\(\int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+(49-28 x+4 x^2) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+(-14 x+4 x^2+(14-4 x) \log (2)) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2)))} \, dx\) [462]

Optimal result

Integrand size = 161, antiderivative size = 23 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=-\frac {x}{7-2 x-2 \log (x)+\log (-2 (-x+\log (2)))} \] Output:

-x/(ln(-2*ln(2)+2*x)-2*ln(x)-2*x+7)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx \] Input:

Integrate[(8*x - 9*Log[2] + (-2*x + 2*Log[2])*Log[x] + (x - Log[2])*Log[2* 
x - 2*Log[2]])/(-49*x + 28*x^2 - 4*x^3 + (49 - 28*x + 4*x^2)*Log[2] + (-4* 
x + 4*Log[2])*Log[x]^2 + (-14*x + 4*x^2 + (14 - 4*x)*Log[2])*Log[2*x - 2*L 
og[2]] + (-x + Log[2])*Log[2*x - 2*Log[2]]^2 + Log[x]*(28*x - 8*x^2 + (-28 
 + 8*x)*Log[2] + (4*x - 4*Log[2])*Log[2*x - 2*Log[2]])),x]
 

Output:

Integrate[(8*x - 9*Log[2] + (-2*x + 2*Log[2])*Log[x] + (x - Log[2])*Log[2* 
x - 2*Log[2]])/(-49*x + 28*x^2 - 4*x^3 + (49 - 28*x + 4*x^2)*Log[2] + (-4* 
x + 4*Log[2])*Log[x]^2 + (-14*x + 4*x^2 + (14 - 4*x)*Log[2])*Log[2*x - 2*L 
og[2]] + (-x + Log[2])*Log[2*x - 2*Log[2]]^2 + Log[x]*(28*x - 8*x^2 + (-28 
 + 8*x)*Log[2] + (4*x - 4*Log[2])*Log[2*x - 2*Log[2]])), x]
 

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.

\(\displaystyle \int \frac {8 x+(2 \log (2)-2 x) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))-9 \log (2)}{-4 x^3+28 x^2+\left (4 x^2-14 x+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+\log (x) \left (-8 x^2+28 x+(4 x-4 \log (2)) \log (2 x-2 \log (2))+(8 x-28) \log (2)\right )+\left (4 x^2-28 x+49\right ) \log (2)-49 x+(4 \log (2)-4 x) \log ^2(x)+(\log (2)-x) \log ^2(2 x-2 \log (2))} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {-8 x+2 (x-\log (2)) \log (x)+(\log (2)-x) \log (2 x-\log (4))+\log (512)}{(x-\log (2)) (-2 x-2 \log (x)+\log (2 x-\log (4))+7)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-2 x^2-x (1-\log (4))+\log (4)}{(x-\log (2)) (-2 x-2 \log (x)+\log (2 x-\log (4))+7)^2}+\frac {1}{2 x+2 \log (x)-\log (2 x-\log (4))-7}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {1}{(2 x+2 \log (x)-\log (2 x-\log (4))-7)^2}dx-2 \int \frac {x}{(2 x+2 \log (x)-\log (2 x-\log (4))-7)^2}dx+\log (2) \int \frac {1}{(x-\log (2)) (2 x+2 \log (x)-\log (2 x-\log (4))-7)^2}dx+\int \frac {1}{2 x+2 \log (x)-\log (2 x-\log (4))-7}dx\)

Input:

Int[(8*x - 9*Log[2] + (-2*x + 2*Log[2])*Log[x] + (x - Log[2])*Log[2*x - 2* 
Log[2]])/(-49*x + 28*x^2 - 4*x^3 + (49 - 28*x + 4*x^2)*Log[2] + (-4*x + 4* 
Log[2])*Log[x]^2 + (-14*x + 4*x^2 + (14 - 4*x)*Log[2])*Log[2*x - 2*Log[2]] 
 + (-x + Log[2])*Log[2*x - 2*Log[2]]^2 + Log[x]*(28*x - 8*x^2 + (-28 + 8*x 
)*Log[2] + (4*x - 4*Log[2])*Log[2*x - 2*Log[2]])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {x}{\ln \left (x -\ln \left (2\right )\right )+7+\ln \left (2\right )-2 \ln \left (x \right )-2 x}\)	\(24\)
risch	\(\frac {x}{2 x +2 \ln \left (x \right )-\ln \left (-2 \ln \left (2\right )+2 x \right )-7}\)	\(25\)
parallelrisch	\(\frac {x}{2 x +2 \ln \left (x \right )-\ln \left (-2 \ln \left (2\right )+2 x \right )-7}\)	\(25\)

Input:

int(((2*ln(2)-2*x)*ln(x)+(x-ln(2))*ln(-2*ln(2)+2*x)-9*ln(2)+8*x)/((4*ln(2) 
-4*x)*ln(x)^2+((-4*ln(2)+4*x)*ln(-2*ln(2)+2*x)+(8*x-28)*ln(2)-8*x^2+28*x)* 
ln(x)+(ln(2)-x)*ln(-2*ln(2)+2*x)^2+((-4*x+14)*ln(2)+4*x^2-14*x)*ln(-2*ln(2 
)+2*x)+(4*x^2-28*x+49)*ln(2)-4*x^3+28*x^2-49*x),x,method=_RETURNVERBOSE)
 

Output:

-x/(ln(x-ln(2))+7+ln(2)-2*ln(x)-2*x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\frac {x}{2 \, x - \log \left (2 \, x - 2 \, \log \left (2\right )\right ) + 2 \, \log \left (x\right ) - 7} \] Input:

integrate(((2*log(2)-2*x)*log(x)+(x-log(2))*log(-2*log(2)+2*x)-9*log(2)+8* 
x)/((4*log(2)-4*x)*log(x)^2+((-4*log(2)+4*x)*log(-2*log(2)+2*x)+(8*x-28)*l 
og(2)-8*x^2+28*x)*log(x)+(log(2)-x)*log(-2*log(2)+2*x)^2+((-4*x+14)*log(2) 
+4*x^2-14*x)*log(-2*log(2)+2*x)+(4*x^2-28*x+49)*log(2)-4*x^3+28*x^2-49*x), 
x, algorithm="fricas")
 

Output:

x/(2*x - log(2*x - 2*log(2)) + 2*log(x) - 7)
 

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=- \frac {x}{- 2 x - 2 \log {\left (x \right )} + \log {\left (2 x - 2 \log {\left (2 \right )} \right )} + 7} \] Input:

integrate(((2*ln(2)-2*x)*ln(x)+(x-ln(2))*ln(-2*ln(2)+2*x)-9*ln(2)+8*x)/((4 
*ln(2)-4*x)*ln(x)**2+((-4*ln(2)+4*x)*ln(-2*ln(2)+2*x)+(8*x-28)*ln(2)-8*x** 
2+28*x)*ln(x)+(ln(2)-x)*ln(-2*ln(2)+2*x)**2+((-4*x+14)*ln(2)+4*x**2-14*x)* 
ln(-2*ln(2)+2*x)+(4*x**2-28*x+49)*ln(2)-4*x**3+28*x**2-49*x),x)
 

Output:

-x/(-2*x - 2*log(x) + log(2*x - 2*log(2)) + 7)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\frac {x}{2 \, x - \log \left (2\right ) - \log \left (x - \log \left (2\right )\right ) + 2 \, \log \left (x\right ) - 7} \] Input:

integrate(((2*log(2)-2*x)*log(x)+(x-log(2))*log(-2*log(2)+2*x)-9*log(2)+8* 
x)/((4*log(2)-4*x)*log(x)^2+((-4*log(2)+4*x)*log(-2*log(2)+2*x)+(8*x-28)*l 
og(2)-8*x^2+28*x)*log(x)+(log(2)-x)*log(-2*log(2)+2*x)^2+((-4*x+14)*log(2) 
+4*x^2-14*x)*log(-2*log(2)+2*x)+(4*x^2-28*x+49)*log(2)-4*x^3+28*x^2-49*x), 
x, algorithm="maxima")
 

Output:

x/(2*x - log(2) - log(x - log(2)) + 2*log(x) - 7)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\frac {x}{2 \, x - \log \left (2\right ) - \log \left (x - \log \left (2\right )\right ) + 2 \, \log \left (x\right ) - 7} \] Input:

integrate(((2*log(2)-2*x)*log(x)+(x-log(2))*log(-2*log(2)+2*x)-9*log(2)+8* 
x)/((4*log(2)-4*x)*log(x)^2+((-4*log(2)+4*x)*log(-2*log(2)+2*x)+(8*x-28)*l 
og(2)-8*x^2+28*x)*log(x)+(log(2)-x)*log(-2*log(2)+2*x)^2+((-4*x+14)*log(2) 
+4*x^2-14*x)*log(-2*log(2)+2*x)+(4*x^2-28*x+49)*log(2)-4*x^3+28*x^2-49*x), 
x, algorithm="giac")
 

Output:

x/(2*x - log(2) - log(x - log(2)) + 2*log(x) - 7)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\int -\frac {8\,x-9\,\ln \left (2\right )-\ln \left (x\right )\,\left (2\,x-2\,\ln \left (2\right )\right )+\ln \left (2\,x-2\,\ln \left (2\right )\right )\,\left (x-\ln \left (2\right )\right )}{49\,x-\ln \left (2\right )\,\left (4\,x^2-28\,x+49\right )+\ln \left (2\,x-2\,\ln \left (2\right )\right )\,\left (14\,x+\ln \left (2\right )\,\left (4\,x-14\right )-4\,x^2\right )-\ln \left (x\right )\,\left (28\,x+\ln \left (2\right )\,\left (8\,x-28\right )+\ln \left (2\,x-2\,\ln \left (2\right )\right )\,\left (4\,x-4\,\ln \left (2\right )\right )-8\,x^2\right )+{\ln \left (x\right )}^2\,\left (4\,x-4\,\ln \left (2\right )\right )+{\ln \left (2\,x-2\,\ln \left (2\right )\right )}^2\,\left (x-\ln \left (2\right )\right )-28\,x^2+4\,x^3} \,d x \] Input:

int(-(8*x - 9*log(2) - log(x)*(2*x - 2*log(2)) + log(2*x - 2*log(2))*(x - 
log(2)))/(49*x - log(2)*(4*x^2 - 28*x + 49) + log(2*x - 2*log(2))*(14*x + 
log(2)*(4*x - 14) - 4*x^2) - log(x)*(28*x + log(2)*(8*x - 28) + log(2*x - 
2*log(2))*(4*x - 4*log(2)) - 8*x^2) + log(x)^2*(4*x - 4*log(2)) + log(2*x 
- 2*log(2))^2*(x - log(2)) - 28*x^2 + 4*x^3),x)
 

Output:

int(-(8*x - 9*log(2) - log(x)*(2*x - 2*log(2)) + log(2*x - 2*log(2))*(x - 
log(2)))/(49*x - log(2)*(4*x^2 - 28*x + 49) + log(2*x - 2*log(2))*(14*x + 
log(2)*(4*x - 14) - 4*x^2) - log(x)*(28*x + log(2)*(8*x - 28) + log(2*x - 
2*log(2))*(4*x - 4*log(2)) - 8*x^2) + log(x)^2*(4*x - 4*log(2)) + log(2*x 
- 2*log(2))^2*(x - log(2)) - 28*x^2 + 4*x^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {8 x-9 \log (2)+(-2 x+2 \log (2)) \log (x)+(x-\log (2)) \log (2 x-2 \log (2))}{-49 x+28 x^2-4 x^3+\left (49-28 x+4 x^2\right ) \log (2)+(-4 x+4 \log (2)) \log ^2(x)+\left (-14 x+4 x^2+(14-4 x) \log (2)\right ) \log (2 x-2 \log (2))+(-x+\log (2)) \log ^2(2 x-2 \log (2))+\log (x) \left (28 x-8 x^2+(-28+8 x) \log (2)+(4 x-4 \log (2)) \log (2 x-2 \log (2))\right )} \, dx=\frac {-\mathrm {log}\left (-2 \,\mathrm {log}\left (2\right )+2 x \right )+2 \,\mathrm {log}\left (x \right )-7}{2 \,\mathrm {log}\left (-2 \,\mathrm {log}\left (2\right )+2 x \right )-4 \,\mathrm {log}\left (x \right )-4 x +14} \] Input:

int(((2*log(2)-2*x)*log(x)+(x-log(2))*log(-2*log(2)+2*x)-9*log(2)+8*x)/((4 
*log(2)-4*x)*log(x)^2+((-4*log(2)+4*x)*log(-2*log(2)+2*x)+(8*x-28)*log(2)- 
8*x^2+28*x)*log(x)+(log(2)-x)*log(-2*log(2)+2*x)^2+((-4*x+14)*log(2)+4*x^2 
-14*x)*log(-2*log(2)+2*x)+(4*x^2-28*x+49)*log(2)-4*x^3+28*x^2-49*x),x)
 

Output:

( - log( - 2*log(2) + 2*x) + 2*log(x) - 7)/(2*(log( - 2*log(2) + 2*x) - 2* 
log(x) - 2*x + 7))