\(\int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+(-1+e^9 x) \log (2) \log (3)}{-1+e^9 x}} (-2+2 e^9 x-2 e^{18} x^2)}{1-2 x+x^2+e^{\frac {2 (e^9 x+(-1+e^9 x) \log (2) \log (3))}{-1+e^9 x}} (1-2 e^9 x+e^{18} x^2)+e^9 (-2 x+4 x^2-2 x^3)+e^{18} (x^2-2 x^3+x^4)+e^{\frac {e^9 x+(-1+e^9 x) \log (2) \log (3)}{-1+e^9 x}} (-2+2 x+e^9 (4 x-4 x^2)+e^{18} (-2 x^2+2 x^3))} \, dx\) [2469]

Optimal result

Integrand size = 219, antiderivative size = 34 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-5+\frac {2 x}{1-e^{-\frac {x}{\frac {1}{e^9}-x}+\log (2) \log (3)}-x} \] Output:

2*x/(1-exp(ln(2)*ln(3)-x/(exp(-9)-x))-x)-5
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 x}{-1+e^{1+\frac {1}{-1+e^9 x}+\log (2) \log (3)}+x} \] Input:

Integrate[(2 - 4*E^9*x + 2*E^18*x^2 + E^((E^9*x + (-1 + E^9*x)*Log[2]*Log[ 
3])/(-1 + E^9*x))*(-2 + 2*E^9*x - 2*E^18*x^2))/(1 - 2*x + x^2 + E^((2*(E^9 
*x + (-1 + E^9*x)*Log[2]*Log[3]))/(-1 + E^9*x))*(1 - 2*E^9*x + E^18*x^2) + 
 E^9*(-2*x + 4*x^2 - 2*x^3) + E^18*(x^2 - 2*x^3 + x^4) + E^((E^9*x + (-1 + 
 E^9*x)*Log[2]*Log[3])/(-1 + E^9*x))*(-2 + 2*x + E^9*(4*x - 4*x^2) + E^18* 
(-2*x^2 + 2*x^3))),x]
 

Output:

(-2*x)/(-1 + E^(1 + (-1 + E^9*x)^(-1) + Log[2]*Log[3]) + 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.

\(\displaystyle \int \frac {\left (-2 e^{18} x^2+2 e^9 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+2 e^{18} x^2-4 e^9 x+2}{\left (e^{18} x^2-2 e^9 x+1\right ) \exp \left (\frac {2 \left (e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)\right )}{e^9 x-1}\right )+\left (e^9 \left (4 x-4 x^2\right )+e^{18} \left (2 x^3-2 x^2\right )+2 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+x^2+e^9 \left (-2 x^3+4 x^2-2 x\right )+e^{18} \left (x^4-2 x^3+x^2\right )-2 x+1} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-2 e^{18} x^2+2 e^9 x-2\right ) \exp \left (\frac {e^9 x+\left (e^9 x-1\right ) \log (2) \log (3)}{e^9 x-1}\right )+2 e^{18} x^2-4 e^9 x+2}{\left (1-e^9 x\right )^2 \left (-x-e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}+1\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x \left (e^{18} x^2-e^9 x-e^9+1\right )}{\left (e^9 x-1\right )^2 \left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}-\frac {2 \left (e^{18} x^2-e^9 x+1\right )}{\left (e^9 x-1\right )^2 \left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}dx}{e^9}+2 \int \frac {x}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2}dx-2 \int \frac {1}{x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1}dx-2 \left (1-\frac {1}{e^9}\right ) \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2 \left (e^9 x-1\right )^2}dx-2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right ) \left (e^9 x-1\right )^2}dx+\frac {2 \left (2-e^9\right ) \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right )^2 \left (e^9 x-1\right )}dx}{e^9}-2 \int \frac {1}{\left (x+e^{\frac {e^9 x}{e^9 x-1}+\log (2) \log (3)}-1\right ) \left (e^9 x-1\right )}dx\)

Input:

Int[(2 - 4*E^9*x + 2*E^18*x^2 + E^((E^9*x + (-1 + E^9*x)*Log[2]*Log[3])/(- 
1 + E^9*x))*(-2 + 2*E^9*x - 2*E^18*x^2))/(1 - 2*x + x^2 + E^((2*(E^9*x + ( 
-1 + E^9*x)*Log[2]*Log[3]))/(-1 + E^9*x))*(1 - 2*E^9*x + E^18*x^2) + E^9*( 
-2*x + 4*x^2 - 2*x^3) + E^18*(x^2 - 2*x^3 + x^4) + E^((E^9*x + (-1 + E^9*x 
)*Log[2]*Log[3])/(-1 + E^9*x))*(-2 + 2*x + E^9*(4*x - 4*x^2) + E^18*(-2*x^ 
2 + 2*x^3))),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.85 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {2 x}{{\mathrm e}^{\frac {{\mathrm e}^{9} \ln \left (3\right ) \ln \left (2\right ) x +x \,{\mathrm e}^{9}-\ln \left (2\right ) \ln \left (3\right )}{x \,{\mathrm e}^{9}-1}}+x -1}\)	\(38\)
norman	\(\frac {2 x -2 \,{\mathrm e}^{9} x^{2}}{\left (x +{\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-1\right ) \left (x \,{\mathrm e}^{9}-1\right )}\)	\(52\)
parallelrisch	\(\frac {\left (2 \,{\mathrm e}^{36} {\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-2 \,{\mathrm e}^{36}\right ) {\mathrm e}^{-36}}{x +{\mathrm e}^{\frac {\left (x \,{\mathrm e}^{9}-1\right ) \ln \left (2\right ) \ln \left (3\right )+x \,{\mathrm e}^{9}}{x \,{\mathrm e}^{9}-1}}-1}\)	\(76\)

Input:

int(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9) 
)/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*exp(9)+1)* 
exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3-2*x^2)*exp 
(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/( 
x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9)+x^2-2*x+1) 
,x,method=_RETURNVERBOSE)
 

Output:

-2*x/(exp((exp(9)*ln(3)*ln(2)*x+x*exp(9)-ln(2)*ln(3))/(x*exp(9)-1))+x-1)
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{x + e^{\left (\frac {{\left (x e^{9} - 1\right )} \log \left (3\right ) \log \left (2\right ) + x e^{9}}{x e^{9} - 1}\right )} - 1} \] Input:

integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ 
x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex 
p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 
2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) 
+x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 
)+x^2-2*x+1),x, algorithm="fricas")
 

Output:

-2*x/(x + e^(((x*e^9 - 1)*log(3)*log(2) + x*e^9)/(x*e^9 - 1)) - 1)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=- \frac {2 x}{x + e^{\frac {x e^{9} + \left (x e^{9} - 1\right ) \log {\left (2 \right )} \log {\left (3 \right )}}{x e^{9} - 1}} - 1} \] Input:

integrate(((-2*x**2*exp(9)**2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*ln(2)*ln(3)+ 
x*exp(9))/(x*exp(9)-1))+2*x**2*exp(9)**2-4*x*exp(9)+2)/((x**2*exp(9)**2-2* 
x*exp(9)+1)*exp(((x*exp(9)-1)*ln(2)*ln(3)+x*exp(9))/(x*exp(9)-1))**2+((2*x 
**3-2*x**2)*exp(9)**2+(-4*x**2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*ln(2)* 
ln(3)+x*exp(9))/(x*exp(9)-1))+(x**4-2*x**3+x**2)*exp(9)**2+(-2*x**3+4*x**2 
-2*x)*exp(9)+x**2-2*x+1),x)
 

Output:

-2*x/(x + exp((x*exp(9) + (x*exp(9) - 1)*log(2)*log(3))/(x*exp(9) - 1)) - 
1)
 

Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{2^{\log \left (3\right )} e^{\left (\frac {1}{x e^{9} - 1} + 1\right )} + x - 1} \] Input:

integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ 
x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex 
p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 
2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) 
+x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 
)+x^2-2*x+1),x, algorithm="maxima")
 

Output:

-2*x/(2^log(3)*e^(1/(x*e^9 - 1) + 1) + x - 1)
 

Giac [A] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2 \, x}{x + e^{\left (\log \left (3\right ) \log \left (2\right ) + \frac {x e^{9}}{x e^{9} - 1}\right )} - 1} \] Input:

integrate(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+ 
x*exp(9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*ex 
p(9)+1)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3- 
2*x^2)*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3) 
+x*exp(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9 
)+x^2-2*x+1),x, algorithm="giac")
 

Output:

-2*x/(x + e^(log(3)*log(2) + x*e^9/(x*e^9 - 1)) - 1)
 

Mupad [B] (verification not implemented)

Time = 2.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=-\frac {2\,x}{x+2^{\ln \left (3\right )}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^9}{x\,{\mathrm {e}}^9-1}}-1} \] Input:

int(-(4*x*exp(9) - 2*x^2*exp(18) + exp((x*exp(9) + log(2)*log(3)*(x*exp(9) 
 - 1))/(x*exp(9) - 1))*(2*x^2*exp(18) - 2*x*exp(9) + 2) - 2)/(exp(18)*(x^2 
 - 2*x^3 + x^4) - 2*x - exp(9)*(2*x - 4*x^2 + 2*x^3) + exp((x*exp(9) + log 
(2)*log(3)*(x*exp(9) - 1))/(x*exp(9) - 1))*(2*x + exp(9)*(4*x - 4*x^2) - e 
xp(18)*(2*x^2 - 2*x^3) - 2) + exp((2*(x*exp(9) + log(2)*log(3)*(x*exp(9) - 
 1)))/(x*exp(9) - 1))*(x^2*exp(18) - 2*x*exp(9) + 1) + x^2 + 1),x)
 

Output:

-(2*x)/(x + 2^log(3)*exp((x*exp(9))/(x*exp(9) - 1)) - 1)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {2-4 e^9 x+2 e^{18} x^2+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 e^9 x-2 e^{18} x^2\right )}{1-2 x+x^2+e^{\frac {2 \left (e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)\right )}{-1+e^9 x}} \left (1-2 e^9 x+e^{18} x^2\right )+e^9 \left (-2 x+4 x^2-2 x^3\right )+e^{18} \left (x^2-2 x^3+x^4\right )+e^{\frac {e^9 x+\left (-1+e^9 x\right ) \log (2) \log (3)}{-1+e^9 x}} \left (-2+2 x+e^9 \left (4 x-4 x^2\right )+e^{18} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {2 e^{\frac {1}{e^{9} x -1}} 3^{\mathrm {log}\left (2\right )} e -2}{e^{\frac {1}{e^{9} x -1}} 3^{\mathrm {log}\left (2\right )} e +x -1} \] Input:

int(((-2*x^2*exp(9)^2+2*x*exp(9)-2)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp( 
9))/(x*exp(9)-1))+2*x^2*exp(9)^2-4*x*exp(9)+2)/((x^2*exp(9)^2-2*x*exp(9)+1 
)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp(9))/(x*exp(9)-1))^2+((2*x^3-2*x^2) 
*exp(9)^2+(-4*x^2+4*x)*exp(9)+2*x-2)*exp(((x*exp(9)-1)*log(2)*log(3)+x*exp 
(9))/(x*exp(9)-1))+(x^4-2*x^3+x^2)*exp(9)^2+(-2*x^3+4*x^2-2*x)*exp(9)+x^2- 
2*x+1),x)
 

Output:

(2*(e**(1/(e**9*x - 1))*3**log(2)*e - 1))/(e**(1/(e**9*x - 1))*3**log(2)*e 
 + x - 1)