\(\int \frac {-2 x^2+(-2-4 x+e^x (-1-x+2 x^2)) \log (-3-6 x)}{(4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x (4-8 x-32 x^2)) \log (-3-6 x)+(-4 x+8 x^2+32 x^3+e^x (-2 x-4 x^2)) \log (-3-6 x) \log (\log (-3-6 x))+(x^2+2 x^3) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx\) [763]

Optimal result

Integrand size = 144, antiderivative size = 24 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x}{-2-e^x+x (8+\log (\log (-3 (1+2 x))))} \] Output:

x/(x*(8+ln(ln(-6*x-3)))-2-exp(x))
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \] Input:

Integrate[(-2*x^2 + (-2 - 4*x + E^x*(-1 - x + 2*x^2))*Log[-3 - 6*x])/((4 - 
 24*x + 128*x^3 + E^(2*x)*(1 + 2*x) + E^x*(4 - 8*x - 32*x^2))*Log[-3 - 6*x 
] + (-4*x + 8*x^2 + 32*x^3 + E^x*(-2*x - 4*x^2))*Log[-3 - 6*x]*Log[Log[-3 
- 6*x]] + (x^2 + 2*x^3)*Log[-3 - 6*x]*Log[Log[-3 - 6*x]]^2),x]
 

Output:

x/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*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.

\(\displaystyle \int \frac {\left (e^x \left (2 x^2-x-1\right )-4 x-2\right ) \log (-6 x-3)-2 x^2}{\left (2 x^3+x^2\right ) \log (-6 x-3) \log ^2(\log (-6 x-3))+\left (32 x^3+8 x^2+e^x \left (-4 x^2-2 x\right )-4 x\right ) \log (-6 x-3) \log (\log (-6 x-3))+\left (128 x^3+e^x \left (-32 x^2-8 x+4\right )-24 x+e^{2 x} (2 x+1)+4\right ) \log (-6 x-3)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (e^x (x-1)-2\right ) (2 x+1) \log (-6 x-3)-2 x^2}{(2 x+1) \log (-6 x-3) \left (-8 x+e^x+x (-\log (\log (-6 x-3)))+2\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x \left (16 x^2 \log (-6 x-3)+2 x^2 \log (-6 x-3) \log (\log (-6 x-3))-2 x-12 x \log (-6 x-3)-x \log (-6 x-3) \log (\log (-6 x-3))-10 \log (-6 x-3)-\log (-6 x-3) \log (\log (-6 x-3))\right )}{(2 x+1) \log (-6 x-3) \left (8 x-e^x+x \log (\log (-6 x-3))-2\right )^2}-\frac {x-1}{8 x-e^x+x \log (\log (-6 x-3))-2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 \int \frac {x^2}{\left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx+\int \frac {x^2 \log (\log (-6 x-3))}{\left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx+\frac {1}{2} \int \frac {1}{\log (-6 x-3) \left (-\log (\log (-6 x-3)) x-8 x+e^x+2\right )^2}dx-\int \frac {1}{-\log (\log (-6 x-3)) x-8 x+e^x+2}dx-10 \int \frac {x}{\left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx-\int \frac {x}{\log (-6 x-3) \left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx-\frac {1}{2} \int \frac {1}{(2 x+1) \log (-6 x-3) \left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx-\int \frac {x \log (\log (-6 x-3))}{\left (\log (\log (-6 x-3)) x+8 x-e^x-2\right )^2}dx-\int \frac {x}{\log (\log (-6 x-3)) x+8 x-e^x-2}dx\)

Input:

Int[(-2*x^2 + (-2 - 4*x + E^x*(-1 - x + 2*x^2))*Log[-3 - 6*x])/((4 - 24*x 
+ 128*x^3 + E^(2*x)*(1 + 2*x) + E^x*(4 - 8*x - 32*x^2))*Log[-3 - 6*x] + (- 
4*x + 8*x^2 + 32*x^3 + E^x*(-2*x - 4*x^2))*Log[-3 - 6*x]*Log[Log[-3 - 6*x] 
] + (x^2 + 2*x^3)*Log[-3 - 6*x]*Log[Log[-3 - 6*x]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 7.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {x}{x \ln \left (\ln \left (-6 x -3\right )\right )+8 x -{\mathrm e}^{x}-2}\)	\(23\)
parallelrisch	\(\frac {x}{x \ln \left (\ln \left (-6 x -3\right )\right )+8 x -{\mathrm e}^{x}-2}\)	\(23\)

Input:

int((((2*x^2-x-1)*exp(x)-4*x-2)*ln(-6*x-3)-2*x^2)/((2*x^3+x^2)*ln(-6*x-3)* 
ln(ln(-6*x-3))^2+((-4*x^2-2*x)*exp(x)+32*x^3+8*x^2-4*x)*ln(-6*x-3)*ln(ln(- 
6*x-3))+((1+2*x)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x+4)*ln(-6*x-3 
)),x,method=_RETURNVERBOSE)
 

Output:

x/(x*ln(ln(-6*x-3))+8*x-exp(x)-2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x}{x \log \left (\log \left (-6 \, x - 3\right )\right ) + 8 \, x - e^{x} - 2} \] Input:

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log( 
-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x 
-3)*log(log(-6*x-3))+((1+2*x)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x 
+4)*log(-6*x-3)),x, algorithm="fricas")
 

Output:

x/(x*log(log(-6*x - 3)) + 8*x - e^x - 2)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=- \frac {x}{- x \log {\left (\log {\left (- 6 x - 3 \right )} \right )} - 8 x + e^{x} + 2} \] Input:

integrate((((2*x**2-x-1)*exp(x)-4*x-2)*ln(-6*x-3)-2*x**2)/((2*x**3+x**2)*l 
n(-6*x-3)*ln(ln(-6*x-3))**2+((-4*x**2-2*x)*exp(x)+32*x**3+8*x**2-4*x)*ln(- 
6*x-3)*ln(ln(-6*x-3))+((1+2*x)*exp(x)**2+(-32*x**2-8*x+4)*exp(x)+128*x**3- 
24*x+4)*ln(-6*x-3)),x)
 

Output:

-x/(-x*log(log(-6*x - 3)) - 8*x + exp(x) + 2)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x}{x \log \left (i \, \pi + \log \left (3\right ) + \log \left (2 \, x + 1\right )\right ) + 8 \, x - e^{x} - 2} \] Input:

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log( 
-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x 
-3)*log(log(-6*x-3))+((1+2*x)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x 
+4)*log(-6*x-3)),x, algorithm="maxima")
 

Output:

x/(x*log(I*pi + log(3) + log(2*x + 1)) + 8*x - e^x - 2)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x}{x \log \left (\log \left (-6 \, x - 3\right )\right ) + 8 \, x - e^{x} - 2} \] Input:

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log( 
-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x 
-3)*log(log(-6*x-3))+((1+2*x)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x 
+4)*log(-6*x-3)),x, algorithm="giac")
 

Output:

x/(x*log(log(-6*x - 3)) + 8*x - e^x - 2)
 

Mupad [B] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 226, normalized size of antiderivative = 9.42 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {x\,{\left (\ln \left (-6\,x-3\right )+2\,x\,\ln \left (-6\,x-3\right )\right )}^2\,\left (2\,\ln \left (-6\,x-3\right )+4\,x\,\ln \left (-6\,x-3\right )+2\,x^2\right )+x\,{\mathrm {e}}^x\,{\left (\ln \left (-6\,x-3\right )+2\,x\,\ln \left (-6\,x-3\right )\right )}^2\,\left (\ln \left (-6\,x-3\right )+x\,\ln \left (-6\,x-3\right )-2\,x^2\,\ln \left (-6\,x-3\right )\right )}{\ln \left (-6\,x-3\right )\,\left (2\,x+1\right )\,\left (8\,x-{\mathrm {e}}^x+x\,\ln \left (\ln \left (-6\,x-3\right )\right )-2\right )\,\left (2\,{\ln \left (-6\,x-3\right )}^2+8\,x^2\,{\ln \left (-6\,x-3\right )}^2+{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2+8\,x\,{\ln \left (-6\,x-3\right )}^2+2\,x^2\,\ln \left (-6\,x-3\right )+4\,x^3\,\ln \left (-6\,x-3\right )-4\,x^3\,{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2+3\,x\,{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2\right )} \] Input:

int(-(log(- 6*x - 3)*(4*x + exp(x)*(x - 2*x^2 + 1) + 2) + 2*x^2)/(log(- 6* 
x - 3)*(exp(2*x)*(2*x + 1) - exp(x)*(8*x + 32*x^2 - 4) - 24*x + 128*x^3 + 
4) - log(log(- 6*x - 3))*log(- 6*x - 3)*(4*x + exp(x)*(2*x + 4*x^2) - 8*x^ 
2 - 32*x^3) + log(log(- 6*x - 3))^2*log(- 6*x - 3)*(x^2 + 2*x^3)),x)
 

Output:

(x*(log(- 6*x - 3) + 2*x*log(- 6*x - 3))^2*(2*log(- 6*x - 3) + 4*x*log(- 6 
*x - 3) + 2*x^2) + x*exp(x)*(log(- 6*x - 3) + 2*x*log(- 6*x - 3))^2*(log(- 
 6*x - 3) + x*log(- 6*x - 3) - 2*x^2*log(- 6*x - 3)))/(log(- 6*x - 3)*(2*x 
 + 1)*(8*x - exp(x) + x*log(log(- 6*x - 3)) - 2)*(2*log(- 6*x - 3)^2 + 8*x 
^2*log(- 6*x - 3)^2 + exp(x)*log(- 6*x - 3)^2 + 8*x*log(- 6*x - 3)^2 + 2*x 
^2*log(- 6*x - 3) + 4*x^3*log(- 6*x - 3) - 4*x^3*exp(x)*log(- 6*x - 3)^2 + 
 3*x*exp(x)*log(- 6*x - 3)^2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx=\frac {-e^{x}+\mathrm {log}\left (\mathrm {log}\left (-6 x -3\right )\right ) x -2}{8 e^{x}-8 \,\mathrm {log}\left (\mathrm {log}\left (-6 x -3\right )\right ) x -64 x +16} \] Input:

int((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log(-6*x-3 
)*log(log(-6*x-3))^2+((-4*x^2-2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x-3)*lo 
g(log(-6*x-3))+((1+2*x)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x+4)*lo 
g(-6*x-3)),x)
 

Output:

( - e**x + log(log( - 6*x - 3))*x - 2)/(8*(e**x - log(log( - 6*x - 3))*x - 
 8*x + 2))