\(\int \frac {-2 e^{2 x} x^5+(-5 x^5+5 x^4 \log (4 e^{e^{2 x}+x})) \log (-x+\log (4 e^{e^{2 x}+x}))+(-x-x \log (x)+\log (4 e^{e^{2 x}+x}) (1+\log (x))) \log ^2(-x+\log (4 e^{e^{2 x}+x}))}{(-x+\log (4 e^{e^{2 x}+x})) \log ^2(-x+\log (4 e^{e^{2 x}+x}))} \, dx\) [1028]

Optimal result

Integrand size = 136, antiderivative size = 28 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \left (\log (x)+\frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \log (x)+\frac {x^5}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 26.61 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(x \ln \left (x \right )+\frac {x^{5}}{\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\)	\(29\)
parallelrisch	\(\frac {600 x^{5}+600 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right ) \ln \left (x \right ) x}{600 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\)	\(46\)

Input:

int((((ln(x)+1)*ln(4*exp(exp(2*x)+x))-x*ln(x)-x)*ln(ln(4*exp(exp(2*x)+x))- 
x)^2+(5*x^4*ln(4*exp(exp(2*x)+x))-5*x^5)*ln(ln(4*exp(exp(2*x)+x))-x)-2*x^5 
*exp(2*x))/(ln(4*exp(exp(2*x)+x))-x)/ln(ln(4*exp(exp(2*x)+x))-x)^2,x,metho 
d=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \] Input:

integrate((((1+log(x))*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(ex 
p(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x 
)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*x) 
+x))-x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \] Input:

integrate((((1+log(x))*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(ex 
p(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x 
)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*x) 
+x))-x)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \] Input:

integrate((((1+log(x))*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(ex 
p(2*x)+x))-x)^2+(5*x^4*log(4*exp(exp(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x 
)+x))-x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*x) 
+x))-x)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^5}{\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (4\right )\right )}+x\,\ln \left (x\right ) \] Input:

int((log(log(4*exp(x + exp(2*x))) - x)^2*(x + x*log(x) - log(4*exp(x + exp 
(2*x)))*(log(x) + 1)) - log(log(4*exp(x + exp(2*x))) - x)*(5*x^4*log(4*exp 
(x + exp(2*x))) - 5*x^5) + 2*x^5*exp(2*x))/(log(log(4*exp(x + exp(2*x))) - 
 x)^2*(x - log(4*exp(x + exp(2*x))))),x)
 

Output:

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

Reduce [F]

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

int((((1+log(x))*log(4*exp(exp(2*x)+x))-x*log(x)-x)*log(log(4*exp(exp(2*x) 
+x))-x)^2+(5*x^4*log(4*exp(exp(2*x)+x))-5*x^5)*log(log(4*exp(exp(2*x)+x))- 
x)-2*x^5*exp(2*x))/(log(4*exp(exp(2*x)+x))-x)/log(log(4*exp(exp(2*x)+x))-x 
)^2,x)
 

Output:

 - 5*int(x**5/(log(log(4*e**(e**(2*x) + x)) - x)*log(4*e**(e**(2*x) + x)) 
- log(log(4*e**(e**(2*x) + x)) - x)*x),x) - 2*int((e**(2*x)*x**5)/(log(log 
(4*e**(e**(2*x) + x)) - x)**2*log(4*e**(e**(2*x) + x)) - log(log(4*e**(e** 
(2*x) + x)) - x)**2*x),x) + 5*int((log(4*e**(e**(2*x) + x))*x**4)/(log(log 
(4*e**(e**(2*x) + x)) - x)*log(4*e**(e**(2*x) + x)) - log(log(4*e**(e**(2* 
x) + x)) - x)*x),x) + log(x)*x