\(\int \frac {e^x (10+e+2 x)+(40+5 e+e^x (-8-e-2 x)+10 x) \log (5-e^x) \log (\log (5-e^x))}{(-10 e^x+2 e^{2 x}) \log (5-e^x)} \, dx\) [684]

Optimal result

Integrand size = 75, antiderivative size = 23 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=e^{-x} \left (5+\frac {e}{2}+x\right ) \log \left (\log \left (5-e^x\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} e^{-x} (10+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \] Input:

Integrate[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log 
[5 - E^x]*Log[Log[5 - E^x]])/((-10*E^x + 2*E^(2*x))*Log[5 - E^x]),x]
 

Output:

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

\(\displaystyle \int \frac {e^x (2 x+e+10)+\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (2 e^{2 x}-10 e^x\right ) \log \left (5-e^x\right )} \, dx\)

\(\Big \downarrow \) 2721

\(\displaystyle \int \frac {e^{-x} \left (e^x (2 x+e+10)+\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (2 e^x-10\right ) \log \left (5-e^x\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-x} \left (-e^x (2 x+e+10)-\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{2 \left (5-e^x\right ) \log \left (5-e^x\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {e^{-x} \left (e^x (2 x+e+10)+\left (10 x-e^x (2 x+e+8)+5 (8+e)\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (5-e^x\right ) \log \left (5-e^x\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {e^{-x} \left (e^x (2 x+e+10)+\left (10 x-e^x (2 x+e+8)+5 (8+e)\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (5-e^x\right ) \log \left (5-e^x\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (\frac {e^{-x} \left (2 \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right ) x-2 x+8 \left (1+\frac {e}{8}\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )-10 \left (1+\frac {e}{10}\right )\right )}{\log \left (5-e^x\right )}-\frac {5 e^{-x} (2 x+e+10)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {1}{5} (10+e) \text {Subst}\left (\int \frac {1}{x \log (5-x)}dx,x,e^x\right )+2 \int \frac {e^{-x} x}{\log \left (5-e^x\right )}dx+10 \int \frac {e^{-x} x}{\left (-5+e^x\right ) \log \left (5-e^x\right )}dx-(8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right )dx-2 \int e^{-x} x \log \left (\log \left (5-e^x\right )\right )dx+\frac {1}{5} (10+e) \log \left (\log \left (5-e^x\right )\right )\right )\)

Input:

Int[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log[5 - E 
^x]*Log[Log[5 - E^x]])/((-10*E^x + 2*E^(2*x))*Log[5 - E^x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left ({\mathrm e}+2 x +10\right ) {\mathrm e}^{-x} \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )}{2}\)	\(22\)
parallelrisch	\(\frac {\left (\ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right ) {\mathrm e}+2 x \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )+10 \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )\right ) {\mathrm e}^{-x}}{2}\)	\(40\)

Input:

int((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5-exp(x) 
))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/ln(5-exp(x)),x,method=_R 
ETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \] Input:

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log 
(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x 
, algorithm="fricas")
 

Output:

1/2*(2*x + e + 10)*e^(-x)*log(log(-e^x + 5))
 

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {\left (2 x + e + 10\right ) e^{- x} \log {\left (\log {\left (5 - e^{x} \right )} \right )}}{2} \] Input:

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5- 
exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)**2-10*exp(x))/ln(5-exp(x)),x)
 

Output:

(2*x + E + 10)*exp(-x)*log(log(5 - exp(x)))/2
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \] Input:

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log 
(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x 
, algorithm="maxima")
 

Output:

1/2*(2*x + e + 10)*e^(-x)*log(log(-e^x + 5))
 

Giac [F]

\[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\int { -\frac {{\left ({\left (2 \, x + e + 8\right )} e^{x} - 10 \, x - 5 \, e - 40\right )} \log \left (-e^{x} + 5\right ) \log \left (\log \left (-e^{x} + 5\right )\right ) - {\left (2 \, x + e + 10\right )} e^{x}}{2 \, {\left (e^{\left (2 \, x\right )} - 5 \, e^{x}\right )} \log \left (-e^{x} + 5\right )} \,d x } \] Input:

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log 
(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x 
, algorithm="giac")
 

Output:

integrate(-1/2*(((2*x + e + 8)*e^x - 10*x - 5*e - 40)*log(-e^x + 5)*log(lo 
g(-e^x + 5)) - (2*x + e + 10)*e^x)/((e^(2*x) - 5*e^x)*log(-e^x + 5)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\int \frac {{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+10\right )+\ln \left (5-{\mathrm {e}}^x\right )\,\ln \left (\ln \left (5-{\mathrm {e}}^x\right )\right )\,\left (10\,x+5\,\mathrm {e}-{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+8\right )+40\right )}{\ln \left (5-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x\right )} \,d x \] Input:

int((exp(x)*(2*x + exp(1) + 10) + log(5 - exp(x))*log(log(5 - exp(x)))*(10 
*x + 5*exp(1) - exp(x)*(2*x + exp(1) + 8) + 40))/(log(5 - exp(x))*(2*exp(2 
*x) - 10*exp(x))),x)
 

Output:

int((exp(x)*(2*x + exp(1) + 10) + log(5 - exp(x))*log(log(5 - exp(x)))*(10 
*x + 5*exp(1) - exp(x)*(2*x + exp(1) + 8) + 40))/(log(5 - exp(x))*(2*exp(2 
*x) - 10*exp(x))), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (-e^{x}+5\right )\right ) \left (e +2 x +10\right )}{2 e^{x}} \] Input:

int((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log(5-exp 
(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x)
 

Output:

(log(log( - e**x + 5))*(e + 2*x + 10))/(2*e**x)