\(\int \frac {-125 x+25 e^x x+e^{4 x} (50 x-10 e^x x)+e^{8 x} (-5 x+e^x x)+(-50 x+10 e^x x+e^{4 x} (10 x-2 e^x x)) \log (5-e^x)+(-5 x+e^x x) \log ^2(5-e^x)+x^{\frac {4}{5-e^{4 x}+\log (5-e^x)}} (-200+40 e^x+e^{4 x} (40-8 e^x)+(-40+8 e^x) \log (5-e^x)+(-8 e^x x+e^{4 x} (-160 x+32 e^x x)) \log (x))}{-125 x+25 e^x x+e^{4 x} (50 x-10 e^x x)+e^{8 x} (-5 x+e^x x)+(-50 x+10 e^x x+e^{4 x} (10 x-2 e^x x)) \log (5-e^x)+(-5 x+e^x x) \log ^2(5-e^x)} \, dx\) [2561]

Optimal result

Integrand size = 281, antiderivative size = 27 \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=x+2 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=x+2 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \] Input:

Integrate[(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + 
 E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*x - 2*E^x*x))*Log[5 - E^x] + (-5 
*x + E^x*x)*Log[5 - E^x]^2 + x^(4/(5 - E^(4*x) + Log[5 - E^x]))*(-200 + 40 
*E^x + E^(4*x)*(40 - 8*E^x) + (-40 + 8*E^x)*Log[5 - E^x] + (-8*E^x*x + E^( 
4*x)*(-160*x + 32*E^x*x))*Log[x]))/(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10 
*E^x*x) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*x - 2*E 
^x*x))*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2),x]
 

Output:

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

\(\displaystyle \int \frac {\left (e^{4 x} \left (40-8 e^x\right )+40 e^x+\left (8 e^x-40\right ) \log \left (5-e^x\right )+\left (e^{4 x} \left (32 e^x x-160 x\right )-8 e^x x\right ) \log (x)-200\right ) x^{\frac {4}{-e^{4 x}+\log \left (5-e^x\right )+5}}+25 e^x x-125 x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (e^x x-5 x\right )+\left (e^x x-5 x\right ) \log ^2\left (5-e^x\right )+\left (10 e^x x-50 x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )}{25 e^x x-125 x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (e^x x-5 x\right )+\left (e^x x-5 x\right ) \log ^2\left (5-e^x\right )+\left (10 e^x x-50 x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left (\left (e^{4 x} \left (40-8 e^x\right )+40 e^x+\left (8 e^x-40\right ) \log \left (5-e^x\right )+\left (e^{4 x} \left (32 e^x x-160 x\right )-8 e^x x\right ) \log (x)-200\right ) x^{\frac {4}{-e^{4 x}+\log \left (5-e^x\right )+5}}\right )-25 e^x x+125 x-e^{4 x} \left (50 x-10 e^x x\right )-e^{8 x} \left (e^x x-5 x\right )-\left (e^x x-5 x\right ) \log ^2\left (5-e^x\right )-\left (10 e^x x-50 x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )}{\left (5-e^x\right ) x \left (-e^{4 x}+\log \left (5-e^x\right )+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {8 \left (5 e^x+5 e^{4 x}-e^{5 x}+e^x \log \left (5-e^x\right )-5 \log \left (5-e^x\right )-e^x x \log (x)-20 e^{4 x} x \log (x)+4 e^{5 x} x \log (x)-25\right ) x^{\frac {4}{-e^{4 x}+\log \left (5-e^x\right )+5}-1}}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {\log ^2\left (5-e^x\right )}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {2 \left (e^{4 x}-5\right ) \log \left (5-e^x\right )}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {10 e^{4 x}}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {e^{8 x}}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {25 e^x}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {125}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {8 \left (5 e^x+5 e^{4 x}-e^{5 x}+e^x \log \left (5-e^x\right )-5 \log \left (5-e^x\right )-e^x x \log (x)-20 e^{4 x} x \log (x)+4 e^{5 x} x \log (x)-25\right ) x^{\frac {4}{-e^{4 x}+\log \left (5-e^x\right )+5}-1}}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {\log ^2\left (5-e^x\right )}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {2 \left (e^{4 x}-5\right ) \log \left (5-e^x\right )}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {10 e^{4 x}}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {e^{8 x}}{\left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}+\frac {25 e^x}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}-\frac {125}{\left (e^x-5\right ) \left (e^{4 x}-\log \left (5-e^x\right )-5\right )^2}\right )dx\)

Input:

Int[(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + E^x*x 
) + (-50*x + 10*E^x*x + E^(4*x)*(10*x - 2*E^x*x))*Log[5 - E^x] + (-5*x + E 
^x*x)*Log[5 - E^x]^2 + x^(4/(5 - E^(4*x) + Log[5 - E^x]))*(-200 + 40*E^x + 
 E^(4*x)*(40 - 8*E^x) + (-40 + 8*E^x)*Log[5 - E^x] + (-8*E^x*x + E^(4*x)*( 
-160*x + 32*E^x*x))*Log[x]))/(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x 
) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*x - 2*E^x*x)) 
*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

\[x +2 x^{\frac {4}{\ln \left (5-{\mathrm e}^{x}\right )-{\mathrm e}^{4 x}+5}}\]

Input:

int((((8*exp(x)-40)*ln(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x) 
*ln(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*ln(x)/(ln(5-exp(x))-ex 
p(4*x)+5))+(exp(x)*x-5*x)*ln(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10*e 
xp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*e 
xp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*ln(5-exp(x))^2+((-2*exp(x)*x+10 
*x)*exp(4*x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10 
*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x)
 

Output:

x+2*x^(4/(ln(5-exp(x))-exp(4*x)+5))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=\frac {x x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}} + 2}{x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}} \] Input:

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*ex 
p(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*log(x)/(log(5- 
exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*ex 
p(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp( 
x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((- 
2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*e 
xp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="fr 
icas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=\text {Timed out} \] Input:

integrate((((8*exp(x)-40)*ln(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp 
(x)*x)*ln(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*ln(x)/(ln(5-exp( 
x))-exp(4*x)+5))+(exp(x)*x-5*x)*ln(5-exp(x))**2+((-2*exp(x)*x+10*x)*exp(4* 
x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)**2+(-10*exp(x)*x 
+50*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*ln(5-exp(x))**2+((-2*ex 
p(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4* 
x)**2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=\int { \frac {{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x - \frac {8 \, {\left ({\left (e^{x} - 5\right )} e^{\left (4 \, x\right )} - {\left (4 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - x e^{x}\right )} \log \left (x\right ) - {\left (e^{x} - 5\right )} \log \left (-e^{x} + 5\right ) - 5 \, e^{x} + 25\right )}}{x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}}}{{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x} \,d x } \] Input:

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*ex 
p(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*log(x)/(log(5- 
exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*ex 
p(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp( 
x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((- 
2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*e 
xp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="ma 
xima")
 

Output:

x + integrate(8*((4*x*log(x) - 1)*e^(5*x) - 5*(4*x*log(x) - 1)*e^(4*x) - ( 
x*log(x) - 5)*e^x + (e^x - 5)*log(-e^x + 5) - 25)/(((x*e^x - 5*x)*log(-e^x 
 + 5)^2 + x*e^(9*x) - 5*x*e^(8*x) - 10*x*e^(5*x) + 50*x*e^(4*x) + 25*x*e^x 
 - 2*(x*e^(5*x) - 5*x*e^(4*x) - 5*x*e^x + 25*x)*log(-e^x + 5) - 125*x)*x^( 
4/(e^(4*x) - log(-e^x + 5) - 5))), x)
 

Giac [F]

\[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=\int { \frac {{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x - \frac {8 \, {\left ({\left (e^{x} - 5\right )} e^{\left (4 \, x\right )} - {\left (4 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - x e^{x}\right )} \log \left (x\right ) - {\left (e^{x} - 5\right )} \log \left (-e^{x} + 5\right ) - 5 \, e^{x} + 25\right )}}{x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}}}{{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x} \,d x } \] Input:

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*ex 
p(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*log(x)/(log(5- 
exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*ex 
p(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp( 
x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((- 
2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*e 
xp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="gi 
ac")
 

Output:

integrate(((x*e^x - 5*x)*log(-e^x + 5)^2 + (x*e^x - 5*x)*e^(8*x) - 10*(x*e 
^x - 5*x)*e^(4*x) + 25*x*e^x - 2*((x*e^x - 5*x)*e^(4*x) - 5*x*e^x + 25*x)* 
log(-e^x + 5) - 125*x - 8*((e^x - 5)*e^(4*x) - (4*(x*e^x - 5*x)*e^(4*x) - 
x*e^x)*log(x) - (e^x - 5)*log(-e^x + 5) - 5*e^x + 25)/x^(4/(e^(4*x) - log( 
-e^x + 5) - 5)))/((x*e^x - 5*x)*log(-e^x + 5)^2 + (x*e^x - 5*x)*e^(8*x) - 
10*(x*e^x - 5*x)*e^(4*x) + 25*x*e^x - 2*((x*e^x - 5*x)*e^(4*x) - 5*x*e^x + 
 25*x)*log(-e^x + 5) - 125*x), x)
 

Mupad [B] (verification not implemented)

Time = 2.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=x+2\,x^{\frac {4}{\ln \left (5-{\mathrm {e}}^x\right )-{\mathrm {e}}^{4\,x}+5}} \] Input:

int((125*x + exp((4*log(x))/(log(5 - exp(x)) - exp(4*x) + 5))*(exp(4*x)*(8 
*exp(x) - 40) - log(5 - exp(x))*(8*exp(x) - 40) - 40*exp(x) + log(x)*(exp( 
4*x)*(160*x - 32*x*exp(x)) + 8*x*exp(x)) + 200) + exp(8*x)*(5*x - x*exp(x) 
) - exp(4*x)*(50*x - 10*x*exp(x)) + log(5 - exp(x))^2*(5*x - x*exp(x)) - 2 
5*x*exp(x) - log(5 - exp(x))*(exp(4*x)*(10*x - 2*x*exp(x)) - 50*x + 10*x*e 
xp(x)))/(125*x + exp(8*x)*(5*x - x*exp(x)) - exp(4*x)*(50*x - 10*x*exp(x)) 
 + log(5 - exp(x))^2*(5*x - x*exp(x)) - 25*x*exp(x) - log(5 - exp(x))*(exp 
(4*x)*(10*x - 2*x*exp(x)) - 50*x + 10*x*exp(x))),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx=\frac {e^{\frac {4 \,\mathrm {log}\left (x \right )}{e^{4 x}-\mathrm {log}\left (-e^{x}+5\right )-5}} x +2}{e^{\frac {4 \,\mathrm {log}\left (x \right )}{e^{4 x}-\mathrm {log}\left (-e^{x}+5\right )-5}}} \] Input:

int((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x 
)*log(x)+(-8*exp(x)+40)*exp(4*x)+40*exp(x)-200)*exp(4*log(x)/(log(5-exp(x) 
)-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x) 
+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+5 
0*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp( 
x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x 
)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x)
 

Output:

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