\(\int \frac {e^{e^x} (5-10 x+e^x (25-5 x+5 x^2)-80 x^2 \log ^3(x^2)+(-30 x^2+10 e^x x^3) \log ^4(x^2)-80 x^3 \log ^7(x^2)+(-20 x^3+5 e^x x^4) \log ^8(x^2))}{25-10 x+11 x^2-2 x^3+x^4+(20 x^3-4 x^4+4 x^5) \log ^4(x^2)+(10 x^4-2 x^5+6 x^6) \log ^8(x^2)+4 x^7 \log ^{12}(x^2)+x^8 \log ^{16}(x^2)} \, dx\) [1339]

Optimal result

Integrand size = 176, antiderivative size = 32 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {e^{e^x}}{1+\frac {1}{5} \left (-x+\left (x+x^2 \log ^4\left (x^2\right )\right )^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5 e^{e^x}}{5-x+x^2+2 x^3 \log ^4\left (x^2\right )+x^4 \log ^8\left (x^2\right )} \] Input:

Integrate[(E^E^x*(5 - 10*x + E^x*(25 - 5*x + 5*x^2) - 80*x^2*Log[x^2]^3 + 
(-30*x^2 + 10*E^x*x^3)*Log[x^2]^4 - 80*x^3*Log[x^2]^7 + (-20*x^3 + 5*E^x*x 
^4)*Log[x^2]^8))/(25 - 10*x + 11*x^2 - 2*x^3 + x^4 + (20*x^3 - 4*x^4 + 4*x 
^5)*Log[x^2]^4 + (10*x^4 - 2*x^5 + 6*x^6)*Log[x^2]^8 + 4*x^7*Log[x^2]^12 + 
 x^8*Log[x^2]^16),x]
 

Output:

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

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.

\(\displaystyle \int \frac {e^{e^x} \left (e^x \left (5 x^2-5 x+25\right )-80 x^2 \log ^3\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (10 e^x x^3-30 x^2\right ) \log ^4\left (x^2\right )+\left (5 e^x x^4-20 x^3\right ) \log ^8\left (x^2\right )-10 x+5\right )}{x^4-2 x^3+11 x^2+x^8 \log ^{16}\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+\left (6 x^6-2 x^5+10 x^4\right ) \log ^8\left (x^2\right )+\left (4 x^5-4 x^4+20 x^3\right ) \log ^4\left (x^2\right )-10 x+25} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{e^x} \left (e^x \left (5 x^2-5 x+25\right )-80 x^2 \log ^3\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (10 e^x x^3-30 x^2\right ) \log ^4\left (x^2\right )+\left (5 e^x x^4-20 x^3\right ) \log ^8\left (x^2\right )-10 x+5\right )}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {20 e^{e^x} x^3 \log ^8\left (x^2\right )}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}-\frac {30 e^{e^x} x^2 \log ^4\left (x^2\right )}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}+\frac {5 e^{x+e^x}}{x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5}+\frac {5 e^{e^x}}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}-\frac {10 e^{e^x} x}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}-\frac {80 e^{e^x} x^3 \log ^7\left (x^2\right )}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}-\frac {80 e^{e^x} x^2 \log ^3\left (x^2\right )}{\left (x^2+x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )-x+5\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -15 \int \frac {e^{e^x}}{\left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx+100 \int \frac {e^{e^x}}{x \left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx+10 \int \frac {e^{e^x} x}{\left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx+10 \int \frac {e^{e^x} x^2 \log ^4\left (x^2\right )}{\left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx+5 \int \frac {e^{x+e^x}}{x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5}dx-20 \int \frac {e^{e^x}}{x \left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )}dx-80 \int \frac {e^{e^x} x^3 \log ^7\left (x^2\right )}{\left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx-80 \int \frac {e^{e^x} x^2 \log ^3\left (x^2\right )}{\left (x^4 \log ^8\left (x^2\right )+2 x^3 \log ^4\left (x^2\right )+x^2-x+5\right )^2}dx\)

Input:

Int[(E^E^x*(5 - 10*x + E^x*(25 - 5*x + 5*x^2) - 80*x^2*Log[x^2]^3 + (-30*x 
^2 + 10*E^x*x^3)*Log[x^2]^4 - 80*x^3*Log[x^2]^7 + (-20*x^3 + 5*E^x*x^4)*Lo 
g[x^2]^8))/(25 - 10*x + 11*x^2 - 2*x^3 + x^4 + (20*x^3 - 4*x^4 + 4*x^5)*Lo 
g[x^2]^4 + (10*x^4 - 2*x^5 + 6*x^6)*Log[x^2]^8 + 4*x^7*Log[x^2]^12 + x^8*L 
og[x^2]^16),x]
 

Output:

$Aborted
 

Maple [F(-1)]

Timed out.

\[\int \frac {\left (\left (5 \,{\mathrm e}^{x} x^{4}-20 x^{3}\right ) \ln \left (x^{2}\right )^{8}-80 x^{3} \ln \left (x^{2}\right )^{7}+\left (10 \,{\mathrm e}^{x} x^{3}-30 x^{2}\right ) \ln \left (x^{2}\right )^{4}-80 x^{2} \ln \left (x^{2}\right )^{3}+\left (5 x^{2}-5 x +25\right ) {\mathrm e}^{x}-10 x +5\right ) {\mathrm e}^{{\mathrm e}^{x}}}{x^{8} \ln \left (x^{2}\right )^{16}+4 x^{7} \ln \left (x^{2}\right )^{12}+\left (6 x^{6}-2 x^{5}+10 x^{4}\right ) \ln \left (x^{2}\right )^{8}+\left (4 x^{5}-4 x^{4}+20 x^{3}\right ) \ln \left (x^{2}\right )^{4}+x^{4}-2 x^{3}+11 x^{2}-10 x +25}d x\]

Input:

int(((5*exp(x)*x^4-20*x^3)*ln(x^2)^8-80*x^3*ln(x^2)^7+(10*exp(x)*x^3-30*x^ 
2)*ln(x^2)^4-80*x^2*ln(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp(exp(x))/(x 
^8*ln(x^2)^16+4*x^7*ln(x^2)^12+(6*x^6-2*x^5+10*x^4)*ln(x^2)^8+(4*x^5-4*x^4 
+20*x^3)*ln(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x)
 

Output:

int(((5*exp(x)*x^4-20*x^3)*ln(x^2)^8-80*x^3*ln(x^2)^7+(10*exp(x)*x^3-30*x^ 
2)*ln(x^2)^4-80*x^2*ln(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp(exp(x))/(x 
^8*ln(x^2)^16+4*x^7*ln(x^2)^12+(6*x^6-2*x^5+10*x^4)*ln(x^2)^8+(4*x^5-4*x^4 
+20*x^3)*ln(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5 \, e^{\left (e^{x}\right )}}{x^{4} \log \left (x^{2}\right )^{8} + 2 \, x^{3} \log \left (x^{2}\right )^{4} + x^{2} - x + 5} \] Input:

integrate(((5*exp(x)*x^4-20*x^3)*log(x^2)^8-80*x^3*log(x^2)^7+(10*exp(x)*x 
^3-30*x^2)*log(x^2)^4-80*x^2*log(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp( 
exp(x))/(x^8*log(x^2)^16+4*x^7*log(x^2)^12+(6*x^6-2*x^5+10*x^4)*log(x^2)^8 
+(4*x^5-4*x^4+20*x^3)*log(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x, algorithm="f 
ricas")
 

Output:

5*e^(e^x)/(x^4*log(x^2)^8 + 2*x^3*log(x^2)^4 + x^2 - x + 5)
 

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5 e^{e^{x}}}{x^{4} \log {\left (x^{2} \right )}^{8} + 2 x^{3} \log {\left (x^{2} \right )}^{4} + x^{2} - x + 5} \] Input:

integrate(((5*exp(x)*x**4-20*x**3)*ln(x**2)**8-80*x**3*ln(x**2)**7+(10*exp 
(x)*x**3-30*x**2)*ln(x**2)**4-80*x**2*ln(x**2)**3+(5*x**2-5*x+25)*exp(x)-1 
0*x+5)*exp(exp(x))/(x**8*ln(x**2)**16+4*x**7*ln(x**2)**12+(6*x**6-2*x**5+1 
0*x**4)*ln(x**2)**8+(4*x**5-4*x**4+20*x**3)*ln(x**2)**4+x**4-2*x**3+11*x** 
2-10*x+25),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5 \, e^{\left (e^{x}\right )}}{256 \, x^{4} \log \left (x\right )^{8} + 32 \, x^{3} \log \left (x\right )^{4} + x^{2} - x + 5} \] Input:

integrate(((5*exp(x)*x^4-20*x^3)*log(x^2)^8-80*x^3*log(x^2)^7+(10*exp(x)*x 
^3-30*x^2)*log(x^2)^4-80*x^2*log(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp( 
exp(x))/(x^8*log(x^2)^16+4*x^7*log(x^2)^12+(6*x^6-2*x^5+10*x^4)*log(x^2)^8 
+(4*x^5-4*x^4+20*x^3)*log(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x, algorithm="m 
axima")
 

Output:

5*e^(e^x)/(256*x^4*log(x)^8 + 32*x^3*log(x)^4 + x^2 - x + 5)
 

Giac [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(((5*exp(x)*x^4-20*x^3)*log(x^2)^8-80*x^3*log(x^2)^7+(10*exp(x)*x 
^3-30*x^2)*log(x^2)^4-80*x^2*log(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp( 
exp(x))/(x^8*log(x^2)^16+4*x^7*log(x^2)^12+(6*x^6-2*x^5+10*x^4)*log(x^2)^8 
+(4*x^5-4*x^4+20*x^3)*log(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x, algorithm="g 
iac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x^4\,{\ln \left (x^2\right )}^8+2\,x^3\,{\ln \left (x^2\right )}^4+x^2-x+5} \] Input:

int((exp(exp(x))*(exp(x)*(5*x^2 - 5*x + 25) - 10*x + log(x^2)^8*(5*x^4*exp 
(x) - 20*x^3) + log(x^2)^4*(10*x^3*exp(x) - 30*x^2) - 80*x^2*log(x^2)^3 - 
80*x^3*log(x^2)^7 + 5))/(log(x^2)^8*(10*x^4 - 2*x^5 + 6*x^6) - 10*x + log( 
x^2)^4*(20*x^3 - 4*x^4 + 4*x^5) + 11*x^2 - 2*x^3 + x^4 + 4*x^7*log(x^2)^12 
 + x^8*log(x^2)^16 + 25),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {e^{e^x} \left (5-10 x+e^x \left (25-5 x+5 x^2\right )-80 x^2 \log ^3\left (x^2\right )+\left (-30 x^2+10 e^x x^3\right ) \log ^4\left (x^2\right )-80 x^3 \log ^7\left (x^2\right )+\left (-20 x^3+5 e^x x^4\right ) \log ^8\left (x^2\right )\right )}{25-10 x+11 x^2-2 x^3+x^4+\left (20 x^3-4 x^4+4 x^5\right ) \log ^4\left (x^2\right )+\left (10 x^4-2 x^5+6 x^6\right ) \log ^8\left (x^2\right )+4 x^7 \log ^{12}\left (x^2\right )+x^8 \log ^{16}\left (x^2\right )} \, dx=\frac {5 e^{e^{x}}}{\mathrm {log}\left (x^{2}\right )^{8} x^{4}+2 \mathrm {log}\left (x^{2}\right )^{4} x^{3}+x^{2}-x +5} \] Input:

int(((5*exp(x)*x^4-20*x^3)*log(x^2)^8-80*x^3*log(x^2)^7+(10*exp(x)*x^3-30* 
x^2)*log(x^2)^4-80*x^2*log(x^2)^3+(5*x^2-5*x+25)*exp(x)-10*x+5)*exp(exp(x) 
)/(x^8*log(x^2)^16+4*x^7*log(x^2)^12+(6*x^6-2*x^5+10*x^4)*log(x^2)^8+(4*x^ 
5-4*x^4+20*x^3)*log(x^2)^4+x^4-2*x^3+11*x^2-10*x+25),x)
 

Output:

(5*e**(e**x))/(log(x**2)**8*x**4 + 2*log(x**2)**4*x**3 + x**2 - x + 5)