\(\int \frac {(-20+104 x-40 x^2+4 x^3+(100 x-40 x^2+4 x^3) \log (4)-20 \log (x)) \log (\frac {5-x+5 x^2-x^3+(5 x^2-x^3) \log (4)-2 x \log (x)}{-25+5 x})}{25-10 x+26 x^2-10 x^3+x^4+(25 x^2-10 x^3+x^4) \log (4)+(-10 x+2 x^2) \log (x)} \, dx\) [2333]

Optimal result

Integrand size = 129, antiderivative size = 33 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2-x^2 \log (4)+\frac {2 x \log (x)}{5-x}\right )\right ) \] Output:

ln(2/5*ln(x)/(5-x)*x-2/5*x^2*ln(2)-1/5*x^2-1/5)^2
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right ) \] Input:

Integrate[((-20 + 104*x - 40*x^2 + 4*x^3 + (100*x - 40*x^2 + 4*x^3)*Log[4] 
 - 20*Log[x])*Log[(5 - x + 5*x^2 - x^3 + (5*x^2 - x^3)*Log[4] - 2*x*Log[x] 
)/(-25 + 5*x)])/(25 - 10*x + 26*x^2 - 10*x^3 + x^4 + (25*x^2 - 10*x^3 + x^ 
4)*Log[4] + (-10*x + 2*x^2)*Log[x]),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (4 x^3-40 x^2+\left (4 x^3-40 x^2+100 x\right ) \log (4)+104 x-20 \log (x)-20\right ) \log \left (\frac {-x^3+5 x^2+\left (5 x^2-x^3\right ) \log (4)-x-2 x \log (x)+5}{5 x-25}\right )}{x^4-10 x^3+26 x^2+\left (2 x^2-10 x\right ) \log (x)+\left (x^4-10 x^3+25 x^2\right ) \log (4)-10 x+25} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {4 \left (x^3 (1+\log (4))-10 x^2 (1+\log (4))+26 x \left (1+\frac {25 \log (4)}{26}\right )-5 \log (x)-5\right ) \log \left (\frac {-x^3+5 x^2+\left (5 x^2-x^3\right ) \log (4)-x-2 x \log (x)+5}{5 x-25}\right )}{(5-x) \left (-\left (x^3 (1+\log (4))\right )+5 x^2 (1+\log (4))-x-2 x \log (x)+5\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int -\frac {\left (-\left ((1+\log (4)) x^3\right )+10 (1+\log (4)) x^2-(26+25 \log (4)) x+5 \log (x)+5\right ) \log \left (-\frac {-x^3+5 x^2-2 \log (x) x-x+\left (5 x^2-x^3\right ) \log (4)+5}{5 (5-x)}\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -4 \int \frac {\left (-\left ((1+\log (4)) x^3\right )+10 (1+\log (4)) x^2-(26+25 \log (4)) x+5 \log (x)+5\right ) \log \left (-\frac {-x^3+5 x^2-2 \log (x) x-x+\left (5 x^2-x^3\right ) \log (4)+5}{5 (5-x)}\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -4 \int \left (\frac {(-1-\log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x^3}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {10 (1+\log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x^2}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {(-26-25 \log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {5 \log (x) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {5 \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \left (-5 (26+25 \log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+125 (1+\log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+5 \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+5 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx-(26+25 \log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx+25 (1+\log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx+5 (1+\log (4)) \int \frac {x \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx-(1+\log (4)) \int \frac {x^2 \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx\right )\)

Input:

Int[((-20 + 104*x - 40*x^2 + 4*x^3 + (100*x - 40*x^2 + 4*x^3)*Log[4] - 20* 
Log[x])*Log[(5 - x + 5*x^2 - x^3 + (5*x^2 - x^3)*Log[4] - 2*x*Log[x])/(-25 
 + 5*x)])/(25 - 10*x + 26*x^2 - 10*x^3 + x^4 + (25*x^2 - 10*x^3 + x^4)*Log 
[4] + (-10*x + 2*x^2)*Log[x]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 6.48 (sec) , antiderivative size = 1052, normalized size of antiderivative = 31.88

method	result	size
default	\(\text {Expression too large to display}\)	\(1052\)

Input:

int((-20*ln(x)+2*(4*x^3-40*x^2+100*x)*ln(2)+4*x^3-40*x^2+104*x-20)*ln((-2* 
x*ln(x)+2*(-x^3+5*x^2)*ln(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-10*x)*ln(x)+ 
2*(x^4-10*x^3+25*x^2)*ln(2)+x^4-10*x^3+26*x^2-10*x+25),x,method=_RETURNVER 
BOSE)
 

Output:

4*ln(5)*(1/2*ln(-5+x)-1/2*ln(2*x^3*ln(2)-10*x^2*ln(2)+x^3+2*x*ln(x)-5*x^2+ 
x-5))+ln((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2)^2-2*ln(-5+x) 
*ln((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2)+ln(-5+x)^2+I*Pi*l 
n(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*csgn(I/(-5+x)*((x^3-5 
*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^3-I*Pi*ln(-5+x)*csgn(I/(-5 
+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^3+2*ln(2)*ln(x^ 
3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))-2*ln(2)*ln(-5+x)+I*Pi*ln( 
x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*csgn(I*((x^3-5*x^2)*ln( 
2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/ 
2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^2-I*Pi*ln(-5+x)*csgn(I*((x^3-5*x^2)*ln(2 
)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2 
*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^2-I*Pi*ln(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2 
))*x-5/(1+2*ln(2)))*csgn(I/(-5+x))*csgn(I*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln( 
x)-5/2*x^2+1/2*x-5/2))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/ 
2*x^2+1/2*x-5/2))-2*I*Pi*ln(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln( 
2)))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^ 
2-2*I*Pi*ln(-5+x)+2*I*Pi*ln(-5+x)*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3 
+x*ln(x)-5/2*x^2+1/2*x-5/2))^2+I*Pi*ln(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x 
-5/(1+2*ln(2)))*csgn(I/(-5+x))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x* 
ln(x)-5/2*x^2+1/2*x-5/2))^2+I*Pi*ln(-5+x)*csgn(I/(-5+x))*csgn(I*((x^3-5...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )^{2} \] Input:

integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) 
*log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 
0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg 
orithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log {\left (\frac {- x^{3} + 5 x^{2} - 2 x \log {\left (x \right )} - x + \left (- 2 x^{3} + 10 x^{2}\right ) \log {\left (2 \right )} + 5}{5 x - 25} \right )}^{2} \] Input:

integrate((-20*ln(x)+2*(4*x**3-40*x**2+100*x)*ln(2)+4*x**3-40*x**2+104*x-2 
0)*ln((-2*x*ln(x)+2*(-x**3+5*x**2)*ln(2)-x**3+5*x**2-x+5)/(5*x-25))/((2*x* 
*2-10*x)*ln(x)+2*(x**4-10*x**3+25*x**2)*ln(2)+x**4-10*x**3+26*x**2-10*x+25 
),x)
 

Output:

log((-x**3 + 5*x**2 - 2*x*log(x) - x + (-2*x**3 + 10*x**2)*log(2) + 5)/(5* 
x - 25))**2
 

Maxima [F]

\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \] Input:

integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) 
*log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 
0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg 
orithm="maxima")
 

Output:

4*integrate((x^3 - 10*x^2 + 2*(x^3 - 10*x^2 + 25*x)*log(2) + 26*x - 5*log( 
x) - 5)*log(-1/5*(x^3 - 5*x^2 + 2*(x^3 - 5*x^2)*log(2) + 2*x*log(x) + x - 
5)/(x - 5))/(x^4 - 10*x^3 + 26*x^2 + 2*(x^4 - 10*x^3 + 25*x^2)*log(2) + 2* 
(x^2 - 5*x)*log(x) - 10*x + 25), x)
 

Giac [F]

\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \] Input:

integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) 
*log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 
0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg 
orithm="giac")
 

Output:

integrate(4*(x^3 - 10*x^2 + 2*(x^3 - 10*x^2 + 25*x)*log(2) + 26*x - 5*log( 
x) - 5)*log(-1/5*(x^3 - 5*x^2 + 2*(x^3 - 5*x^2)*log(2) + 2*x*log(x) + x - 
5)/(x - 5))/(x^4 - 10*x^3 + 26*x^2 + 2*(x^4 - 10*x^3 + 25*x^2)*log(2) + 2* 
(x^2 - 5*x)*log(x) - 10*x + 25), x)
 

Mupad [B] (verification not implemented)

Time = 4.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx={\ln \left (-\frac {x-2\,\ln \left (2\right )\,\left (5\,x^2-x^3\right )+2\,x\,\ln \left (x\right )-5\,x^2+x^3-5}{5\,x-25}\right )}^2 \] Input:

int((log(-(x - 2*log(2)*(5*x^2 - x^3) + 2*x*log(x) - 5*x^2 + x^3 - 5)/(5*x 
 - 25))*(104*x - 20*log(x) + 2*log(2)*(100*x - 40*x^2 + 4*x^3) - 40*x^2 + 
4*x^3 - 20))/(2*log(2)*(25*x^2 - 10*x^3 + x^4) - 10*x - log(x)*(10*x - 2*x 
^2) + 26*x^2 - 10*x^3 + x^4 + 25),x)
 

Output:

log(-(x - 2*log(2)*(5*x^2 - x^3) + 2*x*log(x) - 5*x^2 + x^3 - 5)/(5*x - 25 
))^2
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\mathrm {log}\left (\frac {-2 \,\mathrm {log}\left (x \right ) x -2 \,\mathrm {log}\left (2\right ) x^{3}+10 \,\mathrm {log}\left (2\right ) x^{2}-x^{3}+5 x^{2}-x +5}{-25+5 x}\right )^{2} \] Input:

int((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20)*log(( 
-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-10*x)*l 
og(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x)
 

Output:

log(( - 2*log(x)*x - 2*log(2)*x**3 + 10*log(2)*x**2 - x**3 + 5*x**2 - x + 
5)/(5*x - 25))**2