\(\int \frac {-2 x-x^2+(-2-x-2 x^2-2 x^3-x^4) \log (x)+(-2 x-4 x^2-2 x^3) \log ^2(x)+(-2 x-x^2) \log ^3(x)+(2+3 x+x^2) \log (x) \log ((8+8 x+2 x^2) \log (x))}{(2 x^3+x^4) \log (x)+(4 x^2+2 x^3) \log ^2(x)+(2 x+x^2) \log ^3(x)} \, dx\) [63]

Optimal result

Integrand size = 134, antiderivative size = 23 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=13-x-\frac {\log \left (2 (2+x)^2 \log (x)\right )}{x+\log (x)} \] Output:

13-x-ln(2*ln(x)*(2+x)^2)/(x+ln(x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=-x-\frac {\log \left (2 (2+x)^2 \log (x)\right )}{x+\log (x)} \] Input:

Integrate[(-2*x - x^2 + (-2 - x - 2*x^2 - 2*x^3 - x^4)*Log[x] + (-2*x - 4* 
x^2 - 2*x^3)*Log[x]^2 + (-2*x - x^2)*Log[x]^3 + (2 + 3*x + x^2)*Log[x]*Log 
[(8 + 8*x + 2*x^2)*Log[x]])/((2*x^3 + x^4)*Log[x] + (4*x^2 + 2*x^3)*Log[x] 
^2 + (2*x + x^2)*Log[x]^3),x]
 

Output:

-x - Log[2*(2 + x)^2*Log[x]]/(x + 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.

\(\displaystyle \int \frac {-x^2+\left (-x^2-2 x\right ) \log ^3(x)+\left (x^2+3 x+2\right ) \log \left (\left (2 x^2+8 x+8\right ) \log (x)\right ) \log (x)+\left (-2 x^3-4 x^2-2 x\right ) \log ^2(x)+\left (-x^4-2 x^3-2 x^2-x-2\right ) \log (x)-2 x}{\left (x^2+2 x\right ) \log ^3(x)+\left (x^4+2 x^3\right ) \log (x)+\left (2 x^3+4 x^2\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-x^2+\left (-x^2-2 x\right ) \log ^3(x)+\left (x^2+3 x+2\right ) \log \left (\left (2 x^2+8 x+8\right ) \log (x)\right ) \log (x)+\left (-2 x^3-4 x^2-2 x\right ) \log ^2(x)+\left (-x^4-2 x^3-2 x^2-x-2\right ) \log (x)-2 x}{x (x+2) \log (x) (x+\log (x))^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-2*x - x^2 + (-2 - x - 2*x^2 - 2*x^3 - x^4)*Log[x] + (-2*x - 4*x^2 - 
2*x^3)*Log[x]^2 + (-2*x - x^2)*Log[x]^3 + (2 + 3*x + x^2)*Log[x]*Log[(8 + 
8*x + 2*x^2)*Log[x]])/((2*x^3 + x^4)*Log[x] + (4*x^2 + 2*x^3)*Log[x]^2 + ( 
2*x + x^2)*Log[x]^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87

method	result	size
parallelrisch	\(\frac {-4 x^{2}-4 x \ln \left (x \right )+4 x +4 \ln \left (x \right )-4 \ln \left (\left (2 x^{2}+8 x +8\right ) \ln \left (x \right )\right )}{4 x +4 \ln \left (x \right )}\)	\(43\)
risch	\(-\frac {2 \ln \left (2+x \right )}{x +\ln \left (x \right )}-\frac {-i \pi \operatorname {csgn}\left (i \left (2+x \right )\right )^{2} \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left (2+x \right )\right ) \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )^{3}-i \pi \,\operatorname {csgn}\left (i \left (2+x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (2+x \right )^{2}\right )+i \pi \,\operatorname {csgn}\left (i \left (2+x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (2+x \right )^{2}\right )^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (2+x \right )^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (2+x \right )^{2}\right )^{3}+2 x^{2}+2 x \ln \left (x \right )+2 \ln \left (2\right )+2 \ln \left (\ln \left (x \right )\right )}{2 \left (x +\ln \left (x \right )\right )}\)	\(197\)

Input:

int(((x^2+3*x+2)*ln(x)*ln((2*x^2+8*x+8)*ln(x))+(-x^2-2*x)*ln(x)^3+(-2*x^3- 
4*x^2-2*x)*ln(x)^2+(-x^4-2*x^3-2*x^2-x-2)*ln(x)-x^2-2*x)/((x^2+2*x)*ln(x)^ 
3+(2*x^3+4*x^2)*ln(x)^2+(x^4+2*x^3)*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

1/4*(-4*x^2-4*x*ln(x)+4*x+4*ln(x)-4*ln((2*x^2+8*x+8)*ln(x)))/(x+ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=-\frac {x^{2} + x \log \left (x\right ) + \log \left (2 \, {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right )\right )}{x + \log \left (x\right )} \] Input:

integrate(((x^2+3*x+2)*log(x)*log((2*x^2+8*x+8)*log(x))+(-x^2-2*x)*log(x)^ 
3+(-2*x^3-4*x^2-2*x)*log(x)^2+(-x^4-2*x^3-2*x^2-x-2)*log(x)-x^2-2*x)/((x^2 
+2*x)*log(x)^3+(2*x^3+4*x^2)*log(x)^2+(x^4+2*x^3)*log(x)),x, algorithm="fr 
icas")
 

Output:

-(x^2 + x*log(x) + log(2*(x^2 + 4*x + 4)*log(x)))/(x + log(x))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((x**2+3*x+2)*ln(x)*ln((2*x**2+8*x+8)*ln(x))+(-x**2-2*x)*ln(x)** 
3+(-2*x**3-4*x**2-2*x)*ln(x)**2+(-x**4-2*x**3-2*x**2-x-2)*ln(x)-x**2-2*x)/ 
((x**2+2*x)*ln(x)**3+(2*x**3+4*x**2)*ln(x)**2+(x**4+2*x**3)*ln(x)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=-\frac {x^{2} + x \log \left (x\right ) + \log \left (2\right ) + 2 \, \log \left (x + 2\right ) + \log \left (\log \left (x\right )\right )}{x + \log \left (x\right )} \] Input:

integrate(((x^2+3*x+2)*log(x)*log((2*x^2+8*x+8)*log(x))+(-x^2-2*x)*log(x)^ 
3+(-2*x^3-4*x^2-2*x)*log(x)^2+(-x^4-2*x^3-2*x^2-x-2)*log(x)-x^2-2*x)/((x^2 
+2*x)*log(x)^3+(2*x^3+4*x^2)*log(x)^2+(x^4+2*x^3)*log(x)),x, algorithm="ma 
xima")
 

Output:

-(x^2 + x*log(x) + log(2) + 2*log(x + 2) + log(log(x)))/(x + log(x))
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=-x - \frac {\log \left (2 \, x^{2} \log \left (x\right ) + 8 \, x \log \left (x\right ) + 8 \, \log \left (x\right )\right )}{x + \log \left (x\right )} \] Input:

integrate(((x^2+3*x+2)*log(x)*log((2*x^2+8*x+8)*log(x))+(-x^2-2*x)*log(x)^ 
3+(-2*x^3-4*x^2-2*x)*log(x)^2+(-x^4-2*x^3-2*x^2-x-2)*log(x)-x^2-2*x)/((x^2 
+2*x)*log(x)^3+(2*x^3+4*x^2)*log(x)^2+(x^4+2*x^3)*log(x)),x, algorithm="gi 
ac")
 

Output:

-x - log(2*x^2*log(x) + 8*x*log(x) + 8*log(x))/(x + log(x))
 

Mupad [B] (verification not implemented)

Time = 2.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=-x-\frac {\ln \left (\ln \left (x\right )\,\left (2\,x^2+8\,x+8\right )\right )}{x+\ln \left (x\right )} \] Input:

int(-(2*x + log(x)^2*(2*x + 4*x^2 + 2*x^3) + log(x)*(x + 2*x^2 + 2*x^3 + x 
^4 + 2) + log(x)^3*(2*x + x^2) + x^2 - log(log(x)*(8*x + 2*x^2 + 8))*log(x 
)*(3*x + x^2 + 2))/(log(x)^2*(4*x^2 + 2*x^3) + log(x)^3*(2*x + x^2) + log( 
x)*(2*x^3 + x^4)),x)
 

Output:

- x - log(log(x)*(8*x + 2*x^2 + 8))/(x + log(x))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-2 x-x^2+\left (-2-x-2 x^2-2 x^3-x^4\right ) \log (x)+\left (-2 x-4 x^2-2 x^3\right ) \log ^2(x)+\left (-2 x-x^2\right ) \log ^3(x)+\left (2+3 x+x^2\right ) \log (x) \log \left (\left (8+8 x+2 x^2\right ) \log (x)\right )}{\left (2 x^3+x^4\right ) \log (x)+\left (4 x^2+2 x^3\right ) \log ^2(x)+\left (2 x+x^2\right ) \log ^3(x)} \, dx=\frac {-\mathrm {log}\left (2 \,\mathrm {log}\left (x \right ) x^{2}+8 \,\mathrm {log}\left (x \right ) x +8 \,\mathrm {log}\left (x \right )\right )-\mathrm {log}\left (x \right ) x -x^{2}}{\mathrm {log}\left (x \right )+x} \] Input:

int(((x^2+3*x+2)*log(x)*log((2*x^2+8*x+8)*log(x))+(-x^2-2*x)*log(x)^3+(-2* 
x^3-4*x^2-2*x)*log(x)^2+(-x^4-2*x^3-2*x^2-x-2)*log(x)-x^2-2*x)/((x^2+2*x)* 
log(x)^3+(2*x^3+4*x^2)*log(x)^2+(x^4+2*x^3)*log(x)),x)
 

Output:

( - (log(2*log(x)*x**2 + 8*log(x)*x + 8*log(x)) + log(x)*x + x**2))/(log(x 
) + x)