\(\int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log (x^2)+\log ^2(x^2)}}+\frac {e^{4 e^x+x}}{16-8 \log (x^2)+\log ^2(x^2)}} (16 e^{2 x} x+e^x (4+4 x)+(-e^x x-4 e^{2 x} x) \log (x^2))}{-64 x+48 x \log (x^2)-12 x \log ^2(x^2)+x \log ^3(x^2)} \, dx\) [726]

Optimal result

Integrand size = 131, antiderivative size = 26 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{e^{\frac {e^{4 e^x+x}}{\left (4-\log \left (x^2\right )\right )^2}}} \] Output:

-exp(exp(exp(4*exp(x))/(4-ln(x^2))^2*exp(x)))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}} \] Input:

Integrate[(E^(4*E^x + E^(E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2)) + E 
^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2))*(16*E^(2*x)*x + E^x*(4 + 4*x) 
 + (-(E^x*x) - 4*E^(2*x)*x)*Log[x^2]))/(-64*x + 48*x*Log[x^2] - 12*x*Log[x 
^2]^2 + x*Log[x^2]^3),x]
 

Output:

-E^E^(E^(4*E^x + x)/(-4 + Log[x^2])^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.

\(\displaystyle \int \frac {\left (\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )+16 e^{2 x} x+e^x (4 x+4)\right ) \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+4 e^x\right )}{x \log ^3\left (x^2\right )-12 x \log ^2\left (x^2\right )+48 x \log \left (x^2\right )-64 x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (4 e^x x \log \left (x^2\right )+x \log \left (x^2\right )-16 e^x x-4 x-4\right ) \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+x+4 e^x\right )}{x \left (4-\log \left (x^2\right )\right )^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {\log \left (x^2\right ) \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+x+4 e^x\right )}{\left (\log \left (x^2\right )-4\right )^3}-\frac {4 \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+2 x+4 e^x\right )}{\left (\log \left (x^2\right )-4\right )^2}+\frac {4 \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+x+4 e^x\right )}{\left (\log \left (x^2\right )-4\right )^3}+\frac {4 \exp \left (e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}+x+4 e^x\right )}{x \left (\log \left (x^2\right )-4\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \int \frac {\exp \left (x+4 e^x+e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}\right )}{x \left (\log \left (x^2\right )-4\right )^3}dx-\int \frac {\exp \left (x+4 e^x+e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}\right )}{\left (\log \left (x^2\right )-4\right )^2}dx-4 \int \frac {\exp \left (2 x+4 e^x+e^{\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}}+\frac {e^{x+4 e^x}}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+16}\right )}{\left (\log \left (x^2\right )-4\right )^2}dx\)

Input:

Int[(E^(4*E^x + E^(E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2)) + E^(4*E^ 
x + x)/(16 - 8*Log[x^2] + Log[x^2]^2))*(16*E^(2*x)*x + E^x*(4 + 4*x) + (-( 
E^x*x) - 4*E^(2*x)*x)*Log[x^2]))/(-64*x + 48*x*Log[x^2] - 12*x*Log[x^2]^2 
+ x*Log[x^2]^3),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73

\[-{\mathrm e}^{{\mathrm e}^{\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{x}+x}}{{\left (-i \operatorname {csgn}\left (i x^{2}\right )^{3} \pi +2 i \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi -i \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2} \pi +4 \ln \left (x \right )-8\right )}^{2}}}}\]

Input:

int(((-4*x*exp(x)^2-exp(x)*x)*ln(x^2)+16*x*exp(x)^2+(4+4*x)*exp(x))*exp(4* 
exp(x))*exp(exp(x)*exp(4*exp(x))/(ln(x^2)^2-8*ln(x^2)+16))*exp(exp(exp(x)* 
exp(4*exp(x))/(ln(x^2)^2-8*ln(x^2)+16)))/(x*ln(x^2)^3-12*x*ln(x^2)^2+48*x* 
ln(x^2)-64*x),x)
 

Output:

-exp(exp(4*exp(4*exp(x)+x)/(-I*csgn(I*x^2)^3*Pi+2*I*csgn(I*x^2)^2*csgn(I*x 
)*Pi-I*csgn(I*x^2)*csgn(I*x)^2*Pi+4*ln(x)-8)^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (20) = 40\).

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.62 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{\left (\frac {4 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )} - 32 \, e^{x} \log \left (x^{2}\right ) + e^{\left (x + 4 \, e^{x}\right )} + 64 \, e^{x}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - \frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - 4 \, e^{x}\right )} \] Input:

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4+4*x)*exp(x)) 
*exp(4*exp(x))*exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16))*exp(ex 
p(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*log 
(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="fricas")
 

Output:

-e^((4*e^x*log(x^2)^2 + (log(x^2)^2 - 8*log(x^2) + 16)*e^(e^(x + 4*e^x)/(l 
og(x^2)^2 - 8*log(x^2) + 16)) - 32*e^x*log(x^2) + e^(x + 4*e^x) + 64*e^x)/ 
(log(x^2)^2 - 8*log(x^2) + 16) - e^(x + 4*e^x)/(log(x^2)^2 - 8*log(x^2) + 
16) - 4*e^x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(((-4*x*exp(x)**2-exp(x)*x)*ln(x**2)+16*x*exp(x)**2+(4+4*x)*exp(x 
))*exp(4*exp(x))*exp(exp(x)*exp(4*exp(x))/(ln(x**2)**2-8*ln(x**2)+16))*exp 
(exp(exp(x)*exp(4*exp(x))/(ln(x**2)**2-8*ln(x**2)+16)))/(x*ln(x**2)**3-12* 
x*ln(x**2)**2+48*x*ln(x**2)-64*x),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{\left (e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{4 \, {\left (\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4\right )}}\right )}\right )} \] Input:

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4+4*x)*exp(x)) 
*exp(4*exp(x))*exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16))*exp(ex 
p(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*log 
(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=\int { \frac {{\left (16 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x + 1\right )} e^{x} - {\left (4 \, x e^{\left (2 \, x\right )} + x e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} + 4 \, e^{x} + e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )}\right )}}{x \log \left (x^{2}\right )^{3} - 12 \, x \log \left (x^{2}\right )^{2} + 48 \, x \log \left (x^{2}\right ) - 64 \, x} \,d x } \] Input:

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4+4*x)*exp(x)) 
*exp(4*exp(x))*exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16))*exp(ex 
p(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*log 
(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="giac")
 

Output:

integrate((16*x*e^(2*x) + 4*(x + 1)*e^x - (4*x*e^(2*x) + x*e^x)*log(x^2))* 
e^(e^(x + 4*e^x)/(log(x^2)^2 - 8*log(x^2) + 16) + 4*e^x + e^(e^(x + 4*e^x) 
/(log(x^2)^2 - 8*log(x^2) + 16)))/(x*log(x^2)^3 - 12*x*log(x^2)^2 + 48*x*l 
og(x^2) - 64*x), x)
 

Mupad [B] (verification not implemented)

Time = 4.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^x}{{\ln \left (x^2\right )}^2-8\,\ln \left (x^2\right )+16}}} \] Input:

int(-(exp((exp(4*exp(x))*exp(x))/(log(x^2)^2 - 8*log(x^2) + 16))*exp(4*exp 
(x))*exp(exp((exp(4*exp(x))*exp(x))/(log(x^2)^2 - 8*log(x^2) + 16)))*(16*x 
*exp(2*x) + exp(x)*(4*x + 4) - log(x^2)*(4*x*exp(2*x) + x*exp(x))))/(64*x 
- 48*x*log(x^2) + 12*x*log(x^2)^2 - x*log(x^2)^3),x)
 

Output:

-exp(exp((exp(4*exp(x))*exp(x))/(log(x^2)^2 - 8*log(x^2) + 16)))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}} \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx=-e^{e^{\frac {e^{4 e^{x}+x}}{\mathrm {log}\left (x^{2}\right )^{2}-8 \,\mathrm {log}\left (x^{2}\right )+16}}} \] Input:

int(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4+4*x)*exp(x))*exp(4 
*exp(x))*exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16))*exp(exp(exp( 
x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*log(x^2)^ 
2+48*x*log(x^2)-64*x),x)
 

Output:

 - e**(e**(e**(4*e**x + x)/(log(x**2)**2 - 8*log(x**2) + 16)))