\(\int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log (\frac {3}{\log (x+\log (x))})+(-x-2 x^2+e^x (-x-x^2)+(-1+e^x (-1-x)-2 x) \log (x)) \log (x+\log (x)) \log ^2(\frac {3}{\log (x+\log (x))})}{(x+\log (x)) \log (x+\log (x)) \log ^2(\frac {3}{\log (x+\log (x))})} \, dx\) [900]

Optimal result

Integrand size = 112, antiderivative size = 30 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=5-x+x \left (-e^x-x+\frac {3}{\log \left (\frac {3}{\log (x+\log (x))}\right )}\right ) \] Output:

(3/ln(3/ln(x+ln(x)))-x-exp(x))*x+5-x
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=x \left (-1-e^x-x+\frac {3}{\log \left (\frac {3}{\log (x+\log (x))}\right )}\right ) \] Input:

Integrate[(3 + 3*x + (3*x + 3*Log[x])*Log[x + Log[x]]*Log[3/Log[x + Log[x] 
]] + (-x - 2*x^2 + E^x*(-x - x^2) + (-1 + E^x*(-1 - x) - 2*x)*Log[x])*Log[ 
x + Log[x]]*Log[3/Log[x + Log[x]]]^2)/((x + Log[x])*Log[x + Log[x]]*Log[3/ 
Log[x + Log[x]]]^2),x]
 

Output:

x*(-1 - E^x - x + 3/Log[3/Log[x + Log[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.

\(\displaystyle \int \frac {\left (-2 x^2+e^x \left (-x^2-x\right )-x+\left (e^x (-x-1)-2 x-1\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+3}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {2 x^2}{x+\log (x)}-e^x (x+1)+\frac {3 x}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}+\frac {3}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}-\frac {2 x \log (x)}{x+\log (x)}-\frac {x}{x+\log (x)}-\frac {\log (x)}{x+\log (x)}+\frac {3}{\log \left (\frac {3}{\log (x+\log (x))}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \int \frac {1}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}dx+3 \int \frac {x}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}dx+3 \int \frac {1}{\log \left (\frac {3}{\log (x+\log (x))}\right )}dx-x^2-x+e^x-e^x (x+1)\)

Input:

Int[(3 + 3*x + (3*x + 3*Log[x])*Log[x + Log[x]]*Log[3/Log[x + Log[x]]] + ( 
-x - 2*x^2 + E^x*(-x - x^2) + (-1 + E^x*(-1 - x) - 2*x)*Log[x])*Log[x + Lo 
g[x]]*Log[3/Log[x + Log[x]]]^2)/((x + Log[x])*Log[x + Log[x]]*Log[3/Log[x 
+ Log[x]]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 15.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
default	\(-{\mathrm e}^{x} x -x^{2}-x +\frac {3 x}{\ln \left (3\right )-\ln \left (\ln \left (x +\ln \left (x \right )\right )\right )}\)	\(31\)
parts	\(-{\mathrm e}^{x} x -x^{2}-x +\frac {3 x}{\ln \left (3\right )-\ln \left (\ln \left (x +\ln \left (x \right )\right )\right )}\)	\(31\)
risch	\(-x^{2}-{\mathrm e}^{x} x -x +\frac {6 x}{2 \ln \left (3\right )-2 \ln \left (\ln \left (x +\ln \left (x \right )\right )\right )}\)	\(33\)
parallelrisch	\(\frac {-2 \ln \left (\frac {3}{\ln \left (x +\ln \left (x \right )\right )}\right ) x^{2}-2 \ln \left (\frac {3}{\ln \left (x +\ln \left (x \right )\right )}\right ) x \,{\mathrm e}^{x}-2 \ln \left (\frac {3}{\ln \left (x +\ln \left (x \right )\right )}\right ) x +6 x}{2 \ln \left (\frac {3}{\ln \left (x +\ln \left (x \right )\right )}\right )}\)	\(62\)

Input:

int(((((-1-x)*exp(x)-2*x-1)*ln(x)+(-x^2-x)*exp(x)-2*x^2-x)*ln(x+ln(x))*ln( 
3/ln(x+ln(x)))^2+(3*x+3*ln(x))*ln(x+ln(x))*ln(3/ln(x+ln(x)))+3*x+3)/(x+ln( 
x))/ln(x+ln(x))/ln(3/ln(x+ln(x)))^2,x,method=_RETURNVERBOSE)
 

Output:

-exp(x)*x-x^2-x+3*x/(ln(3)-ln(ln(x+ln(x))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=-\frac {{\left (x^{2} + x e^{x} + x\right )} \log \left (\frac {3}{\log \left (x + \log \left (x\right )\right )}\right ) - 3 \, x}{\log \left (\frac {3}{\log \left (x + \log \left (x\right )\right )}\right )} \] Input:

integrate(((((-1-x)*exp(x)-2*x-1)*log(x)+(-x^2-x)*exp(x)-2*x^2-x)*log(x+lo 
g(x))*log(3/log(x+log(x)))^2+(3*x+3*log(x))*log(x+log(x))*log(3/log(x+log( 
x)))+3*x+3)/(x+log(x))/log(x+log(x))/log(3/log(x+log(x)))^2,x, algorithm=" 
fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=- x^{2} - x e^{x} - x + \frac {3 x}{\log {\left (\frac {3}{\log {\left (x + \log {\left (x \right )} \right )}} \right )}} \] Input:

integrate(((((-1-x)*exp(x)-2*x-1)*ln(x)+(-x**2-x)*exp(x)-2*x**2-x)*ln(x+ln 
(x))*ln(3/ln(x+ln(x)))**2+(3*x+3*ln(x))*ln(x+ln(x))*ln(3/ln(x+ln(x)))+3*x+ 
3)/(x+ln(x))/ln(x+ln(x))/ln(3/ln(x+ln(x)))**2,x)
 

Output:

-x**2 - x*exp(x) - x + 3*x/log(3/log(x + log(x)))
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=-\frac {x^{2} \log \left (3\right ) + x e^{x} \log \left (3\right ) + x {\left (\log \left (3\right ) - 3\right )} - {\left (x^{2} + x e^{x} + x\right )} \log \left (\log \left (x + \log \left (x\right )\right )\right )}{\log \left (3\right ) - \log \left (\log \left (x + \log \left (x\right )\right )\right )} \] Input:

integrate(((((-1-x)*exp(x)-2*x-1)*log(x)+(-x^2-x)*exp(x)-2*x^2-x)*log(x+lo 
g(x))*log(3/log(x+log(x)))^2+(3*x+3*log(x))*log(x+log(x))*log(3/log(x+log( 
x)))+3*x+3)/(x+log(x))/log(x+log(x))/log(3/log(x+log(x)))^2,x, algorithm=" 
maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.20 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=-\frac {x^{2} \log \left (3\right ) + x e^{x} \log \left (3\right ) - x^{2} \log \left (\log \left (x + \log \left (x\right )\right )\right ) - x e^{x} \log \left (\log \left (x + \log \left (x\right )\right )\right ) + x \log \left (3\right ) - x \log \left (\log \left (x + \log \left (x\right )\right )\right ) - 3 \, x}{\log \left (3\right ) - \log \left (\log \left (x + \log \left (x\right )\right )\right )} \] Input:

integrate(((((-1-x)*exp(x)-2*x-1)*log(x)+(-x^2-x)*exp(x)-2*x^2-x)*log(x+lo 
g(x))*log(3/log(x+log(x)))^2+(3*x+3*log(x))*log(x+log(x))*log(3/log(x+log( 
x)))+3*x+3)/(x+log(x))/log(x+log(x))/log(3/log(x+log(x)))^2,x, algorithm=" 
giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 8.41 (sec) , antiderivative size = 204, normalized size of antiderivative = 6.80 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=\frac {3\,x+\frac {3\,x\,\ln \left (x+\ln \left (x\right )\right )\,\ln \left (\frac {3}{\ln \left (x+\ln \left (x\right )\right )}\right )\,\left (x+\ln \left (x\right )\right )}{x+1}}{\ln \left (\frac {3}{\ln \left (x+\ln \left (x\right )\right )}\right )}-\ln \left (x+\ln \left (x\right )\right )\,\left (\ln \left (x\right )\,\left (\frac {3\,\left (x^3+2\,x^2+x\right )}{x\,{\left (x+1\right )}^2}-\frac {3\,x^2+3\,x}{x\,{\left (x+1\right )}^2}\right )-\frac {3\,x^4+12\,x^3+12\,x^2+3\,x}{x\,{\left (x+1\right )}^2}+\frac {6\,x^4+12\,x^3+6\,x^2}{x\,{\left (x+1\right )}^2}+\frac {3\,\left (x^3+3\,x^2+2\,x\right )}{x\,{\left (x+1\right )}^2}-\frac {3\,x^2+3\,x}{x\,{\left (x+1\right )}^2}\right )-x-x\,{\mathrm {e}}^x-x^2 \] Input:

int((3*x + log(x + log(x))*log(3/log(x + log(x)))*(3*x + 3*log(x)) - log(x 
 + log(x))*log(3/log(x + log(x)))^2*(x + log(x)*(2*x + exp(x)*(x + 1) + 1) 
 + 2*x^2 + exp(x)*(x + x^2)) + 3)/(log(x + log(x))*log(3/log(x + log(x)))^ 
2*(x + log(x))),x)
 

Output:

(3*x + (3*x*log(x + log(x))*log(3/log(x + log(x)))*(x + log(x)))/(x + 1))/ 
log(3/log(x + log(x))) - log(x + log(x))*(log(x)*((3*(x + 2*x^2 + x^3))/(x 
*(x + 1)^2) - (3*x + 3*x^2)/(x*(x + 1)^2)) - (3*x + 12*x^2 + 12*x^3 + 3*x^ 
4)/(x*(x + 1)^2) + (6*x^2 + 12*x^3 + 6*x^4)/(x*(x + 1)^2) + (3*(2*x + 3*x^ 
2 + x^3))/(x*(x + 1)^2) - (3*x + 3*x^2)/(x*(x + 1)^2)) - x - x*exp(x) - x^ 
2
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {3+3 x+(3 x+3 \log (x)) \log (x+\log (x)) \log \left (\frac {3}{\log (x+\log (x))}\right )+\left (-x-2 x^2+e^x \left (-x-x^2\right )+\left (-1+e^x (-1-x)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )}{(x+\log (x)) \log (x+\log (x)) \log ^2\left (\frac {3}{\log (x+\log (x))}\right )} \, dx=\frac {x \left (-e^{x} \mathrm {log}\left (\frac {3}{\mathrm {log}\left (\mathrm {log}\left (x \right )+x \right )}\right )-\mathrm {log}\left (\frac {3}{\mathrm {log}\left (\mathrm {log}\left (x \right )+x \right )}\right ) x -\mathrm {log}\left (\frac {3}{\mathrm {log}\left (\mathrm {log}\left (x \right )+x \right )}\right )+3\right )}{\mathrm {log}\left (\frac {3}{\mathrm {log}\left (\mathrm {log}\left (x \right )+x \right )}\right )} \] Input:

int(((((-1-x)*exp(x)-2*x-1)*log(x)+(-x^2-x)*exp(x)-2*x^2-x)*log(x+log(x))* 
log(3/log(x+log(x)))^2+(3*x+3*log(x))*log(x+log(x))*log(3/log(x+log(x)))+3 
*x+3)/(x+log(x))/log(x+log(x))/log(3/log(x+log(x)))^2,x)
 

Output:

(x*( - e**x*log(3/log(log(x) + x)) - log(3/log(log(x) + x))*x - log(3/log( 
log(x) + x)) + 3))/log(3/log(log(x) + x))