\(\int \frac {e^{5+e^x+x} (2-\log (\frac {\log ^2(-2+x)}{x^2})) (2 x+(8-2 x^2+e^x (4 x-2 x^2)) \log (-2+x)+(-2-x+x^2+e^x (-2 x+x^2)) \log (-2+x) \log (\frac {\log ^2(-2+x)}{x^2}))}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log (\frac {\log ^2(-2+x)}{x^2})} \, dx\) [232]

Optimal result

Integrand size = 120, antiderivative size = 25 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=e^{5+e^x+x} x \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 5.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=-e^{5+e^x+x} x \left (-2+\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \] Input:

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

Output:

-(E^(5 + E^x + x)*x*(-2 + Log[Log[-2 + x]^2/x^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.

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.09 (sec) , antiderivative size = 627, normalized size of antiderivative = 25.08

\[\text {Expression too large to display}\]

Input:

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

Output:

2*I*(-1/2*I*Pi*x*csgn(I/x^2)*csgn(I/x^2*ln(-2+x)^2)^2-1/2*I*Pi*x*csgn(I*ln 
(-2+x)^2)*csgn(I/x^2*ln(-2+x)^2)^2+2*x+1/2*I*Pi*x*csgn(I/x^2)*csgn(I*ln(-2 
+x)^2)*csgn(I/x^2*ln(-2+x)^2)+1/2*I*Pi*x*csgn(I*ln(-2+x))^2*csgn(I*ln(-2+x 
)^2)-I*Pi*x*csgn(I*ln(-2+x))*csgn(I*ln(-2+x)^2)^2-1/2*I*Pi*x*csgn(I*x)^2*c 
sgn(I*x^2)+I*Pi*x*csgn(I*x)*csgn(I*x^2)^2+1/2*I*Pi*x*csgn(I/x^2*ln(-2+x)^2 
)^3+1/2*I*Pi*x*csgn(I*ln(-2+x)^2)^3-1/2*I*Pi*x*csgn(I*x^2)^3+2*x*ln(x)-2*x 
*ln(ln(-2+x)))/(4*I*ln(x)+4*I-Pi*csgn(I*ln(-2+x))^2*csgn(I*ln(-2+x)^2)+2*P 
i*csgn(I*ln(-2+x))*csgn(I*ln(-2+x)^2)^2+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*cs 
gn(I*x)*csgn(I*x^2)^2+Pi*csgn(I/x^2)*csgn(I/x^2*ln(-2+x)^2)^2+Pi*csgn(I*ln 
(-2+x)^2)*csgn(I/x^2*ln(-2+x)^2)^2-Pi*csgn(I/x^2)*csgn(I*ln(-2+x)^2)*csgn( 
I/x^2*ln(-2+x)^2)-4*I*ln(ln(-2+x))-Pi*csgn(I/x^2*ln(-2+x)^2)^3+Pi*csgn(I*x 
^2)^3-Pi*csgn(I*ln(-2+x)^2)^3)*(2*ln(x)-2*ln(ln(-2+x))+1/2*I*Pi*csgn(I*ln( 
-2+x)^2)*(-csgn(I*ln(-2+x)^2)+csgn(I*ln(-2+x)))^2-1/2*I*Pi*csgn(I*x^2)*(-c 
sgn(I*x^2)+csgn(I*x))^2+1/2*I*Pi*csgn(I/x^2*ln(-2+x)^2)*(-csgn(I/x^2*ln(-2 
+x)^2)+csgn(I/x^2))*(-csgn(I/x^2*ln(-2+x)^2)+csgn(I*ln(-2+x)^2))+2)*exp(ex 
p(x)+5+x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=x e^{\left (x + e^{x} + \log \left (-\log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) + 2\right ) + 5\right )} \] Input:

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(-2+x)*log(log(-2+x)^2/x^2)+((-2* 
x^2+4*x)*exp(x)-2*x^2+8)*log(-2+x)+2*x)*exp(log(-log(log(-2+x)^2/x^2)+2)+e 
xp(x)+5+x)/((-2+x)*log(-2+x)*log(log(-2+x)^2/x^2)+(4-2*x)*log(-2+x)),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=-2 \, x e^{\left (x + e^{x} + 5\right )} \log \left (\log \left (x - 2\right )\right ) + 2 \, {\left (x e^{5} \log \left (x\right ) + x e^{5}\right )} e^{\left (x + e^{x}\right )} \] Input:

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(-2+x)*log(log(-2+x)^2/x^2)+((-2* 
x^2+4*x)*exp(x)-2*x^2+8)*log(-2+x)+2*x)*exp(log(-log(log(-2+x)^2/x^2)+2)+e 
xp(x)+5+x)/((-2+x)*log(-2+x)*log(log(-2+x)^2/x^2)+(4-2*x)*log(-2+x)),x, al 
gorithm="maxima")
 

Output:

-2*x*e^(x + e^x + 5)*log(log(x - 2)) + 2*(x*e^5*log(x) + x*e^5)*e^(x + e^x 
)
 

Giac [F]

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

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(-2+x)*log(log(-2+x)^2/x^2)+((-2* 
x^2+4*x)*exp(x)-2*x^2+8)*log(-2+x)+2*x)*exp(log(-log(log(-2+x)^2/x^2)+2)+e 
xp(x)+5+x)/((-2+x)*log(-2+x)*log(log(-2+x)^2/x^2)+(4-2*x)*log(-2+x)),x, al 
gorithm="giac")
 

Output:

undef
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx=e^{e^{x}+x} e^{5} x \left (-\mathrm {log}\left (\frac {\mathrm {log}\left (x -2\right )^{2}}{x^{2}}\right )+2\right ) \] Input:

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

Output:

e**(e**x + x)*e**5*x*( - log(log(x - 2)**2/x**2) + 2)