\(\int \frac {-2-2 e^x+(1+e^x (1-x)) \log (x)+(1+e^x+(-1-e^x) \log (x)) \log (\frac {5}{x+e^x x})}{(-10 x-10 e^x x) \log (x)+(5 x+5 e^x x) \log (x) \log (\frac {5}{x+e^x x})+((-2 x-2 e^x x) \log (x)+(x+e^x x) \log (x) \log (\frac {5}{x+e^x x})) \log (\frac {2 \log (x)-\log (x) \log (\frac {5}{x+e^x x})}{x})} \, dx\) [2991]

Optimal result

Integrand size = 154, antiderivative size = 26 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log \left (5+\log \left (\frac {\log (x) \left (2-\log \left (\frac {5}{x+e^x x}\right )\right )}{x}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log \left (-5-\log \left (-\frac {\log (x) \left (-2+\log \left (\frac {5}{x+e^x x}\right )\right )}{x}\right )\right ) \] Input:

Integrate[(-2 - 2*E^x + (1 + E^x*(1 - x))*Log[x] + (1 + E^x + (-1 - E^x)*L 
og[x])*Log[5/(x + E^x*x)])/((-10*x - 10*E^x*x)*Log[x] + (5*x + 5*E^x*x)*Lo 
g[x]*Log[5/(x + E^x*x)] + ((-2*x - 2*E^x*x)*Log[x] + (x + E^x*x)*Log[x]*Lo 
g[5/(x + E^x*x)])*Log[(2*Log[x] - Log[x]*Log[5/(x + E^x*x)])/x]),x]
 

Output:

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

\(\displaystyle \int \frac {-2 e^x+\left (e^x (1-x)+1\right ) \log (x)+\left (e^x+\left (-e^x-1\right ) \log (x)+1\right ) \log \left (\frac {5}{e^x x+x}\right )-2}{\left (-10 e^x x-10 x\right ) \log (x)+\left (5 e^x x+5 x\right ) \log \left (\frac {5}{e^x x+x}\right ) \log (x)+\left (\left (-2 e^x x-2 x\right ) \log (x)+\left (e^x x+x\right ) \log \left (\frac {5}{e^x x+x}\right ) \log (x)\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{e^x x+x}\right )}{x}\right )} \, dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-2 - 2*E^x + (1 + E^x*(1 - x))*Log[x] + (1 + E^x + (-1 - E^x)*Log[x]) 
*Log[5/(x + E^x*x)])/((-10*x - 10*E^x*x)*Log[x] + (5*x + 5*E^x*x)*Log[x]*L 
og[5/(x + E^x*x)] + ((-2*x - 2*E^x*x)*Log[x] + (x + E^x*x)*Log[x]*Log[5/(x 
 + E^x*x)])*Log[(2*Log[x] - Log[x]*Log[5/(x + E^x*x)])/x]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.33 (sec) , antiderivative size = 15838, normalized size of antiderivative = 609.15

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

Input:

int((((-exp(x)-1)*ln(x)+exp(x)+1)*ln(5/(exp(x)*x+x))+((1-x)*exp(x)+1)*ln(x 
)-2*exp(x)-2)/(((exp(x)*x+x)*ln(x)*ln(5/(exp(x)*x+x))+(-2*exp(x)*x-2*x)*ln 
(x))*ln((-ln(x)*ln(5/(exp(x)*x+x))+2*ln(x))/x)+(5*exp(x)*x+5*x)*ln(x)*ln(5 
/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*ln(x)),x)
 

Output:

result too large to display
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log \left (\log \left (-\frac {\log \left (x\right ) \log \left (\frac {5}{x e^{x} + x}\right ) - 2 \, \log \left (x\right )}{x}\right ) + 5\right ) \] Input:

integrate((((-1-exp(x))*log(x)+exp(x)+1)*log(5/(exp(x)*x+x))+((1-x)*exp(x) 
+1)*log(x)-2*exp(x)-2)/(((exp(x)*x+x)*log(x)*log(5/(exp(x)*x+x))+(-2*exp(x 
)*x-2*x)*log(x))*log((-log(x)*log(5/(exp(x)*x+x))+2*log(x))/x)+(5*exp(x)*x 
+5*x)*log(x)*log(5/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*log(x)),x, algorithm= 
"fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 21.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log {\left (\log {\left (\frac {- \log {\left (x \right )} \log {\left (\frac {5}{x e^{x} + x} \right )} + 2 \log {\left (x \right )}}{x} \right )} + 5 \right )} \] Input:

integrate((((-1-exp(x))*ln(x)+exp(x)+1)*ln(5/(exp(x)*x+x))+((1-x)*exp(x)+1 
)*ln(x)-2*exp(x)-2)/(((exp(x)*x+x)*ln(x)*ln(5/(exp(x)*x+x))+(-2*exp(x)*x-2 
*x)*ln(x))*ln((-ln(x)*ln(5/(exp(x)*x+x))+2*ln(x))/x)+(5*exp(x)*x+5*x)*ln(x 
)*ln(5/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*ln(x)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log \left (-\log \left (x\right ) + \log \left (-\log \left (5\right ) + \log \left (x\right ) + \log \left (e^{x} + 1\right ) + 2\right ) + \log \left (\log \left (x\right )\right ) + 5\right ) \] Input:

integrate((((-1-exp(x))*log(x)+exp(x)+1)*log(5/(exp(x)*x+x))+((1-x)*exp(x) 
+1)*log(x)-2*exp(x)-2)/(((exp(x)*x+x)*log(x)*log(5/(exp(x)*x+x))+(-2*exp(x 
)*x-2*x)*log(x))*log((-log(x)*log(5/(exp(x)*x+x))+2*log(x))/x)+(5*exp(x)*x 
+5*x)*log(x)*log(5/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*log(x)),x, algorithm= 
"maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\log \left (\log \left (-\log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2} + \log \left (x\right ) \log \left (e^{x} + 1\right ) + 2 \, \log \left (x\right )\right ) - \log \left (x\right ) + 5\right ) \] Input:

integrate((((-1-exp(x))*log(x)+exp(x)+1)*log(5/(exp(x)*x+x))+((1-x)*exp(x) 
+1)*log(x)-2*exp(x)-2)/(((exp(x)*x+x)*log(x)*log(5/(exp(x)*x+x))+(-2*exp(x 
)*x-2*x)*log(x))*log((-log(x)*log(5/(exp(x)*x+x))+2*log(x))/x)+(5*exp(x)*x 
+5*x)*log(x)*log(5/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*log(x)),x, algorithm= 
"giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\ln \left (\ln \left (\frac {2\,\ln \left (x\right )-\ln \left (\frac {5}{x+x\,{\mathrm {e}}^x}\right )\,\ln \left (x\right )}{x}\right )+5\right ) \] Input:

int((2*exp(x) + log(x)*(exp(x)*(x - 1) - 1) - log(5/(x + x*exp(x)))*(exp(x 
) - log(x)*(exp(x) + 1) + 1) + 2)/(log(x)*(10*x + 10*x*exp(x)) + log((2*lo 
g(x) - log(5/(x + x*exp(x)))*log(x))/x)*(log(x)*(2*x + 2*x*exp(x)) - log(5 
/(x + x*exp(x)))*log(x)*(x + x*exp(x))) - log(5/(x + x*exp(x)))*log(x)*(5* 
x + 5*x*exp(x))),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2-2 e^x+\left (1+e^x (1-x)\right ) \log (x)+\left (1+e^x+\left (-1-e^x\right ) \log (x)\right ) \log \left (\frac {5}{x+e^x x}\right )}{\left (-10 x-10 e^x x\right ) \log (x)+\left (5 x+5 e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )+\left (\left (-2 x-2 e^x x\right ) \log (x)+\left (x+e^x x\right ) \log (x) \log \left (\frac {5}{x+e^x x}\right )\right ) \log \left (\frac {2 \log (x)-\log (x) \log \left (\frac {5}{x+e^x x}\right )}{x}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {-\mathrm {log}\left (\frac {5}{e^{x} x +x}\right ) \mathrm {log}\left (x \right )+2 \,\mathrm {log}\left (x \right )}{x}\right )+5\right ) \] Input:

int((((-1-exp(x))*log(x)+exp(x)+1)*log(5/(exp(x)*x+x))+((1-x)*exp(x)+1)*lo 
g(x)-2*exp(x)-2)/(((exp(x)*x+x)*log(x)*log(5/(exp(x)*x+x))+(-2*exp(x)*x-2* 
x)*log(x))*log((-log(x)*log(5/(exp(x)*x+x))+2*log(x))/x)+(5*exp(x)*x+5*x)* 
log(x)*log(5/(exp(x)*x+x))+(-10*exp(x)*x-10*x)*log(x)),x)
 

Output:

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