\(\int \frac {(\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))})^x (5-x+e^x (-1-\log (4))+(-5+2 x+e^x (1+x+(1+x) \log (4))) \log (x \log (4))+(-5+x+e^x (1+\log (4))) \log (x \log (4)) \log (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}))}{(-5+x+e^x (1+\log (4))) \log (x \log (4))} \, dx\) [1440]

Optimal result

Integrand size = 130, antiderivative size = 23 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x \left (-5+e^x+x+e^x \log (4)\right )}{\log (x \log (4))}\right )^x \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x \left (-5+x+e^x (1+\log (4))\right )}{\log (x \log (4))}\right )^x \] Input:

Integrate[(((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x*(5 - x + E^ 
x*(-1 - Log[4]) + (-5 + 2*x + E^x*(1 + x + (1 + x)*Log[4]))*Log[x*Log[4]] 
+ (-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]*Log[(-5*x + x^2 + E^x*(x + x*L 
og[4]))/Log[x*Log[4]]]))/((-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]),x]
 

Output:

((x*(-5 + x + E^x*(1 + Log[4])))/Log[x*Log[4]])^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.

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (x \log (x \log (4))+\log \left (\frac {x \left (x+e^x (1+\log (4))-5\right )}{\log (x \log (4))}\right ) \log (x \log (4))+\log (x \log (4))-1\right ) \left (\frac {x^2-5 x+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{\log (x \log (4))}+\frac {(x-6) x \left (\frac {x^2-5 x+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x}{-x-e^x (1+\log (4))+5}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^xdx+\int x \left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^xdx+\int \frac {x^2 \left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^x}{-x-e^x (1+\log (4))+5}dx+6 \int \frac {x \left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^x}{x+e^x (1+\log (4))-5}dx-\int \frac {\left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^x}{\log (x \log (4))}dx+\int \left (\frac {x^2-5 x+e^x (\log (4) x+x)}{\log (x \log (4))}\right )^x \log \left (\frac {x \left (x+e^x (1+\log (4))-5\right )}{\log (x \log (4))}\right )dx\)

Input:

Int[(((-5*x + x^2 + E^x*(x + x*Log[4]))/Log[x*Log[4]])^x*(5 - x + E^x*(-1 
- Log[4]) + (-5 + 2*x + E^x*(1 + x + (1 + x)*Log[4]))*Log[x*Log[4]] + (-5 
+ x + E^x*(1 + Log[4]))*Log[x*Log[4]]*Log[(-5*x + x^2 + E^x*(x + x*Log[4]) 
)/Log[x*Log[4]]]))/((-5 + x + E^x*(1 + Log[4]))*Log[x*Log[4]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 55.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
parallelrisch	\({\mathrm e}^{\ln \left (\frac {x \left ({\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (2\right )-5+x \right )}{\ln \left (2 x \ln \left (2\right )\right )}\right ) x}\)	\(26\)

Input:

int((((1+2*ln(2))*exp(x)+x-5)*ln(2*x*ln(2))*ln(((x+2*x*ln(2))*exp(x)+x^2-5 
*x)/ln(2*x*ln(2)))+((2*ln(2)*(1+x)+x+1)*exp(x)+2*x-5)*ln(2*x*ln(2))+(-1-2* 
ln(2))*exp(x)+5-x)*exp(x*ln(((x+2*x*ln(2))*exp(x)+x^2-5*x)/ln(2*x*ln(2)))) 
/((1+2*ln(2))*exp(x)+x-5)/ln(2*x*ln(2)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\left (\frac {x^{2} + {\left (2 \, x \log \left (2\right ) + x\right )} e^{x} - 5 \, x}{\log \left (2 \, x \log \left (2\right )\right )}\right )^{x} \] Input:

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*e 
xp(x)+x^2-5*x)/log(2*x*log(2)))+((2*log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2* 
x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2-5 
*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorit 
hm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\text {Timed out} \] Input:

integrate((((1+2*ln(2))*exp(x)+x-5)*ln(2*x*ln(2))*ln(((x+2*x*ln(2))*exp(x) 
+x**2-5*x)/ln(2*x*ln(2)))+((2*ln(2)*(1+x)+x+1)*exp(x)+2*x-5)*ln(2*x*ln(2)) 
+(-1-2*ln(2))*exp(x)+5-x)*exp(x*ln(((x+2*x*ln(2))*exp(x)+x**2-5*x)/ln(2*x* 
ln(2))))/((1+2*ln(2))*exp(x)+x-5)/ln(2*x*ln(2)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=e^{\left (x \log \left ({\left (2 \, \log \left (2\right ) + 1\right )} e^{x} + x - 5\right ) + x \log \left (x\right ) - x \log \left (\log \left (2\right ) + \log \left (x\right ) + \log \left (\log \left (2\right )\right )\right )\right )} \] Input:

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*e 
xp(x)+x^2-5*x)/log(2*x*log(2)))+((2*log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2* 
x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2-5 
*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorit 
hm="maxima")
 

Output:

e^(x*log((2*log(2) + 1)*e^x + x - 5) + x*log(x) - x*log(log(2) + log(x) + 
log(log(2))))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*e 
xp(x)+x^2-5*x)/log(2*x*log(2)))+((2*log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2* 
x*log(2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2-5 
*x)/log(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x, algorit 
hm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,2,0,0]%%%}+%%%{2,[0,1,1,1,1]%%%}+%%%{2,[0,1,1,1,0]% 
%%} / %%%
 

Mupad [B] (verification not implemented)

Time = 3.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx={\left (\frac {x\,{\mathrm {e}}^x-5\,x+x^2+2\,x\,{\mathrm {e}}^x\,\ln \left (2\right )}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}\right )}^x \] Input:

int((exp(x*log((exp(x)*(x + 2*x*log(2)) - 5*x + x^2)/log(2*x*log(2))))*(lo 
g(2*x*log(2))*(2*x + exp(x)*(x + 2*log(2)*(x + 1) + 1) - 5) - exp(x)*(2*lo 
g(2) + 1) - x + log((exp(x)*(x + 2*x*log(2)) - 5*x + x^2)/log(2*x*log(2))) 
*log(2*x*log(2))*(x + exp(x)*(2*log(2) + 1) - 5) + 5))/(log(2*x*log(2))*(x 
 + exp(x)*(2*log(2) + 1) - 5)),x)
 

Output:

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

Reduce [F]

\[ \int \frac {\left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )^x \left (5-x+e^x (-1-\log (4))+\left (-5+2 x+e^x (1+x+(1+x) \log (4))\right ) \log (x \log (4))+\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4)) \log \left (\frac {-5 x+x^2+e^x (x+x \log (4))}{\log (x \log (4))}\right )\right )}{\left (-5+x+e^x (1+\log (4))\right ) \log (x \log (4))} \, dx =\text {Too large to display} \] Input:

int((((1+2*log(2))*exp(x)+x-5)*log(2*x*log(2))*log(((x+2*x*log(2))*exp(x)+ 
x^2-5*x)/log(2*x*log(2)))+((2*log(2)*(1+x)+x+1)*exp(x)+2*x-5)*log(2*x*log( 
2))+(-1-2*log(2))*exp(x)+5-x)*exp(x*log(((x+2*x*log(2))*exp(x)+x^2-5*x)/lo 
g(2*x*log(2))))/((1+2*log(2))*exp(x)+x-5)/log(2*x*log(2)),x)
 

Output:

5*int((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x/(2*e**x*log(2*log(2)*x)** 
x*log(2*log(2)*x)*log(2) + e**x*log(2*log(2)*x)**x*log(2*log(2)*x) + log(2 
*log(2)*x)**x*log(2*log(2)*x)*x - 5*log(2*log(2)*x)**x*log(2*log(2)*x)),x) 
 - 5*int((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x/(2*e**x*log(2*log(2)*x 
)**x*log(2) + e**x*log(2*log(2)*x)**x + log(2*log(2)*x)**x*x - 5*log(2*log 
(2)*x)**x),x) + int(((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x*log((2*e** 
x*log(2)*x + e**x*x + x**2 - 5*x)/log(2*log(2)*x))*x)/(2*e**x*log(2*log(2) 
*x)**x*log(2) + e**x*log(2*log(2)*x)**x + log(2*log(2)*x)**x*x - 5*log(2*l 
og(2)*x)**x),x) - 5*int(((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x*log((2 
*e**x*log(2)*x + e**x*x + x**2 - 5*x)/log(2*log(2)*x)))/(2*e**x*log(2*log( 
2)*x)**x*log(2) + e**x*log(2*log(2)*x)**x + log(2*log(2)*x)**x*x - 5*log(2 
*log(2)*x)**x),x) - int(((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x*x)/(2* 
e**x*log(2*log(2)*x)**x*log(2*log(2)*x)*log(2) + e**x*log(2*log(2)*x)**x*l 
og(2*log(2)*x) + log(2*log(2)*x)**x*log(2*log(2)*x)*x - 5*log(2*log(2)*x)* 
*x*log(2*log(2)*x)),x) + 2*int(((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)**x 
*x)/(2*e**x*log(2*log(2)*x)**x*log(2) + e**x*log(2*log(2)*x)**x + log(2*lo 
g(2)*x)**x*x - 5*log(2*log(2)*x)**x),x) + 2*int((e**x*(2*e**x*log(2)*x + e 
**x*x + x**2 - 5*x)**x*log((2*e**x*log(2)*x + e**x*x + x**2 - 5*x)/log(2*l 
og(2)*x)))/(2*e**x*log(2*log(2)*x)**x*log(2) + e**x*log(2*log(2)*x)**x + l 
og(2*log(2)*x)**x*x - 5*log(2*log(2)*x)**x),x)*log(2) + int((e**x*(2*e*...