\(\int \frac {(-8 x+8 x^2+(3 x-4 x^2+x^3) \log (x)) \log (\frac {1}{3} (8+(-3+x) \log (x)))+(-6+8 x-2 x^2+(2 x-2 x^2) \log (x)) \log (\log (\frac {1}{3} (8+(-3+x) \log (x))))+(-8 x+(3 x-x^2) \log (x)) \log (\frac {1}{3} (8+(-3+x) \log (x))) \log ^2(\log (\frac {1}{3} (8+(-3+x) \log (x))))+(-6+2 x+2 x \log (x)) \log ^3(\log (\frac {1}{3} (8+(-3+x) \log (x))))}{(16 x+(-6 x+2 x^2) \log (x)) \log (\frac {1}{3} (8+(-3+x) \log (x)))} \, dx\) [2388]

Optimal result

Integrand size = 183, antiderivative size = 27 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \left (1-x+\log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )\right )^2 \] Output:

1/2*(1+ln(ln(1/3*ln(x)*(-3+x)+8/3))^2-x)*(1/2+1/2*ln(ln(1/3*ln(x)*(-3+x)+8 
/3))^2-1/2*x)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \left ((-2+x) x-2 (-1+x) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\log ^4\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )\right ) \] Input:

Integrate[((-8*x + 8*x^2 + (3*x - 4*x^2 + x^3)*Log[x])*Log[(8 + (-3 + x)*L 
og[x])/3] + (-6 + 8*x - 2*x^2 + (2*x - 2*x^2)*Log[x])*Log[Log[(8 + (-3 + x 
)*Log[x])/3]] + (-8*x + (3*x - x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x])/3]*L 
og[Log[(8 + (-3 + x)*Log[x])/3]]^2 + (-6 + 2*x + 2*x*Log[x])*Log[Log[(8 + 
(-3 + x)*Log[x])/3]]^3)/((16*x + (-6*x + 2*x^2)*Log[x])*Log[(8 + (-3 + x)* 
Log[x])/3]),x]
 

Output:

((-2 + x)*x - 2*(-1 + x)*Log[Log[(8 + (-3 + x)*Log[x])/3]]^2 + Log[Log[(8 
+ (-3 + x)*Log[x])/3]]^4)/4
 

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.

\(\displaystyle \int \frac {\left (\left (3 x-x^2\right ) \log (x)-8 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right ) \log ^2\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )+\left (-2 x^2+\left (2 x-2 x^2\right ) \log (x)+8 x-6\right ) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )+\left (8 x^2+\left (x^3-4 x^2+3 x\right ) \log (x)-8 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )+(2 x+2 x \log (x)-6) \log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{\left (\left (2 x^2-6 x\right ) \log (x)+16 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[((-8*x + 8*x^2 + (3*x - 4*x^2 + x^3)*Log[x])*Log[(8 + (-3 + x)*Log[x]) 
/3] + (-6 + 8*x - 2*x^2 + (2*x - 2*x^2)*Log[x])*Log[Log[(8 + (-3 + x)*Log[ 
x])/3]] + (-8*x + (3*x - x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x])/3]*Log[Log 
[(8 + (-3 + x)*Log[x])/3]]^2 + (-6 + 2*x + 2*x*Log[x])*Log[Log[(8 + (-3 + 
x)*Log[x])/3]]^3)/((16*x + (-6*x + 2*x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x] 
)/3]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 6.58 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63

method	result	size
risch	\(\frac {\ln \left (\ln \left (\frac {\ln \left (x \right ) \left (-3+x \right )}{3}+\frac {8}{3}\right )\right )^{4}}{4}+\left (-\frac {x}{2}+\frac {1}{2}\right ) \ln \left (\ln \left (\frac {\ln \left (x \right ) \left (-3+x \right )}{3}+\frac {8}{3}\right )\right )^{2}+\frac {x^{2}}{4}-\frac {x}{2}\)	\(44\)

Input:

int(((2*x*ln(x)+2*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))^3+((-x^2+3*x)*ln(x)-8* 
x)*ln(1/3*ln(x)*(-3+x)+8/3)*ln(ln(1/3*ln(x)*(-3+x)+8/3))^2+((-2*x^2+2*x)*l 
n(x)-2*x^2+8*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))+((x^3-4*x^2+3*x)*ln(x)+8*x^ 
2-8*x)*ln(1/3*ln(x)*(-3+x)+8/3))/((2*x^2-6*x)*ln(x)+16*x)/ln(1/3*ln(x)*(-3 
+x)+8/3),x,method=_RETURNVERBOSE)
 

Output:

1/4*ln(ln(1/3*ln(x)*(-3+x)+8/3))^4+(-1/2*x+1/2)*ln(ln(1/3*ln(x)*(-3+x)+8/3 
))^2+1/4*x^2-1/2*x
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (\log \left (\frac {1}{3} \, {\left (x - 3\right )} \log \left (x\right ) + \frac {8}{3}\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (\log \left (\frac {1}{3} \, {\left (x - 3\right )} \log \left (x\right ) + \frac {8}{3}\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \] Input:

integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x 
)*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 
+((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 
*x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x 
)+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((2*x*ln(x)+2*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))**3+((-x**2+3*x)* 
ln(x)-8*x)*ln(1/3*ln(x)*(-3+x)+8/3)*ln(ln(1/3*ln(x)*(-3+x)+8/3))**2+((-2*x 
**2+2*x)*ln(x)-2*x**2+8*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))+((x**3-4*x**2+3* 
x)*ln(x)+8*x**2-8*x)*ln(1/3*ln(x)*(-3+x)+8/3))/((2*x**2-6*x)*ln(x)+16*x)/l 
n(1/3*ln(x)*(-3+x)+8/3),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).

Time = 0.17 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (3\right ) + \log \left ({\left (x - 3\right )} \log \left (x\right ) + 8\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (-\log \left (3\right ) + \log \left ({\left (x - 3\right )} \log \left (x\right ) + 8\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \] Input:

integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x 
)*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 
+((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 
*x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x 
)+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm="maxima")
 

Output:

1/4*log(-log(3) + log((x - 3)*log(x) + 8))^4 - 1/2*(x - 1)*log(-log(3) + l 
og((x - 3)*log(x) + 8))^2 + 1/4*x^2 - 1/2*x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 1.54 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (3\right ) + \log \left (x \log \left (x\right ) - 3 \, \log \left (x\right ) + 8\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (-\log \left (3\right ) + \log \left (x \log \left (x\right ) - 3 \, \log \left (x\right ) + 8\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \] Input:

integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x 
)*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 
+((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 
*x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x 
)+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx={\ln \left (\ln \left (\frac {\ln \left (x\right )\,\left (x-3\right )}{3}+\frac {8}{3}\right )\right )}^2\,\left (\frac {\frac {3\,x^2}{2}-\frac {x^3}{2}}{x\,\left (x-3\right )}+\frac {1}{2}\right )-\frac {x}{2}+\frac {{\ln \left (\ln \left (\frac {\ln \left (x\right )\,\left (x-3\right )}{3}+\frac {8}{3}\right )\right )}^4}{4}+\frac {x^2}{4} \] Input:

int((log(log((log(x)*(x - 3))/3 + 8/3))^3*(2*x + 2*x*log(x) - 6) + log((lo 
g(x)*(x - 3))/3 + 8/3)*(log(x)*(3*x - 4*x^2 + x^3) - 8*x + 8*x^2) + log(lo 
g((log(x)*(x - 3))/3 + 8/3))*(8*x + log(x)*(2*x - 2*x^2) - 2*x^2 - 6) - lo 
g(log((log(x)*(x - 3))/3 + 8/3))^2*log((log(x)*(x - 3))/3 + 8/3)*(8*x - lo 
g(x)*(3*x - x^2)))/(log((log(x)*(x - 3))/3 + 8/3)*(16*x - log(x)*(6*x - 2* 
x^2))),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (x \right ) x}{3}-\mathrm {log}\left (x \right )+\frac {8}{3}\right )\right )^{4}}{4}-\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (x \right ) x}{3}-\mathrm {log}\left (x \right )+\frac {8}{3}\right )\right )^{2} x}{2}+\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (x \right ) x}{3}-\mathrm {log}\left (x \right )+\frac {8}{3}\right )\right )^{2}}{2}+\frac {x^{2}}{4}-\frac {x}{2} \] Input:

int(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x)*log( 
x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2+((-2* 
x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4*x^2+3 
*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x)+16*x 
)/log(1/3*log(x)*(-3+x)+8/3),x)
 

Output:

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