2)^2 - 1)*x - (4*log(2)^2 - 1)*e^(e^x) + 100*log(2)^2 - 25)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (33) = 66\).

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.33 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=\frac {16 \, x^{3} e^{2} \log \left (2\right )^{2} + 80 \, x^{2} e^{2} \log \left (2\right )^{2} - 4 \, x^{3} e^{2} + 100 \, x e^{2} \log \left (2\right )^{2} - 4 \, x e^{\left (e^{x} + 2\right )} \log \left (2\right )^{2} - 20 \, x^{2} e^{2} - 25 \, x e^{2} + x e^{\left (e^{x} + 2\right )} - x}{16 \, x^{2} \log \left (2\right )^{2} + 80 \, x \log \left (2\right )^{2} - 4 \, e^{\left (e^{x}\right )} \log \left (2\right )^{2} - 4 \, x^{2} + 100 \, \log \left (2\right )^{2} - 20 \, x + e^{\left (e^{x}\right )} - 25} \] Input:

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-4 
0*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+16 
0*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x- 
625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*lo 
g(2)^2+8*x^2+40*x+50)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*lo 
g(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="giac")
 

Output:

(16*x^3*e^2*log(2)^2 + 80*x^2*e^2*log(2)^2 - 4*x^3*e^2 + 100*x*e^2*log(2)^ 
2 - 4*x*e^(e^x + 2)*log(2)^2 - 20*x^2*e^2 - 25*x*e^2 + x*e^(e^x + 2) - x)/ 
(16*x^2*log(2)^2 + 80*x*log(2)^2 - 4*e^(e^x)*log(2)^2 - 4*x^2 + 100*log(2) 
^2 - 20*x + e^(e^x) - 25)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=-\int -\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2-4\,{\mathrm {e}}^2\,{\ln \left (2\right )}^2\right )+{\mathrm {e}}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-4\,x^2-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2\,\left (8\,x^2+40\,x+50\right )-x\,{\mathrm {e}}^x-4\,{\mathrm {e}}^2\,{\ln \left (2\right )}^2\,\left (8\,x^2+40\,x+50\right )+1\right )-4\,{\mathrm {e}}^2\,{\ln \left (2\right )}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )+25}{1000\,x-4\,{\ln \left (2\right )}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (4\,{\ln \left (2\right )}^2-1\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (40\,x-4\,{\ln \left (2\right )}^2\,\left (8\,x^2+40\,x+50\right )+8\,x^2+50\right )+600\,x^2+160\,x^3+16\,x^4+625} \,d x \] Input:

int((exp(2*exp(x))*(exp(2) - 4*exp(2)*log(2)^2) + exp(2)*(1000*x + 600*x^2 
 + 160*x^3 + 16*x^4 + 625) - 4*x^2 - exp(exp(x))*(exp(2)*(40*x + 8*x^2 + 5 
0) - x*exp(x) - 4*exp(2)*log(2)^2*(40*x + 8*x^2 + 50) + 1) - 4*exp(2)*log( 
2)^2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) + 25)/(1000*x - 4*log(2)^ 
2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) - exp(2*exp(x))*(4*log(2)^2 
- 1) - exp(exp(x))*(40*x - 4*log(2)^2*(40*x + 8*x^2 + 50) + 8*x^2 + 50) + 
600*x^2 + 160*x^3 + 16*x^4 + 625),x)
 

Output:

-int(-(exp(2*exp(x))*(exp(2) - 4*exp(2)*log(2)^2) + exp(2)*(1000*x + 600*x 
^2 + 160*x^3 + 16*x^4 + 625) - 4*x^2 - exp(exp(x))*(exp(2)*(40*x + 8*x^2 + 
 50) - x*exp(x) - 4*exp(2)*log(2)^2*(40*x + 8*x^2 + 50) + 1) - 4*exp(2)*lo 
g(2)^2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) + 25)/(1000*x - 4*log(2 
)^2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) - exp(2*exp(x))*(4*log(2)^ 
2 - 1) - exp(exp(x))*(40*x - 4*log(2)^2*(40*x + 8*x^2 + 50) + 8*x^2 + 50) 
+ 600*x^2 + 160*x^3 + 16*x^4 + 625), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 4.31 \[ \int \frac {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx=\frac {4 e^{e^{x}} \mathrm {log}\left (2\right )^{2} e^{2} x -20 e^{e^{x}} \mathrm {log}\left (2\right )^{2} e^{2}-e^{e^{x}} e^{2} x +5 e^{e^{x}} e^{2}-16 \mathrm {log}\left (2\right )^{2} e^{2} x^{3}+300 \mathrm {log}\left (2\right )^{2} e^{2} x +500 \mathrm {log}\left (2\right )^{2} e^{2}+4 e^{2} x^{3}-75 e^{2} x -125 e^{2}+x}{4 e^{e^{x}} \mathrm {log}\left (2\right )^{2}-e^{e^{x}}-16 \mathrm {log}\left (2\right )^{2} x^{2}-80 \mathrm {log}\left (2\right )^{2} x -100 \mathrm {log}\left (2\right )^{2}+4 x^{2}+20 x +25} \] Input:

int(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50 
)*exp(2)*log(2)^2+(8*x^2+40*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+ 
600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x-625)*e 
xp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*log(2)^2 
+8*x^2+40*x+50)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*log(2)^2 
-16*x^4-160*x^3-600*x^2-1000*x-625),x)
 

Output:

(4*e**(e**x)*log(2)**2*e**2*x - 20*e**(e**x)*log(2)**2*e**2 - e**(e**x)*e* 
*2*x + 5*e**(e**x)*e**2 - 16*log(2)**2*e**2*x**3 + 300*log(2)**2*e**2*x + 
500*log(2)**2*e**2 + 4*e**2*x**3 - 75*e**2*x - 125*e**2 + x)/(4*e**(e**x)* 
log(2)**2 - e**(e**x) - 16*log(2)**2*x**2 - 80*log(2)**2*x - 100*log(2)**2 
 + 4*x**2 + 20*x + 25)