\(\int \frac {(80-40 x-2 x^2+x^3) \log (-2+x)+e^{3 e^x+x} (-60 x^2+30 x^3) \log (-2+x)+(-40 x-10 e^{3 e^x} x^2-x^3) \log (\frac {40+10 e^{3 e^x} x+x^2}{x})}{e^{3 e^x} (-20 x^2+10 x^3) \log ^2(-2+x)+(-80 x+40 x^2-2 x^3+x^4) \log ^2(-2+x)} \, dx\) [2523]

Optimal result

Integrand size = 139, antiderivative size = 27 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=-5+\frac {\log \left (10 \left (e^{3 e^x}+\frac {4}{x}\right )+x\right )}{\log (-2+x)} \] Output:

ln(10*exp(3*exp(x))+40/x+x)/ln(-2+x)-5
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{\log (-2+x)} \] Input:

Integrate[((80 - 40*x - 2*x^2 + x^3)*Log[-2 + x] + E^(3*E^x + x)*(-60*x^2 
+ 30*x^3)*Log[-2 + x] + (-40*x - 10*E^(3*E^x)*x^2 - x^3)*Log[(40 + 10*E^(3 
*E^x)*x + x^2)/x])/(E^(3*E^x)*(-20*x^2 + 10*x^3)*Log[-2 + x]^2 + (-80*x + 
40*x^2 - 2*x^3 + x^4)*Log[-2 + x]^2),x]
 

Output:

Log[10*E^(3*E^x) + 40/x + x]/Log[-2 + 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.

\(\displaystyle \int \frac {\left (x^3-2 x^2-40 x+80\right ) \log (x-2)+e^{x+3 e^x} \left (30 x^3-60 x^2\right ) \log (x-2)+\left (-x^3-10 e^{3 e^x} x^2-40 x\right ) \log \left (\frac {x^2+10 e^{3 e^x} x+40}{x}\right )}{e^{3 e^x} \left (10 x^3-20 x^2\right ) \log ^2(x-2)+\left (x^4-2 x^3+40 x^2-80 x\right ) \log ^2(x-2)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {\left (\left (30 e^{x+3 e^x}+1\right ) x^2-40\right ) \log (x-2)}{x \left (x^2+10 e^{3 e^x} x+40\right )}-\frac {\log \left (x+10 e^{3 e^x}+\frac {40}{x}\right )}{x-2}}{\log ^2(x-2)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {30 e^{x+3 e^x} x}{\left (x^2+10 e^{3 e^x} x+40\right ) \log (x-2)}+\frac {x^3 \log (x-2)-x^3 \log \left (x+10 e^{3 e^x}+\frac {40}{x}\right )-2 x^2 \log (x-2)-10 e^{3 e^x} x^2 \log \left (x+10 e^{3 e^x}+\frac {40}{x}\right )-40 x \log (x-2)-40 x \log \left (x+10 e^{3 e^x}+\frac {40}{x}\right )+80 \log (x-2)}{(x-2) \left (x^2+10 e^{3 e^x} x+40\right ) x \log ^2(x-2)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -40 \int \frac {1}{x \left (x^2+10 e^{3 e^x} x+40\right ) \log (x-2)}dx+\int \frac {x}{\left (x^2+10 e^{3 e^x} x+40\right ) \log (x-2)}dx+30 \int \frac {e^{x+3 e^x} x}{\left (x^2+10 e^{3 e^x} x+40\right ) \log (x-2)}dx-\int \frac {\log \left (x+10 e^{3 e^x}+\frac {40}{x}\right )}{(x-2) \log ^2(x-2)}dx\)

Input:

Int[((80 - 40*x - 2*x^2 + x^3)*Log[-2 + x] + E^(3*E^x + x)*(-60*x^2 + 30*x 
^3)*Log[-2 + x] + (-40*x - 10*E^(3*E^x)*x^2 - x^3)*Log[(40 + 10*E^(3*E^x)* 
x + x^2)/x])/(E^(3*E^x)*(-20*x^2 + 10*x^3)*Log[-2 + x]^2 + (-80*x + 40*x^2 
 - 2*x^3 + x^4)*Log[-2 + x]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.02 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.89

\[\frac {\ln \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{\ln \left (-2+x \right )}-\frac {-i \pi \,\operatorname {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) \operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{3}-i \pi {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \ln \left (x \right )}{2 \ln \left (-2+x \right )}\]

Input:

int(((-10*x^2*exp(3*exp(x))-x^3-40*x)*ln((10*x*exp(3*exp(x))+x^2+40)/x)+(3 
0*x^3-60*x^2)*exp(x)*ln(-2+x)*exp(3*exp(x))+(x^3-2*x^2-40*x+80)*ln(-2+x))/ 
((10*x^3-20*x^2)*ln(-2+x)^2*exp(3*exp(x))+(x^4-2*x^3+40*x^2-80*x)*ln(-2+x) 
^2),x)
 

Output:

1/ln(-2+x)*ln(10*x*exp(3*exp(x))+x^2+40)-1/2*(-I*Pi*csgn(I*(10*x*exp(3*exp 
(x))+x^2+40))*csgn(I/x*(10*x*exp(3*exp(x))+x^2+40))^2+I*Pi*csgn(I*(10*x*ex 
p(3*exp(x))+x^2+40))*csgn(I/x*(10*x*exp(3*exp(x))+x^2+40))*csgn(I/x)+I*Pi* 
csgn(I/x*(10*x*exp(3*exp(x))+x^2+40))^3-I*Pi*csgn(I/x*(10*x*exp(3*exp(x))+ 
x^2+40))^2*csgn(I/x)+2*ln(x))/ln(-2+x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (\frac {{\left (10 \, x e^{\left (x + 3 \, e^{x}\right )} + {\left (x^{2} + 40\right )} e^{x}\right )} e^{\left (-x\right )}}{x}\right )}{\log \left (x - 2\right )} \] Input:

integrate(((-10*x^2*exp(3*exp(x))-x^3-40*x)*log((10*x*exp(3*exp(x))+x^2+40 
)/x)+(30*x^3-60*x^2)*exp(x)*log(-2+x)*exp(3*exp(x))+(x^3-2*x^2-40*x+80)*lo 
g(-2+x))/((10*x^3-20*x^2)*log(-2+x)^2*exp(3*exp(x))+(x^4-2*x^3+40*x^2-80*x 
)*log(-2+x)^2),x, algorithm="fricas")
 

Output:

log((10*x*e^(x + 3*e^x) + (x^2 + 40)*e^x)*e^(-x)/x)/log(x - 2)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-10*x**2*exp(3*exp(x))-x**3-40*x)*ln((10*x*exp(3*exp(x))+x**2+ 
40)/x)+(30*x**3-60*x**2)*exp(x)*ln(-2+x)*exp(3*exp(x))+(x**3-2*x**2-40*x+8 
0)*ln(-2+x))/((10*x**3-20*x**2)*ln(-2+x)**2*exp(3*exp(x))+(x**4-2*x**3+40* 
x**2-80*x)*ln(-2+x)**2),x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \left (x\right )}{\log \left (x - 2\right )} \] Input:

integrate(((-10*x^2*exp(3*exp(x))-x^3-40*x)*log((10*x*exp(3*exp(x))+x^2+40 
)/x)+(30*x^3-60*x^2)*exp(x)*log(-2+x)*exp(3*exp(x))+(x^3-2*x^2-40*x+80)*lo 
g(-2+x))/((10*x^3-20*x^2)*log(-2+x)^2*exp(3*exp(x))+(x^4-2*x^3+40*x^2-80*x 
)*log(-2+x)^2),x, algorithm="maxima")
 

Output:

(log(x^2 + 10*x*e^(3*e^x) + 40) - log(x))/log(x - 2)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \left (x\right )}{\log \left (x - 2\right )} \] Input:

integrate(((-10*x^2*exp(3*exp(x))-x^3-40*x)*log((10*x*exp(3*exp(x))+x^2+40 
)/x)+(30*x^3-60*x^2)*exp(x)*log(-2+x)*exp(3*exp(x))+(x^3-2*x^2-40*x+80)*lo 
g(-2+x))/((10*x^3-20*x^2)*log(-2+x)^2*exp(3*exp(x))+(x^4-2*x^3+40*x^2-80*x 
)*log(-2+x)^2),x, algorithm="giac")
 

Output:

(log(x^2 + 10*x*e^(3*e^x) + 40) - log(x))/log(x - 2)
 

Mupad [B] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\ln \left (\frac {10\,x\,{\mathrm {e}}^{3\,{\mathrm {e}}^x}+x^2+40}{x}\right )}{\ln \left (x-2\right )} \] Input:

int((log(x - 2)*(40*x + 2*x^2 - x^3 - 80) + log((10*x*exp(3*exp(x)) + x^2 
+ 40)/x)*(40*x + x^3 + 10*x^2*exp(3*exp(x))) + log(x - 2)*exp(3*exp(x))*ex 
p(x)*(60*x^2 - 30*x^3))/(log(x - 2)^2*(80*x - 40*x^2 + 2*x^3 - x^4) + log( 
x - 2)^2*exp(3*exp(x))*(20*x^2 - 10*x^3)),x)
                                                                                    
                                                                                    
 

Output:

log((10*x*exp(3*exp(x)) + x^2 + 40)/x)/log(x - 2)
 

Reduce [F]

\[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx =\text {Too large to display} \] Input:

int(((-10*x^2*exp(3*exp(x))-x^3-40*x)*log((10*x*exp(3*exp(x))+x^2+40)/x)+( 
30*x^3-60*x^2)*exp(x)*log(-2+x)*exp(3*exp(x))+(x^3-2*x^2-40*x+80)*log(-2+x 
))/((10*x^3-20*x^2)*log(-2+x)^2*exp(3*exp(x))+(x^4-2*x^3+40*x^2-80*x)*log( 
-2+x)^2),x)
 

Output:

int(x**2/(10*e**(3*e**x)*log(x - 2)*x**2 - 20*e**(3*e**x)*log(x - 2)*x + l 
og(x - 2)*x**3 - 2*log(x - 2)*x**2 + 40*log(x - 2)*x - 80*log(x - 2)),x) - 
 40*int(log((10*e**(3*e**x)*x + x**2 + 40)/x)/(10*e**(3*e**x)*log(x - 2)** 
2*x**2 - 20*e**(3*e**x)*log(x - 2)**2*x + log(x - 2)**2*x**3 - 2*log(x - 2 
)**2*x**2 + 40*log(x - 2)**2*x - 80*log(x - 2)**2),x) + 30*int((e**(3*e**x 
 + x)*x**2)/(10*e**(3*e**x)*log(x - 2)*x**2 - 20*e**(3*e**x)*log(x - 2)*x 
+ log(x - 2)*x**3 - 2*log(x - 2)*x**2 + 40*log(x - 2)*x - 80*log(x - 2)),x 
) - 60*int((e**(3*e**x + x)*x)/(10*e**(3*e**x)*log(x - 2)*x**2 - 20*e**(3* 
e**x)*log(x - 2)*x + log(x - 2)*x**3 - 2*log(x - 2)*x**2 + 40*log(x - 2)*x 
 - 80*log(x - 2)),x) - 10*int((e**(3*e**x)*log((10*e**(3*e**x)*x + x**2 + 
40)/x)*x)/(10*e**(3*e**x)*log(x - 2)**2*x**2 - 20*e**(3*e**x)*log(x - 2)** 
2*x + log(x - 2)**2*x**3 - 2*log(x - 2)**2*x**2 + 40*log(x - 2)**2*x - 80* 
log(x - 2)**2),x) - int((log((10*e**(3*e**x)*x + x**2 + 40)/x)*x**2)/(10*e 
**(3*e**x)*log(x - 2)**2*x**2 - 20*e**(3*e**x)*log(x - 2)**2*x + log(x - 2 
)**2*x**3 - 2*log(x - 2)**2*x**2 + 40*log(x - 2)**2*x - 80*log(x - 2)**2), 
x) - 2*int(x/(10*e**(3*e**x)*log(x - 2)*x**2 - 20*e**(3*e**x)*log(x - 2)*x 
 + log(x - 2)*x**3 - 2*log(x - 2)*x**2 + 40*log(x - 2)*x - 80*log(x - 2)), 
x) + 80*int(1/(10*e**(3*e**x)*log(x - 2)*x**3 - 20*e**(3*e**x)*log(x - 2)* 
x**2 + log(x - 2)*x**4 - 2*log(x - 2)*x**3 + 40*log(x - 2)*x**2 - 80*log(x 
 - 2)*x),x) - 40*int(1/(10*e**(3*e**x)*log(x - 2)*x**2 - 20*e**(3*e**x)...