\(\int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} (-500 x-200 x \log (3)-20 x \log ^2(3))+(-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} (-250-100 \log (3)-10 \log ^2(3))) \log (6+e^{e^x}-x)}{(6 x^2+e^{e^x} x^2-x^3) \log ^3(6+e^{e^x}-x)} \, dx\) [1783]

Optimal result

Integrand size = 129, antiderivative size = 24 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 (5+\log (3))^2}{x \log ^2\left (6+e^{e^x}-x\right )} \] Output:

10/ln(exp(exp(x))-x+6)^2*(5+ln(3))^2/x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 (5+\log (3))^2}{x \log ^2\left (6+e^{e^x}-x\right )} \] Input:

Integrate[(500*x + 200*x*Log[3] + 20*x*Log[3]^2 + E^(E^x + x)*(-500*x - 20 
0*x*Log[3] - 20*x*Log[3]^2) + (-1500 + 250*x + (-600 + 100*x)*Log[3] + (-6 
0 + 10*x)*Log[3]^2 + E^E^x*(-250 - 100*Log[3] - 10*Log[3]^2))*Log[6 + E^E^ 
x - x])/((6*x^2 + E^E^x*x^2 - x^3)*Log[6 + E^E^x - x]^3),x]
 

Output:

(10*(5 + Log[3])^2)/(x*Log[6 + E^E^x - x]^2)
 

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.

\(\displaystyle \int \frac {500 x+20 x \log ^2(3)+e^{x+e^x} \left (-500 x-20 x \log ^2(3)-200 x \log (3)\right )+\left (250 x+e^{e^x} \left (-250-10 \log ^2(3)-100 \log (3)\right )+(10 x-60) \log ^2(3)+(100 x-600) \log (3)-1500\right ) \log \left (-x+e^{e^x}+6\right )+200 x \log (3)}{\left (-x^3+e^{e^x} x^2+6 x^2\right ) \log ^3\left (-x+e^{e^x}+6\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {20 x \log ^2(3)+e^{x+e^x} \left (-500 x-20 x \log ^2(3)-200 x \log (3)\right )+\left (250 x+e^{e^x} \left (-250-10 \log ^2(3)-100 \log (3)\right )+(10 x-60) \log ^2(3)+(100 x-600) \log (3)-1500\right ) \log \left (-x+e^{e^x}+6\right )+x (500+200 \log (3))}{\left (-x^3+e^{e^x} x^2+6 x^2\right ) \log ^3\left (-x+e^{e^x}+6\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {x \left (500+20 \log ^2(3)+200 \log (3)\right )+e^{x+e^x} \left (-500 x-20 x \log ^2(3)-200 x \log (3)\right )+\left (250 x+e^{e^x} \left (-250-10 \log ^2(3)-100 \log (3)\right )+(10 x-60) \log ^2(3)+(100 x-600) \log (3)-1500\right ) \log \left (-x+e^{e^x}+6\right )}{\left (-x^3+e^{e^x} x^2+6 x^2\right ) \log ^3\left (-x+e^{e^x}+6\right )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {10 (5+\log (3))^2 \left (-2 \left (e^{x+e^x}-1\right ) x-\left (-x+e^{e^x}+6\right ) \log \left (-x+e^{e^x}+6\right )\right )}{\left (-x+e^{e^x}+6\right ) x^2 \log ^3\left (-x+e^{e^x}+6\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 10 (5+\log (3))^2 \int \frac {2 \left (1-e^{x+e^x}\right ) x-\left (-x+e^{e^x}+6\right ) \log \left (-x+e^{e^x}+6\right )}{\left (-x+e^{e^x}+6\right ) x^2 \log ^3\left (-x+e^{e^x}+6\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 10 (5+\log (3))^2 \int \left (-\frac {-\log \left (-x+e^{e^x}+6\right ) x-2 x+e^{e^x} \log \left (-x+e^{e^x}+6\right )+6 \log \left (-x+e^{e^x}+6\right )}{\left (-x+e^{e^x}+6\right ) x^2 \log ^3\left (-x+e^{e^x}+6\right )}-\frac {2 e^{x+e^x}}{\left (-x+e^{e^x}+6\right ) x \log ^3\left (-x+e^{e^x}+6\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 10 (5+\log (3))^2 \left (-\int \frac {1}{x^2 \log ^2\left (-x+e^{e^x}+6\right )}dx+2 \int \frac {1}{\left (-x+e^{e^x}+6\right ) x \log ^3\left (-x+e^{e^x}+6\right )}dx-2 \int \frac {e^{x+e^x}}{\left (-x+e^{e^x}+6\right ) x \log ^3\left (-x+e^{e^x}+6\right )}dx\right )\)

Input:

Int[(500*x + 200*x*Log[3] + 20*x*Log[3]^2 + E^(E^x + x)*(-500*x - 200*x*Lo 
g[3] - 20*x*Log[3]^2) + (-1500 + 250*x + (-600 + 100*x)*Log[3] + (-60 + 10 
*x)*Log[3]^2 + E^E^x*(-250 - 100*Log[3] - 10*Log[3]^2))*Log[6 + E^E^x - x] 
)/((6*x^2 + E^E^x*x^2 - x^3)*Log[6 + E^E^x - x]^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 5.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {10 \ln \left (3\right )^{2}+100 \ln \left (3\right )+250}{\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-x +6\right )^{2} x}\)	\(27\)
parallelrisch	\(\frac {10 \ln \left (3\right )^{2}+100 \ln \left (3\right )+250}{\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-x +6\right )^{2} x}\)	\(28\)

Input:

int((((-10*ln(3)^2-100*ln(3)-250)*exp(exp(x))+(10*x-60)*ln(3)^2+(100*x-600 
)*ln(3)+250*x-1500)*ln(exp(exp(x))-x+6)+(-20*x*ln(3)^2-200*x*ln(3)-500*x)* 
exp(x)*exp(exp(x))+20*x*ln(3)^2+200*x*ln(3)+500*x)/(exp(exp(x))*x^2-x^3+6* 
x^2)/ln(exp(exp(x))-x+6)^3,x,method=_RETURNVERBOSE)
 

Output:

10*(ln(3)^2+10*ln(3)+25)/x/ln(exp(exp(x))-x+6)^2
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 \, {\left (\log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25\right )}}{x \log \left (-{\left ({\left (x - 6\right )} e^{x} - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )^{2}} \] Input:

integrate((((-10*log(3)^2-100*log(3)-250)*exp(exp(x))+(10*x-60)*log(3)^2+( 
100*x-600)*log(3)+250*x-1500)*log(exp(exp(x))-x+6)+(-20*x*log(3)^2-200*x*l 
og(3)-500*x)*exp(x)*exp(exp(x))+20*x*log(3)^2+200*x*log(3)+500*x)/(exp(exp 
(x))*x^2-x^3+6*x^2)/log(exp(exp(x))-x+6)^3,x, algorithm="fricas")
 

Output:

10*(log(3)^2 + 10*log(3) + 25)/(x*log(-((x - 6)*e^x - e^(x + e^x))*e^(-x)) 
^2)
 

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 \log {\left (3 \right )}^{2} + 100 \log {\left (3 \right )} + 250}{x \log {\left (- x + e^{e^{x}} + 6 \right )}^{2}} \] Input:

integrate((((-10*ln(3)**2-100*ln(3)-250)*exp(exp(x))+(10*x-60)*ln(3)**2+(1 
00*x-600)*ln(3)+250*x-1500)*ln(exp(exp(x))-x+6)+(-20*x*ln(3)**2-200*x*ln(3 
)-500*x)*exp(x)*exp(exp(x))+20*x*ln(3)**2+200*x*ln(3)+500*x)/(exp(exp(x))* 
x**2-x**3+6*x**2)/ln(exp(exp(x))-x+6)**3,x)
 

Output:

(10*log(3)**2 + 100*log(3) + 250)/(x*log(-x + exp(exp(x)) + 6)**2)
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 \, {\left (\log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25\right )}}{x \log \left (-x + e^{\left (e^{x}\right )} + 6\right )^{2}} \] Input:

integrate((((-10*log(3)^2-100*log(3)-250)*exp(exp(x))+(10*x-60)*log(3)^2+( 
100*x-600)*log(3)+250*x-1500)*log(exp(exp(x))-x+6)+(-20*x*log(3)^2-200*x*l 
og(3)-500*x)*exp(x)*exp(exp(x))+20*x*log(3)^2+200*x*log(3)+500*x)/(exp(exp 
(x))*x^2-x^3+6*x^2)/log(exp(exp(x))-x+6)^3,x, algorithm="maxima")
 

Output:

10*(log(3)^2 + 10*log(3) + 25)/(x*log(-x + e^(e^x) + 6)^2)
 

Giac [F]

\[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\int { -\frac {10 \, {\left (2 \, x \log \left (3\right )^{2} - 2 \, {\left (x \log \left (3\right )^{2} + 10 \, x \log \left (3\right ) + 25 \, x\right )} e^{\left (x + e^{x}\right )} + 20 \, x \log \left (3\right ) + {\left ({\left (x - 6\right )} \log \left (3\right )^{2} - {\left (\log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25\right )} e^{\left (e^{x}\right )} + 10 \, {\left (x - 6\right )} \log \left (3\right ) + 25 \, x - 150\right )} \log \left (-x + e^{\left (e^{x}\right )} + 6\right ) + 50 \, x\right )}}{{\left (x^{3} - x^{2} e^{\left (e^{x}\right )} - 6 \, x^{2}\right )} \log \left (-x + e^{\left (e^{x}\right )} + 6\right )^{3}} \,d x } \] Input:

integrate((((-10*log(3)^2-100*log(3)-250)*exp(exp(x))+(10*x-60)*log(3)^2+( 
100*x-600)*log(3)+250*x-1500)*log(exp(exp(x))-x+6)+(-20*x*log(3)^2-200*x*l 
og(3)-500*x)*exp(x)*exp(exp(x))+20*x*log(3)^2+200*x*log(3)+500*x)/(exp(exp 
(x))*x^2-x^3+6*x^2)/log(exp(exp(x))-x+6)^3,x, algorithm="giac")
 

Output:

integrate(-10*(2*x*log(3)^2 - 2*(x*log(3)^2 + 10*x*log(3) + 25*x)*e^(x + e 
^x) + 20*x*log(3) + ((x - 6)*log(3)^2 - (log(3)^2 + 10*log(3) + 25)*e^(e^x 
) + 10*(x - 6)*log(3) + 25*x - 150)*log(-x + e^(e^x) + 6) + 50*x)/((x^3 - 
x^2*e^(e^x) - 6*x^2)*log(-x + e^(e^x) + 6)^3), x)
 

Mupad [B] (verification not implemented)

Time = 4.58 (sec) , antiderivative size = 1482, normalized size of antiderivative = 61.75 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\text {Too large to display} \] Input:

int((500*x + 200*x*log(3) + 20*x*log(3)^2 + log(exp(exp(x)) - x + 6)*(250* 
x + log(3)*(100*x - 600) + log(3)^2*(10*x - 60) - exp(exp(x))*(100*log(3) 
+ 10*log(3)^2 + 250) - 1500) - exp(exp(x))*exp(x)*(500*x + 200*x*log(3) + 
20*x*log(3)^2))/(log(exp(exp(x)) - x + 6)^3*(x^2*exp(exp(x)) + 6*x^2 - x^3 
)),x)
 

Output:

((10*(log(3) + 5)^2)/x + (5*log(exp(exp(x)) - x + 6)*(log(3) + 5)^2*(exp(e 
xp(x)) - x + 6))/(x^2*(exp(x + exp(x)) - 1)))/log(exp(exp(x)) - x + 6)^2 - 
 ((5*(log(3) + 5)^2*(exp(exp(x)) - x + 6))/(x^2*(exp(x + exp(x)) - 1)) + ( 
5*log(exp(exp(x)) - x + 6)*(log(3) + 5)^2*(exp(exp(x)) - x + 6)*(x + 12*ex 
p(x + exp(x)) - 2*exp(exp(x)) + 2*exp(x + 2*exp(x)) + x*exp(x + 2*exp(x)) 
+ 6*x*exp(2*x + exp(x)) - x^2*exp(x + exp(x)) - x^2*exp(2*x + exp(x)) + 6* 
x*exp(x + exp(x)) - 12))/(x^3*(exp(x + exp(x)) - 1)^3))/log(exp(exp(x)) - 
x + 6) + (exp(-2*x)*(100*log(3) + x*(50*log(3) + 5*log(3)^2 + 125) + 10*lo 
g(3)^2 + 250))/x^3 + (5*exp(-x)*(100*x + 3*x^2*log(3)^2 + 600*x*exp(2*x) + 
 325*x^2*exp(x) - 50*x^3*exp(x) + 40*x*log(3) + 400*x^2*exp(2*x) + 150*x^2 
*exp(3*x) - 75*x^3*exp(2*x) - 25*x^3*exp(3*x) + 4*x*log(3)^2 + 30*x^2*log( 
3) + 700*x*exp(x) + 75*x^2 + 280*x*exp(x)*log(3) + 16*x^2*exp(2*x)*log(3)^ 
2 + 6*x^2*exp(3*x)*log(3)^2 - 3*x^3*exp(2*x)*log(3)^2 - x^3*exp(3*x)*log(3 
)^2 + 240*x*exp(2*x)*log(3) + 28*x*exp(x)*log(3)^2 + 130*x^2*exp(x)*log(3) 
 - 20*x^3*exp(x)*log(3) + 24*x*exp(2*x)*log(3)^2 + 160*x^2*exp(2*x)*log(3) 
 + 60*x^2*exp(3*x)*log(3) - 30*x^3*exp(2*x)*log(3) - 10*x^3*exp(3*x)*log(3 
) + 13*x^2*exp(x)*log(3)^2 - 2*x^3*exp(x)*log(3)^2))/(x^4*(exp(2*x) + exp( 
x))*(exp(x + exp(x)) - 1)) + (5*exp(-x)*(50*x + 3*x^2*log(3)^2 + 2400*x*ex 
p(2*x) + 1800*x*exp(3*x) + 650*x^2*exp(x) - 100*x^3*exp(x) + 20*x*log(3) + 
 1475*x^2*exp(2*x) + 1800*x^2*exp(3*x) - 450*x^3*exp(2*x) + 900*x^2*exp...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx=\frac {10 \mathrm {log}\left (3\right )^{2}+100 \,\mathrm {log}\left (3\right )+250}{\mathrm {log}\left (e^{e^{x}}-x +6\right )^{2} x} \] Input:

int((((-10*log(3)^2-100*log(3)-250)*exp(exp(x))+(10*x-60)*log(3)^2+(100*x- 
600)*log(3)+250*x-1500)*log(exp(exp(x))-x+6)+(-20*x*log(3)^2-200*x*log(3)- 
500*x)*exp(x)*exp(exp(x))+20*x*log(3)^2+200*x*log(3)+500*x)/(exp(exp(x))*x 
^2-x^3+6*x^2)/log(exp(exp(x))-x+6)^3,x)
 

Output:

(10*(log(3)**2 + 10*log(3) + 25))/(log(e**(e**x) - x + 6)**2*x)