\(\int \frac {(\frac {x+4 \log (\frac {(-4+4 x) \log (4)}{5 x})}{\log (\frac {(-4+4 x) \log (4)}{5 x})})^{\frac {1}{\log (x^2)}} (-x \log (x^2)+(-x+x^2) \log (x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+((2 x-2 x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+(8-8 x) \log ^2(\frac {(-4+4 x) \log (4)}{5 x})) \log (\frac {x+4 \log (\frac {(-4+4 x) \log (4)}{5 x})}{\log (\frac {(-4+4 x) \log (4)}{5 x})}))}{(-x^2+x^3) \log ^2(x^2) \log (\frac {(-4+4 x) \log (4)}{5 x})+(-4 x+4 x^2) \log ^2(x^2) \log ^2(\frac {(-4+4 x) \log (4)}{5 x})} \, dx\) [1049]

Optimal result

Integrand size = 235, antiderivative size = 26 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \] Output:

exp(ln(4+x/ln(8/5*(-1+x)/x*ln(2)))/ln(x^2))
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \] Input:

Integrate[(((x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4]) 
/(5*x)])^Log[x^2]^(-1)*(-(x*Log[x^2]) + (-x + x^2)*Log[x^2]*Log[((-4 + 4*x 
)*Log[4])/(5*x)] + ((2*x - 2*x^2)*Log[((-4 + 4*x)*Log[4])/(5*x)] + (8 - 8* 
x)*Log[((-4 + 4*x)*Log[4])/(5*x)]^2)*Log[(x + 4*Log[((-4 + 4*x)*Log[4])/(5 
*x)])/Log[((-4 + 4*x)*Log[4])/(5*x)]]))/((-x^2 + x^3)*Log[x^2]^2*Log[((-4 
+ 4*x)*Log[4])/(5*x)] + (-4*x + 4*x^2)*Log[x^2]^2*Log[((-4 + 4*x)*Log[4])/ 
(5*x)]^2),x]
 

Output:

(4 + x/Log[(4*(-1 + x)*Log[4])/(5*x)])^Log[x^2]^(-1)
 

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.

\(\displaystyle \int \frac {\left (\frac {x+4 \log \left (\frac {(4 x-4) \log (4)}{5 x}\right )}{\log \left (\frac {(4 x-4) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (\left (\left (2 x-2 x^2\right ) \log \left (\frac {(4 x-4) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(4 x-4) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(4 x-4) \log (4)}{5 x}\right )}{\log \left (\frac {(4 x-4) \log (4)}{5 x}\right )}\right )-x \log \left (x^2\right )+\left (x^2-x\right ) \log \left (\frac {(4 x-4) \log (4)}{5 x}\right ) \log \left (x^2\right )\right )}{\left (4 x^2-4 x\right ) \log ^2\left (\frac {(4 x-4) \log (4)}{5 x}\right ) \log ^2\left (x^2\right )+\left (x^3-x^2\right ) \log \left (\frac {(4 x-4) \log (4)}{5 x}\right ) \log ^2\left (x^2\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}} \left (\frac {\log \left (x^2\right ) \left ((x-1) \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )-1\right )}{(x-1) \log \left (\frac {4 (x-1) \log (4)}{5 x}\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right )}-\frac {2 \log \left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )}{x}\right )}{\log ^2\left (x^2\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )+\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )+1\right )}{(1-x) \log \left (x^2\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right ) \log \left (\frac {4 \log (4)}{5}-\frac {4 \log (4)}{5 x}\right )}-\frac {2 \left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}} \log \left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )}{x \log ^2\left (x^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}} \log \left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )}{x \log ^2\left (x^2\right )}dx+\int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}}}{\log \left (x^2\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right )}dx+\int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}}}{(1-x) \log \left (x^2\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right ) \log \left (\frac {4 \log (4)}{5}-\frac {4 \log (4)}{5 x}\right )}dx\)

Input:

Int[(((x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/Log[((-4 + 4*x)*Log[4])/(5*x) 
])^Log[x^2]^(-1)*(-(x*Log[x^2]) + (-x + x^2)*Log[x^2]*Log[((-4 + 4*x)*Log[ 
4])/(5*x)] + ((2*x - 2*x^2)*Log[((-4 + 4*x)*Log[4])/(5*x)] + (8 - 8*x)*Log 
[((-4 + 4*x)*Log[4])/(5*x)]^2)*Log[(x + 4*Log[((-4 + 4*x)*Log[4])/(5*x)])/ 
Log[((-4 + 4*x)*Log[4])/(5*x)]]))/((-x^2 + x^3)*Log[x^2]^2*Log[((-4 + 4*x) 
*Log[4])/(5*x)] + (-4*x + 4*x^2)*Log[x^2]^2*Log[((-4 + 4*x)*Log[4])/(5*x)] 
^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.18 (sec) , antiderivative size = 1263, normalized size of antiderivative = 48.58

\[\text {Expression too large to display}\]

Input:

int((((-8*x+8)*ln(2/5*(-4+4*x)*ln(2)/x)^2+(-2*x^2+2*x)*ln(2/5*(-4+4*x)*ln( 
2)/x))*ln((4*ln(2/5*(-4+4*x)*ln(2)/x)+x)/ln(2/5*(-4+4*x)*ln(2)/x))+(x^2-x) 
*ln(x^2)*ln(2/5*(-4+4*x)*ln(2)/x)-x*ln(x^2))*exp(ln((4*ln(2/5*(-4+4*x)*ln( 
2)/x)+x)/ln(2/5*(-4+4*x)*ln(2)/x))/ln(x^2))/((4*x^2-4*x)*ln(x^2)^2*ln(2/5* 
(-4+4*x)*ln(2)/x)^2+(x^3-x^2)*ln(x^2)^2*ln(2/5*(-4+4*x)*ln(2)/x)),x)
 

Output:

16^(1/2/(-I*Pi*csgn(I*x^2)+I*Pi*csgn(I*x)+2*ln(x)))*(3*ln(2)-ln(5)+ln(ln(2 
))-ln(x)+ln(-1+x)-1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)+1/2*I 
*Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2+1/2*I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^ 
2-1/2*I*Pi*csgn(I*(-1+x)/x)^3)^(1/(I*Pi*csgn(I*x^2)-I*Pi*csgn(I*x)-2*ln(x) 
))*(-1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)+1/2*I*Pi*csgn(I/x) 
*csgn(I*(-1+x)/x)^2+1/2*I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2-1/2*I*Pi*cs 
gn(I*(-1+x)/x)^3+3*ln(2)-ln(5)+ln(ln(2))+1/4*x-ln(x)+ln(-1+x))^(1/(-I*Pi*c 
sgn(I*x^2)+I*Pi*csgn(I*x)+2*ln(x)))*exp(-I*Pi*csgn(I*(1/2*I*Pi*csgn(I/x)*c 
sgn(I*(-1+x))*csgn(I*(-1+x)/x)-1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-1/2*I 
*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+1/2*I*Pi*csgn(I*(-1+x)/x)^3-ln(8/5*l 
n(2))-1/4*x+ln(x)-ln(-1+x))/(-ln(8/5*ln(2))+ln(x)-ln(-1+x)+1/2*I*Pi*csgn(I 
/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2- 
1/2*I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+1/2*I*Pi*csgn(I*(-1+x)/x)^3))*( 
csgn(I*(1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-1/2*I*Pi*csgn(I 
/x)*csgn(I*(-1+x)/x)^2-1/2*I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+1/2*I*Pi 
*csgn(I*(-1+x)/x)^3-ln(8/5*ln(2))-1/4*x+ln(x)-ln(-1+x))/(-ln(8/5*ln(2))+ln 
(x)-ln(-1+x)+1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-1/2*I*Pi*c 
sgn(I/x)*csgn(I*(-1+x)/x)^2-1/2*I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+1/2 
*I*Pi*csgn(I*(-1+x)/x)^3))+csgn(I*(1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn( 
I*(-1+x)/x)-1/2*I*Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-1/2*I*Pi*csgn(I*(-1+x...
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (\frac {x + 4 \, \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}{\log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}\right )^{\left (\frac {1}{\log \left (x^{2}\right )}\right )} \] Input:

integrate((((-8*x+8)*log(2/5*(-4+4*x)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(-4 
+4*x)*log(2)/x))*log((4*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log 
(2)/x))+(x^2-x)*log(x^2)*log(2/5*(-4+4*x)*log(2)/x)-x*log(x^2))*exp(log((4 
*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log(2)/x))/log(x^2))/((4*x 
^2-4*x)*log(x^2)^2*log(2/5*(-4+4*x)*log(2)/x)^2+(x^3-x^2)*log(x^2)^2*log(2 
/5*(-4+4*x)*log(2)/x)),x, algorithm="fricas")
 

Output:

((x + 4*log(8/5*(x - 1)*log(2)/x))/log(8/5*(x - 1)*log(2)/x))^(1/log(x^2))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((-8*x+8)*ln(2/5*(-4+4*x)*ln(2)/x)**2+(-2*x**2+2*x)*ln(2/5*(-4+ 
4*x)*ln(2)/x))*ln((4*ln(2/5*(-4+4*x)*ln(2)/x)+x)/ln(2/5*(-4+4*x)*ln(2)/x)) 
+(x**2-x)*ln(x**2)*ln(2/5*(-4+4*x)*ln(2)/x)-x*ln(x**2))*exp(ln((4*ln(2/5*( 
-4+4*x)*ln(2)/x)+x)/ln(2/5*(-4+4*x)*ln(2)/x))/ln(x**2))/((4*x**2-4*x)*ln(x 
**2)**2*ln(2/5*(-4+4*x)*ln(2)/x)**2+(x**3-x**2)*ln(x**2)**2*ln(2/5*(-4+4*x 
)*ln(2)/x)),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. 61 vs. \(2 (24) = 48\).

Time = 0.54 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=e^{\left (\frac {\log \left (x - 4 \, \log \left (5\right ) + 12 \, \log \left (2\right ) + 4 \, \log \left (x - 1\right ) - 4 \, \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right )\right )}{2 \, \log \left (x\right )} - \frac {\log \left (-\log \left (5\right ) + 3 \, \log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right ) + \log \left (\log \left (2\right )\right )\right )}{2 \, \log \left (x\right )}\right )} \] Input:

integrate((((-8*x+8)*log(2/5*(-4+4*x)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(-4 
+4*x)*log(2)/x))*log((4*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log 
(2)/x))+(x^2-x)*log(x^2)*log(2/5*(-4+4*x)*log(2)/x)-x*log(x^2))*exp(log((4 
*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log(2)/x))/log(x^2))/((4*x 
^2-4*x)*log(x^2)^2*log(2/5*(-4+4*x)*log(2)/x)^2+(x^3-x^2)*log(x^2)^2*log(2 
/5*(-4+4*x)*log(2)/x)),x, algorithm="maxima")
 

Output:

e^(1/2*log(x - 4*log(5) + 12*log(2) + 4*log(x - 1) - 4*log(x) + 4*log(log( 
2)))/log(x) - 1/2*log(-log(5) + 3*log(2) + log(x - 1) - log(x) + log(log(2 
)))/log(x))
 

Giac [F]

\[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\int { \frac {{\left ({\left (x^{2} - x\right )} \log \left (x^{2}\right ) \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right ) - x \log \left (x^{2}\right ) - 2 \, {\left (4 \, {\left (x - 1\right )} \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )^{2} + {\left (x^{2} - x\right )} \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )\right )} \log \left (\frac {x + 4 \, \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}{\log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}\right )\right )} \left (\frac {x + 4 \, \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}{\log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}\right )^{\left (\frac {1}{\log \left (x^{2}\right )}\right )}}{4 \, {\left (x^{2} - x\right )} \log \left (x^{2}\right )^{2} \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )^{2} + {\left (x^{3} - x^{2}\right )} \log \left (x^{2}\right )^{2} \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )} \,d x } \] Input:

integrate((((-8*x+8)*log(2/5*(-4+4*x)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(-4 
+4*x)*log(2)/x))*log((4*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log 
(2)/x))+(x^2-x)*log(x^2)*log(2/5*(-4+4*x)*log(2)/x)-x*log(x^2))*exp(log((4 
*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log(2)/x))/log(x^2))/((4*x 
^2-4*x)*log(x^2)^2*log(2/5*(-4+4*x)*log(2)/x)^2+(x^3-x^2)*log(x^2)^2*log(2 
/5*(-4+4*x)*log(2)/x)),x, algorithm="giac")
 

Output:

integrate(((x^2 - x)*log(x^2)*log(8/5*(x - 1)*log(2)/x) - x*log(x^2) - 2*( 
4*(x - 1)*log(8/5*(x - 1)*log(2)/x)^2 + (x^2 - x)*log(8/5*(x - 1)*log(2)/x 
))*log((x + 4*log(8/5*(x - 1)*log(2)/x))/log(8/5*(x - 1)*log(2)/x)))*((x + 
 4*log(8/5*(x - 1)*log(2)/x))/log(8/5*(x - 1)*log(2)/x))^(1/log(x^2))/(4*( 
x^2 - x)*log(x^2)^2*log(8/5*(x - 1)*log(2)/x)^2 + (x^3 - x^2)*log(x^2)^2*l 
og(8/5*(x - 1)*log(2)/x)), x)
 

Mupad [B] (verification not implemented)

Time = 2.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx={\left (\frac {x}{\ln \left (-\frac {8\,\ln \left (2\right )-8\,x\,\ln \left (2\right )}{5\,x}\right )}+4\right )}^{\frac {1}{\ln \left (x^2\right )}} \] Input:

int((exp(log((x + 4*log((2*log(2)*(4*x - 4))/(5*x)))/log((2*log(2)*(4*x - 
4))/(5*x)))/log(x^2))*(x*log(x^2) - log((x + 4*log((2*log(2)*(4*x - 4))/(5 
*x)))/log((2*log(2)*(4*x - 4))/(5*x)))*(log((2*log(2)*(4*x - 4))/(5*x))*(2 
*x - 2*x^2) - log((2*log(2)*(4*x - 4))/(5*x))^2*(8*x - 8)) + log(x^2)*log( 
(2*log(2)*(4*x - 4))/(5*x))*(x - x^2)))/(log(x^2)^2*log((2*log(2)*(4*x - 4 
))/(5*x))^2*(4*x - 4*x^2) + log(x^2)^2*log((2*log(2)*(4*x - 4))/(5*x))*(x^ 
2 - x^3)),x)
 

Output:

(x/log(-(8*log(2) - 8*x*log(2))/(5*x)) + 4)^(1/log(x^2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=e^{\frac {\mathrm {log}\left (\frac {4 \,\mathrm {log}\left (\frac {8 \,\mathrm {log}\left (2\right ) x -8 \,\mathrm {log}\left (2\right )}{5 x}\right )+x}{\mathrm {log}\left (\frac {8 \,\mathrm {log}\left (2\right ) x -8 \,\mathrm {log}\left (2\right )}{5 x}\right )}\right )}{\mathrm {log}\left (x^{2}\right )}} \] Input:

int((((-8*x+8)*log(2/5*(-4+4*x)*log(2)/x)^2+(-2*x^2+2*x)*log(2/5*(-4+4*x)* 
log(2)/x))*log((4*log(2/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log(2)/x) 
)+(x^2-x)*log(x^2)*log(2/5*(-4+4*x)*log(2)/x)-x*log(x^2))*exp(log((4*log(2 
/5*(-4+4*x)*log(2)/x)+x)/log(2/5*(-4+4*x)*log(2)/x))/log(x^2))/((4*x^2-4*x 
)*log(x^2)^2*log(2/5*(-4+4*x)*log(2)/x)^2+(x^3-x^2)*log(x^2)^2*log(2/5*(-4 
+4*x)*log(2)/x)),x)
 

Output:

e**(log((4*log((8*log(2)*x - 8*log(2))/(5*x)) + x)/log((8*log(2)*x - 8*log 
(2))/(5*x)))/log(x**2))