Integrand size = 235, antiderivative size = 26 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \]
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (4+\frac {x}{\log \left (\frac {4 (-1+x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \]
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]
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 |
| \(\displaystyle -2 \int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}} \log \left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )}{x \log ^2\left (x^2\right )}dx+\int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}}}{\log \left (x^2\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right )}dx+\int \frac {\left (\frac {x}{\log \left (\frac {4 (x-1) \log (4)}{5 x}\right )}+4\right )^{\frac {1}{\log \left (x^2\right )}}}{(1-x) \log \left (x^2\right ) \left (x+4 \log \left (\frac {4 (x-1) \log (4)}{5 x}\right )\right ) \log \left (\frac {4 \log (4)}{5}-\frac {4 \log (4)}{5 x}\right )}dx\) |
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]
3.11.49.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.20 (sec) , antiderivative size = 1267, normalized size of antiderivative = 48.73
\[\text {Expression too large to display}\]
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)
16^(-1/2/(I*Pi*csgn(I*x^2)-I*Pi*csgn(I*x)-2*ln(x)))*(-2*I*ln(x)+2*I*ln(-1+ x)-Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+Pi *csgn(I*(-1+x)/x)^3+Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)+2*I*(3*ln (2)-ln(5)+ln(ln(2))))^(1/(I*Pi*csgn(I*x^2)-I*Pi*csgn(I*x)-2*ln(x)))*(1/2*I *x-2*I*ln(x)+2*I*ln(-1+x)-Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-Pi*csgn(I*(-1+x) )*csgn(I*(-1+x)/x)^2+Pi*csgn(I*(-1+x)/x)^3+Pi*csgn(I/x)*csgn(I*(-1+x))*csg n(I*(-1+x)/x)+2*I*(3*ln(2)-ln(5)+ln(ln(2))))^(-1/(I*Pi*csgn(I*x^2)-I*Pi*cs gn(I*x)-2*ln(x)))*exp(I*Pi*csgn(I*(-1/2*I*x+2*I*ln(x)-2*I*ln(-1+x)+Pi*csgn (I/x)*csgn(I*(-1+x)/x)^2+Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2-Pi*csgn(I*(- 1+x)/x)^3-Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-2*I*ln(8/5*ln(2)))/ (Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x) /x)-Pi*csgn(I*(-1+x)/x)^3+Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+2*I*ln(x)-2 *I*ln(8/5*ln(2))-2*I*ln(-1+x)))*(-csgn(I*(-1/2*I*x+2*I*ln(x)-2*I*ln(-1+x)+ Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2+Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2-Pi*cs gn(I*(-1+x)/x)^3-Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-2*I*ln(8/5*l n(2)))/(Pi*csgn(I/x)*csgn(I*(-1+x)/x)^2-Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I *(-1+x)/x)-Pi*csgn(I*(-1+x)/x)^3+Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+2*I* ln(x)-2*I*ln(8/5*ln(2))-2*I*ln(-1+x)))+csgn(-1/2*x+2*ln(x)-2*ln(-1+x)-I*Pi *csgn(I/x)*csgn(I*(-1+x)/x)^2-I*Pi*csgn(I*(-1+x))*csgn(I*(-1+x)/x)^2+I*Pi* csgn(I*(-1+x)/x)^3+I*Pi*csgn(I/x)*csgn(I*(-1+x))*csgn(I*(-1+x)/x)-2*ln(...
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\left (\frac {x + 4 \, \log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}{\log \left (\frac {8 \, {\left (x - 1\right )} \log \left (2\right )}{5 \, x}\right )}\right )^{\left (\frac {1}{\log \left (x^{2}\right )}\right )} \]
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=\
Exception generated. \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=\text {Exception raised: TypeError} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).
Time = 0.67 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx=e^{\left (\frac {\log \left (x - 4 \, \log \left (5\right ) + 12 \, \log \left (2\right ) + 4 \, \log \left (x - 1\right ) - 4 \, \log \left (x\right ) + 4 \, \log \left (\log \left (2\right )\right )\right )}{2 \, \log \left (x\right )} - \frac {\log \left (-\log \left (5\right ) + 3 \, \log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right ) + \log \left (\log \left (2\right )\right )\right )}{2 \, \log \left (x\right )}\right )} \]
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=\
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))
\[ \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 } \]
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=\
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)
Time = 9.77 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )^{\frac {1}{\log \left (x^2\right )}} \left (-x \log \left (x^2\right )+\left (-x+x^2\right ) \log \left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (\left (2 x-2 x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+(8-8 x) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )\right ) \log \left (\frac {x+4 \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}{\log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )}\right )\right )}{\left (-x^2+x^3\right ) \log ^2\left (x^2\right ) \log \left (\frac {(-4+4 x) \log (4)}{5 x}\right )+\left (-4 x+4 x^2\right ) \log ^2\left (x^2\right ) \log ^2\left (\frac {(-4+4 x) \log (4)}{5 x}\right )} \, dx={\left (\frac {x}{\ln \left (-\frac {8\,\ln \left (2\right )-8\,x\,\ln \left (2\right )}{5\,x}\right )}+4\right )}^{\frac {1}{\ln \left (x^2\right )}} \]
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)