\(\int \frac {(-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)) \log (\log (\frac {54-18 x-27 x^2+9 x^3+(-27 x+9 x^2) \log (x)}{x}))}{(54 x-18 x^2-27 x^3+9 x^4+(-27 x^2+9 x^3) \log (x)) \log (\frac {54-18 x-27 x^2+9 x^3+(-27 x+9 x^2) \log (x)}{x})} \, dx\) [2256]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 125, antiderivative size = 24 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {1}{9} \log ^2\left (\log \left (\frac {9 (-3+x) (-2+x (x+\log (x)))}{x}\right )\right ) \] Output:

1/9*ln(ln(3*(-3+x)/x*(-6+3*(x+ln(x))*x)))^2
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {1}{9} \log ^2\left (\log \left (\frac {9 (-3+x) \left (-2+x^2+x \log (x)\right )}{x}\right )\right ) \] Input:

Integrate[((-12 - 6*x - 4*x^2 + 4*x^3 + 2*x^2*Log[x])*Log[Log[(54 - 18*x - 
 27*x^2 + 9*x^3 + (-27*x + 9*x^2)*Log[x])/x]])/((54*x - 18*x^2 - 27*x^3 + 
9*x^4 + (-27*x^2 + 9*x^3)*Log[x])*Log[(54 - 18*x - 27*x^2 + 9*x^3 + (-27*x 
 + 9*x^2)*Log[x])/x]),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (4 x^3-4 x^2+2 x^2 \log (x)-6 x-12\right ) \log \left (\log \left (\frac {9 x^3-27 x^2+\left (9 x^2-27 x\right ) \log (x)-18 x+54}{x}\right )\right )}{\left (9 x^4-27 x^3-18 x^2+\left (9 x^3-27 x^2\right ) \log (x)+54 x\right ) \log \left (\frac {9 x^3-27 x^2+\left (9 x^2-27 x\right ) \log (x)-18 x+54}{x}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {2 \left (2 x^3-2 x^2+x^2 \log (x)-3 x-6\right ) \log \left (\log \left (\frac {9 (x-3) \left (x^2+x \log (x)-2\right )}{x}\right )\right )}{9 (3-x) x \left (-x^2-x \log (x)+2\right ) \log \left (\frac {9 (x-3) \left (x^2+x \log (x)-2\right )}{x}\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{9} \int -\frac {\left (-2 x^3-\log (x) x^2+2 x^2+3 x+6\right ) \log \left (\log \left (\frac {9 (3-x) \left (-x^2-\log (x) x+2\right )}{x}\right )\right )}{(3-x) x \left (-x^2-\log (x) x+2\right ) \log \left (\frac {9 (3-x) \left (-x^2-\log (x) x+2\right )}{x}\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2}{9} \int \frac {\left (-2 x^3-\log (x) x^2+2 x^2+3 x+6\right ) \log \left (\log \left (\frac {9 (3-x) \left (-x^2-\log (x) x+2\right )}{x}\right )\right )}{(3-x) x \left (-x^2-\log (x) x+2\right ) \log \left (\frac {9 (3-x) \left (-x^2-\log (x) x+2\right )}{x}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {2}{9} \int \left (\frac {\left (2 x^3+\log (x) x^2-2 x^2-3 x-6\right ) \log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{3 x \left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}-\frac {\left (2 x^3+\log (x) x^2-2 x^2-3 x-6\right ) \log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{3 (x-3) \left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2}{9} \left (-4 \int \frac {\log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{\left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx-7 \int \frac {\log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{(x-3) \left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx-2 \int \frac {\log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{x \left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx-2 \int \frac {x \log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{\left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx-\int \frac {\log (x) \log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{\left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx-3 \int \frac {\log (x) \log \left (\log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )\right )}{(x-3) \left (x^2+\log (x) x-2\right ) \log \left (\frac {9 (x-3) \left (x^2+\log (x) x-2\right )}{x}\right )}dx\right )\)

Input:

Int[((-12 - 6*x - 4*x^2 + 4*x^3 + 2*x^2*Log[x])*Log[Log[(54 - 18*x - 27*x^ 
2 + 9*x^3 + (-27*x + 9*x^2)*Log[x])/x]])/((54*x - 18*x^2 - 27*x^3 + 9*x^4 
+ (-27*x^2 + 9*x^3)*Log[x])*Log[(54 - 18*x - 27*x^2 + 9*x^3 + (-27*x + 9*x 
^2)*Log[x])/x]),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 30.89 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67

method result size
default \(\frac {{\ln \left (2 \ln \left (3\right )+\ln \left (\frac {x^{2} \ln \left (x \right )+x^{3}-3 x \ln \left (x \right )-3 x^{2}-2 x +6}{x}\right )\right )}^{2}}{9}\) \(40\)
risch \(\frac {{\ln \left (2 \ln \left (3\right )-\ln \left (x \right )+\ln \left (6+x^{3}+\left (\ln \left (x \right )-3\right ) x^{2}+\left (-3 \ln \left (x \right )-2\right ) x \right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (6+x^{3}+\left (\ln \left (x \right )-3\right ) x^{2}+\left (-3 \ln \left (x \right )-2\right ) x \right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (6+x^{3}+\left (\ln \left (x \right )-3\right ) x^{2}+\left (-3 \ln \left (x \right )-2\right ) x \right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (6+x^{3}+\left (\ln \left (x \right )-3\right ) x^{2}+\left (-3 \ln \left (x \right )-2\right ) x \right )}{x}\right )+\operatorname {csgn}\left (i \left (6+x^{3}+\left (\ln \left (x \right )-3\right ) x^{2}+\left (-3 \ln \left (x \right )-2\right ) x \right )\right )\right )}{2}\right )}^{2}}{9}\) \(163\)

Input:

int((2*x^2*ln(x)+4*x^3-4*x^2-6*x-12)*ln(ln(((9*x^2-27*x)*ln(x)+9*x^3-27*x^ 
2-18*x+54)/x))/((9*x^3-27*x^2)*ln(x)+9*x^4-27*x^3-18*x^2+54*x)/ln(((9*x^2- 
27*x)*ln(x)+9*x^3-27*x^2-18*x+54)/x),x,method=_RETURNVERBOSE)
 

Output:

1/9*ln(2*ln(3)+ln((x^2*ln(x)+x^3-3*x*ln(x)-3*x^2-2*x+6)/x))^2
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {1}{9} \, \log \left (\log \left (\frac {9 \, {\left (x^{3} - 3 \, x^{2} + {\left (x^{2} - 3 \, x\right )} \log \left (x\right ) - 2 \, x + 6\right )}}{x}\right )\right )^{2} \] Input:

integrate((2*x^2*log(x)+4*x^3-4*x^2-6*x-12)*log(log(((9*x^2-27*x)*log(x)+9 
*x^3-27*x^2-18*x+54)/x))/((9*x^3-27*x^2)*log(x)+9*x^4-27*x^3-18*x^2+54*x)/ 
log(((9*x^2-27*x)*log(x)+9*x^3-27*x^2-18*x+54)/x),x, algorithm="fricas")
 

Output:

1/9*log(log(9*(x^3 - 3*x^2 + (x^2 - 3*x)*log(x) - 2*x + 6)/x))^2
 

Sympy [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {\log {\left (\log {\left (\frac {9 x^{3} - 27 x^{2} - 18 x + \left (9 x^{2} - 27 x\right ) \log {\left (x \right )} + 54}{x} \right )} \right )}^{2}}{9} \] Input:

integrate((2*x**2*ln(x)+4*x**3-4*x**2-6*x-12)*ln(ln(((9*x**2-27*x)*ln(x)+9 
*x**3-27*x**2-18*x+54)/x))/((9*x**3-27*x**2)*ln(x)+9*x**4-27*x**3-18*x**2+ 
54*x)/ln(((9*x**2-27*x)*ln(x)+9*x**3-27*x**2-18*x+54)/x),x)
 

Output:

log(log((9*x**3 - 27*x**2 - 18*x + (9*x**2 - 27*x)*log(x) + 54)/x))**2/9
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {1}{9} \, \log \left (2 \, \log \left (3\right ) + \log \left (x^{2} + x \log \left (x\right ) - 2\right ) + \log \left (x - 3\right ) - \log \left (x\right )\right )^{2} \] Input:

integrate((2*x^2*log(x)+4*x^3-4*x^2-6*x-12)*log(log(((9*x^2-27*x)*log(x)+9 
*x^3-27*x^2-18*x+54)/x))/((9*x^3-27*x^2)*log(x)+9*x^4-27*x^3-18*x^2+54*x)/ 
log(((9*x^2-27*x)*log(x)+9*x^3-27*x^2-18*x+54)/x),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {1}{9} \, \log \left (\log \left (9 \, x^{3} + 9 \, x^{2} \log \left (x\right ) - 27 \, x^{2} - 27 \, x \log \left (x\right ) - 18 \, x + 54\right ) - \log \left (x\right )\right )^{2} \] Input:

integrate((2*x^2*log(x)+4*x^3-4*x^2-6*x-12)*log(log(((9*x^2-27*x)*log(x)+9 
*x^3-27*x^2-18*x+54)/x))/((9*x^3-27*x^2)*log(x)+9*x^4-27*x^3-18*x^2+54*x)/ 
log(((9*x^2-27*x)*log(x)+9*x^3-27*x^2-18*x+54)/x),x, algorithm="giac")
 

Output:

1/9*log(log(9*x^3 + 9*x^2*log(x) - 27*x^2 - 27*x*log(x) - 18*x + 54) - log 
(x))^2
 

Mupad [B] (verification not implemented)

Time = 3.39 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {{\ln \left (\ln \left (-\frac {18\,x+\ln \left (x\right )\,\left (27\,x-9\,x^2\right )+27\,x^2-9\,x^3-54}{x}\right )\right )}^2}{9} \] Input:

int((log(log(-(18*x + log(x)*(27*x - 9*x^2) + 27*x^2 - 9*x^3 - 54)/x))*(6* 
x - 2*x^2*log(x) + 4*x^2 - 4*x^3 + 12))/(log(-(18*x + log(x)*(27*x - 9*x^2 
) + 27*x^2 - 9*x^3 - 54)/x)*(log(x)*(27*x^2 - 9*x^3) - 54*x + 18*x^2 + 27* 
x^3 - 9*x^4)),x)
 

Output:

log(log(-(18*x + log(x)*(27*x - 9*x^2) + 27*x^2 - 9*x^3 - 54)/x))^2/9
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-12-6 x-4 x^2+4 x^3+2 x^2 \log (x)\right ) \log \left (\log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )\right )}{\left (54 x-18 x^2-27 x^3+9 x^4+\left (-27 x^2+9 x^3\right ) \log (x)\right ) \log \left (\frac {54-18 x-27 x^2+9 x^3+\left (-27 x+9 x^2\right ) \log (x)}{x}\right )} \, dx=\frac {{\mathrm {log}\left (\mathrm {log}\left (\frac {9 \,\mathrm {log}\left (x \right ) x^{2}-27 \,\mathrm {log}\left (x \right ) x +9 x^{3}-27 x^{2}-18 x +54}{x}\right )\right )}^{2}}{9} \] Input:

int((2*x^2*log(x)+4*x^3-4*x^2-6*x-12)*log(log(((9*x^2-27*x)*log(x)+9*x^3-2 
7*x^2-18*x+54)/x))/((9*x^3-27*x^2)*log(x)+9*x^4-27*x^3-18*x^2+54*x)/log((( 
9*x^2-27*x)*log(x)+9*x^3-27*x^2-18*x+54)/x),x)
 

Output:

log(log((9*log(x)*x**2 - 27*log(x)*x + 9*x**3 - 27*x**2 - 18*x + 54)/x))** 
2/9