\(\int \frac {-24 x^2+3 x \log (4)+(-12 x^2+3 x \log (4)) \log (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)})}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)})+(-4 x+\log (4)) \log ^2(\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)})} \, dx\) [896]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 118, antiderivative size = 31 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=3 \left (5+\frac {x^2}{2 \left (3+\log \left (\frac {x^2}{\left (4-\frac {\log (4)}{x}\right )^2}\right )\right )}\right ) \] Output:

15+3/2*x^2/(ln(x^2/(4-2*ln(2)/x)^2)+3)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=\frac {3 x^2}{2 \left (3+\log \left (\frac {x^4}{(-4 x+\log (4))^2}\right )\right )} \] Input:

Integrate[(-24*x^2 + 3*x*Log[4] + (-12*x^2 + 3*x*Log[4])*Log[x^4/(16*x^2 - 
 8*x*Log[4] + Log[4]^2)])/(-36*x + 9*Log[4] + (-24*x + 6*Log[4])*Log[x^4/( 
16*x^2 - 8*x*Log[4] + Log[4]^2)] + (-4*x + Log[4])*Log[x^4/(16*x^2 - 8*x*L 
og[4] + Log[4]^2)]^2),x]
 

Output:

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

\(\displaystyle \int \frac {-24 x^2+\left (3 x \log (4)-12 x^2\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+3 x \log (4)}{(\log (4)-4 x) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(6 \log (4)-24 x) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )-36 x+9 \log (4)} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-24*x^2 + 3*x*Log[4] + (-12*x^2 + 3*x*Log[4])*Log[x^4/(16*x^2 - 8*x*L 
og[4] + Log[4]^2)])/(-36*x + 9*Log[4] + (-24*x + 6*Log[4])*Log[x^4/(16*x^2 
 - 8*x*Log[4] + Log[4]^2)] + (-4*x + Log[4])*Log[x^4/(16*x^2 - 8*x*Log[4] 
+ Log[4]^2)]^2),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method result size
parallelrisch \(\frac {3 x^{2}}{2 \left (\ln \left (\frac {x^{4}}{4 \ln \left (2\right )^{2}-16 x \ln \left (2\right )+16 x^{2}}\right )+3\right )}\) \(33\)
norman \(\frac {3 x^{2}}{2 \left (\ln \left (\frac {x^{4}}{4 \ln \left (2\right )^{2}-16 x \ln \left (2\right )+16 x^{2}}\right )+3\right )}\) \(34\)
risch \(\frac {3 x^{2}}{2 \left (\ln \left (\frac {x^{4}}{4 \ln \left (2\right )^{2}-16 x \ln \left (2\right )+16 x^{2}}\right )+3\right )}\) \(34\)

Input:

int(((6*x*ln(2)-12*x^2)*ln(x^4/(4*ln(2)^2-16*x*ln(2)+16*x^2))+6*x*ln(2)-24 
*x^2)/((2*ln(2)-4*x)*ln(x^4/(4*ln(2)^2-16*x*ln(2)+16*x^2))^2+(12*ln(2)-24* 
x)*ln(x^4/(4*ln(2)^2-16*x*ln(2)+16*x^2))+18*ln(2)-36*x),x,method=_RETURNVE 
RBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=\frac {3 \, x^{2}}{2 \, {\left (\log \left (\frac {x^{4}}{4 \, {\left (4 \, x^{2} - 4 \, x \log \left (2\right ) + \log \left (2\right )^{2}\right )}}\right ) + 3\right )}} \] Input:

integrate(((6*x*log(2)-12*x^2)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+6* 
x*log(2)-24*x^2)/((2*log(2)-4*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))^ 
2+(12*log(2)-24*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+18*log(2)-36*x 
),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=\frac {3 x^{2}}{2 \log {\left (\frac {x^{4}}{16 x^{2} - 16 x \log {\left (2 \right )} + 4 \log {\left (2 \right )}^{2}} \right )} + 6} \] Input:

integrate(((6*x*ln(2)-12*x**2)*ln(x**4/(4*ln(2)**2-16*x*ln(2)+16*x**2))+6* 
x*ln(2)-24*x**2)/((2*ln(2)-4*x)*ln(x**4/(4*ln(2)**2-16*x*ln(2)+16*x**2))** 
2+(12*ln(2)-24*x)*ln(x**4/(4*ln(2)**2-16*x*ln(2)+16*x**2))+18*ln(2)-36*x), 
x)
 

Output:

3*x**2/(2*log(x**4/(16*x**2 - 16*x*log(2) + 4*log(2)**2)) + 6)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=-\frac {3 \, x^{2}}{2 \, {\left (2 \, \log \left (2\right ) + 2 \, \log \left (2 \, x - \log \left (2\right )\right ) - 4 \, \log \left (x\right ) - 3\right )}} \] Input:

integrate(((6*x*log(2)-12*x^2)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+6* 
x*log(2)-24*x^2)/((2*log(2)-4*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))^ 
2+(12*log(2)-24*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+18*log(2)-36*x 
),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=-\frac {3 \, x^{2}}{2 \, {\left (2 \, \log \left (2\right ) - \log \left (x^{4}\right ) + \log \left (4 \, x^{2} - 4 \, x \log \left (2\right ) + \log \left (2\right )^{2}\right ) - 3\right )}} \] Input:

integrate(((6*x*log(2)-12*x^2)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+6* 
x*log(2)-24*x^2)/((2*log(2)-4*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))^ 
2+(12*log(2)-24*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+18*log(2)-36*x 
),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=-\int \frac {\ln \left (\frac {x^4}{16\,x^2-16\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2}\right )\,\left (6\,x\,\ln \left (2\right )-12\,x^2\right )+6\,x\,\ln \left (2\right )-24\,x^2}{\left (4\,x-2\,\ln \left (2\right )\right )\,{\ln \left (\frac {x^4}{16\,x^2-16\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2}\right )}^2+\left (24\,x-12\,\ln \left (2\right )\right )\,\ln \left (\frac {x^4}{16\,x^2-16\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2}\right )+36\,x-18\,\ln \left (2\right )} \,d x \] Input:

int(-(log(x^4/(4*log(2)^2 - 16*x*log(2) + 16*x^2))*(6*x*log(2) - 12*x^2) + 
 6*x*log(2) - 24*x^2)/(36*x - 18*log(2) + log(x^4/(4*log(2)^2 - 16*x*log(2 
) + 16*x^2))*(24*x - 12*log(2)) + log(x^4/(4*log(2)^2 - 16*x*log(2) + 16*x 
^2))^2*(4*x - 2*log(2))),x)
 

Output:

-int((log(x^4/(4*log(2)^2 - 16*x*log(2) + 16*x^2))*(6*x*log(2) - 12*x^2) + 
 6*x*log(2) - 24*x^2)/(36*x - 18*log(2) + log(x^4/(4*log(2)^2 - 16*x*log(2 
) + 16*x^2))*(24*x - 12*log(2)) + log(x^4/(4*log(2)^2 - 16*x*log(2) + 16*x 
^2))^2*(4*x - 2*log(2))), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-24 x^2+3 x \log (4)+\left (-12 x^2+3 x \log (4)\right ) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )}{-36 x+9 \log (4)+(-24 x+6 \log (4)) \log \left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )+(-4 x+\log (4)) \log ^2\left (\frac {x^4}{16 x^2-8 x \log (4)+\log ^2(4)}\right )} \, dx=\frac {3 x^{2}}{2 \,\mathrm {log}\left (\frac {x^{4}}{4 \mathrm {log}\left (2\right )^{2}-16 \,\mathrm {log}\left (2\right ) x +16 x^{2}}\right )+6} \] Input:

int(((6*x*log(2)-12*x^2)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+6*x*log( 
2)-24*x^2)/((2*log(2)-4*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))^2+(12* 
log(2)-24*x)*log(x^4/(4*log(2)^2-16*x*log(2)+16*x^2))+18*log(2)-36*x),x)
 

Output:

(3*x**2)/(2*(log(x**4/(4*log(2)**2 - 16*log(2)*x + 16*x**2)) + 3))