\(\int \frac {-10+2 e^3-2 x+e^x (-50+e^3 (10-10 x)+35 x+10 x^2)+e^x (10-7 x-2 x^2+e^3 (-2+2 x)) \log (x)+(-50+10 e^3-15 x+(10-2 e^3+3 x) \log (x)) \log (5-\log (x))}{e^x (25 x-5 e^3 x+5 x^2)+e^x (-5 x+e^3 x-x^2) \log (x)+(25 x-5 e^3 x+5 x^2+(-5 x+e^3 x-x^2) \log (x)) \log (5-\log (x))} \, dx\) [564]

Optimal result

Integrand size = 172, antiderivative size = 28 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=\log \left (\frac {\left (e^x+\log (5-\log (x))\right )^2}{\left (-5+e^3-x\right ) x^2}\right ) \] Output:

ln(1/x^2/(exp(3)-5-x)*(exp(x)+ln(5-ln(x)))^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=-2 \log (x)-\log \left (5-e^3+x\right )+2 \log \left (e^x+\log (5-\log (x))\right ) \] Input:

Integrate[(-10 + 2*E^3 - 2*x + E^x*(-50 + E^3*(10 - 10*x) + 35*x + 10*x^2) 
 + E^x*(10 - 7*x - 2*x^2 + E^3*(-2 + 2*x))*Log[x] + (-50 + 10*E^3 - 15*x + 
 (10 - 2*E^3 + 3*x)*Log[x])*Log[5 - Log[x]])/(E^x*(25*x - 5*E^3*x + 5*x^2) 
 + E^x*(-5*x + E^3*x - x^2)*Log[x] + (25*x - 5*E^3*x + 5*x^2 + (-5*x + E^3 
*x - x^2)*Log[x])*Log[5 - Log[x]]),x]
 

Output:

-2*Log[x] - Log[5 - E^3 + x] + 2*Log[E^x + Log[5 - 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.

\(\displaystyle \int \frac {e^x \left (10 x^2+35 x+e^3 (10-10 x)-50\right )+e^x \left (-2 x^2-7 x+e^3 (2 x-2)+10\right ) \log (x)-2 x+\left (-15 x+\left (3 x-2 e^3+10\right ) \log (x)+10 e^3-50\right ) \log (5-\log (x))+2 e^3-10}{e^x \left (5 x^2-5 e^3 x+25 x\right )+e^x \left (-x^2+e^3 x-5 x\right ) \log (x)+\left (5 x^2+\left (-x^2+e^3 x-5 x\right ) \log (x)-5 e^3 x+25 x\right ) \log (5-\log (x))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (10 x^2+35 x+e^3 (10-10 x)-50\right )+e^x \left (-2 x^2-7 x+e^3 (2 x-2)+10\right ) \log (x)-2 x+\left (-15 x+\left (3 x-2 e^3+10\right ) \log (x)+10 e^3-50\right ) \log (5-\log (x))-10 \left (1-\frac {e^3}{5}\right )}{x \left (x-e^3+5\right ) (5-\log (x)) \left (e^x+\log (5-\log (x))\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x^2+\left (7-2 e^3\right ) x-2 \left (5-e^3\right )}{x \left (x-e^3+5\right )}-\frac {2 (-5 x \log (5-\log (x))+x \log (x) \log (5-\log (x))-1)}{x (\log (x)-5) \left (e^x+\log (5-\log (x))\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \frac {1}{x (\log (x)-5) \left (\log (5-\log (x))+e^x\right )}dx+10 \int \frac {\log (5-\log (x))}{(\log (x)-5) \left (\log (5-\log (x))+e^x\right )}dx-2 \int \frac {\log (x) \log (5-\log (x))}{(\log (x)-5) \left (\log (5-\log (x))+e^x\right )}dx+2 x-2 \log (x)-\log \left (x-e^3+5\right )\)

Input:

Int[(-10 + 2*E^3 - 2*x + E^x*(-50 + E^3*(10 - 10*x) + 35*x + 10*x^2) + E^x 
*(10 - 7*x - 2*x^2 + E^3*(-2 + 2*x))*Log[x] + (-50 + 10*E^3 - 15*x + (10 - 
 2*E^3 + 3*x)*Log[x])*Log[5 - Log[x]])/(E^x*(25*x - 5*E^3*x + 5*x^2) + E^x 
*(-5*x + E^3*x - x^2)*Log[x] + (25*x - 5*E^3*x + 5*x^2 + (-5*x + E^3*x - x 
^2)*Log[x])*Log[5 - Log[x]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 7.96 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(-2 \ln \left (x \right )-\ln \left (-{\mathrm e}^{3}+x +5\right )+2 \ln \left ({\mathrm e}^{x}+\ln \left (5-\ln \left (x \right )\right )\right )\)	\(29\)
parallelrisch	\(-20+2 \ln \left ({\mathrm e}^{x}+\ln \left (5-\ln \left (x \right )\right )\right )-\ln \left (-{\mathrm e}^{3}+x +5\right )-2 \ln \left (x \right )\)	\(30\)

Input:

int((((-2*exp(3)+3*x+10)*ln(x)+10*exp(3)-15*x-50)*ln(5-ln(x))+((-2+2*x)*ex 
p(3)-2*x^2-7*x+10)*exp(x)*ln(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50)*exp(x)+ 
2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*ln(x)-5*x*exp(3)+5*x^2+25*x)*ln(5-ln 
(x))+(x*exp(3)-x^2-5*x)*exp(x)*ln(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x)),x,me 
thod=_RETURNVERBOSE)
 

Output:

-2*ln(x)-ln(-exp(3)+x+5)+2*ln(exp(x)+ln(5-ln(x)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=-\log \left (x - e^{3} + 5\right ) - 2 \, \log \left (x\right ) + 2 \, \log \left (e^{x} + \log \left (-\log \left (x\right ) + 5\right )\right ) \] Input:

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2 
*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*log(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50 
)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+25* 
x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x) 
*exp(x)),x, algorithm="fricas")
 

Output:

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))
 

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=- 2 \log {\left (x \right )} + 2 \log {\left (e^{x} + \log {\left (5 - \log {\left (x \right )} \right )} \right )} - \log {\left (x - e^{3} + 5 \right )} \] Input:

integrate((((-2*exp(3)+3*x+10)*ln(x)+10*exp(3)-15*x-50)*ln(5-ln(x))+((2*x- 
2)*exp(3)-2*x**2-7*x+10)*exp(x)*ln(x)+((-10*x+10)*exp(3)+10*x**2+35*x-50)* 
exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x**2-5*x)*ln(x)-5*x*exp(3)+5*x**2+25*x 
)*ln(5-ln(x))+(x*exp(3)-x**2-5*x)*exp(x)*ln(x)+(-5*x*exp(3)+5*x**2+25*x)*e 
xp(x)),x)
 

Output:

-2*log(x) + 2*log(exp(x) + log(5 - log(x))) - log(x - exp(3) + 5)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=-\log \left (x - e^{3} + 5\right ) - 2 \, \log \left (x\right ) + 2 \, \log \left (e^{x} + \log \left (-\log \left (x\right ) + 5\right )\right ) \] Input:

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2 
*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*log(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50 
)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+25* 
x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x) 
*exp(x)),x, algorithm="maxima")
 

Output:

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=-\log \left (x - e^{3} + 5\right ) - 2 \, \log \left (x\right ) + 2 \, \log \left (e^{x} + \log \left (-\log \left (x\right ) + 5\right )\right ) \] Input:

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2 
*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*log(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50 
)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+25* 
x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x) 
*exp(x)),x, algorithm="giac")
 

Output:

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))
 

Mupad [B] (verification not implemented)

Time = 4.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=2\,\ln \left (\ln \left (5-\ln \left (x\right )\right )+{\mathrm {e}}^x\right )-\ln \left (x-{\mathrm {e}}^3+5\right )-2\,\ln \left (x\right ) \] Input:

int(-(2*x - 2*exp(3) - exp(x)*(35*x + 10*x^2 - exp(3)*(10*x - 10) - 50) + 
log(5 - log(x))*(15*x - 10*exp(3) - log(x)*(3*x - 2*exp(3) + 10) + 50) + e 
xp(x)*log(x)*(7*x + 2*x^2 - exp(3)*(2*x - 2) - 10) + 10)/(log(5 - log(x))* 
(25*x - 5*x*exp(3) - log(x)*(5*x - x*exp(3) + x^2) + 5*x^2) + exp(x)*(25*x 
 - 5*x*exp(3) + 5*x^2) - exp(x)*log(x)*(5*x - x*exp(3) + x^2)),x)
 

Output:

2*log(log(5 - log(x)) + exp(x)) - log(x - exp(3) + 5) - 2*log(x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx=2 \,\mathrm {log}\left (e^{x}+\mathrm {log}\left (-\mathrm {log}\left (x \right )+5\right )\right )-\mathrm {log}\left (e^{3}-x -5\right )-2 \,\mathrm {log}\left (x \right ) \] Input:

int((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2*x-2)* 
exp(3)-2*x^2-7*x+10)*exp(x)*log(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50)*exp( 
x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+25*x)*log 
(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x 
)),x)
 

Output:

2*log(e**x + log( - log(x) + 5)) - log(e**3 - x - 5) - 2*log(x)