\(\int \frac {((-5 x^2+x^3) \log (4)+(5-x) \log ^2(4) \log (5-x)+((15 x^2-3 x^3) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)) \log (x)) \log (\log (2))}{-5 x^6+x^7+(10 x^4-2 x^5) \log (4) \log (5-x)+(-5 x^2+x^3) \log ^2(4) \log ^2(5-x)} \, dx\) [943]

Optimal result

Integrand size = 124, antiderivative size = 31 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\left (6-\frac {\log (x)}{x \left (-\frac {x^2}{\log (4)}+\log (5-x)\right )}\right ) \log (\log (2)) \] Output:

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

Mathematica [A] (verified)

Time = 6.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\frac {\log (4) \log (x) \log (\log (2))}{x^3-x \log (4) \log (5-x)} \] Input:

Integrate[(((-5*x^2 + x^3)*Log[4] + (5 - x)*Log[4]^2*Log[5 - x] + ((15*x^2 
 - 3*x^3)*Log[4] + x*Log[4]^2 + (-5 + x)*Log[4]^2*Log[5 - x])*Log[x])*Log[ 
Log[2]])/(-5*x^6 + x^7 + (10*x^4 - 2*x^5)*Log[4]*Log[5 - x] + (-5*x^2 + x^ 
3)*Log[4]^2*Log[5 - x]^2),x]
 

Output:

(Log[4]*Log[x]*Log[Log[2]])/(x^3 - x*Log[4]*Log[5 - 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.

\(\displaystyle \int \frac {\log (\log (2)) \left (\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(x-5) \log ^2(4) \log (5-x)\right ) \log (x)+\left (x^3-5 x^2\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)\right )}{x^7-5 x^6+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (x^3-5 x^2\right ) \log ^2(4) \log ^2(5-x)} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \log (\log (2)) \int \frac {\log (4) \left (5 x^2-x^3\right )-(5-x) \log ^2(4) \log (5-x)-\left (\log ^2(4) x-(5-x) \log ^2(4) \log (5-x)+3 \left (5 x^2-x^3\right ) \log (4)\right ) \log (x)}{-x^7+5 x^6+\left (5 x^2-x^3\right ) \log ^2(4) \log ^2(5-x)-2 \left (5 x^4-x^5\right ) \log (4) \log (5-x)}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \log (\log (2)) \int \frac {\log (4) \left (-((x-5) \log (4) \log (5-x) (\log (x)-1))-x \left ((x-5) x+\left (-3 x^2+15 x+\log (4)\right ) \log (x)\right )\right )}{(5-x) \left (x^3-x \log (4) \log (5-x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \log (4) \log (\log (2)) \int -\frac {(5-x) \log (4) \log (5-x) (1-\log (x))-x \left ((5-x) x-\left (-3 x^2+15 x+\log (4)\right ) \log (x)\right )}{(5-x) \left (x^3-x \log (4) \log (5-x)\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\log (4) \log (\log (2)) \int \frac {(5-x) \log (4) \log (5-x) (1-\log (x))-x \left ((5-x) x-\left (-3 x^2+15 x+\log (4)\right ) \log (x)\right )}{(5-x) \left (x^3-x \log (4) \log (5-x)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\log (4) \log (\log (2)) \int \left (\frac {\left (3 x^3-15 x^2-\log (4) \log (5-x) x-\log (4) x+5 \log (4) \log (5-x)\right ) \log (x)}{(x-5) x^2 \left (x^2-\log (4) \log (5-x)\right )^2}-\frac {1}{x^2 \left (x^2-\log (4) \log (5-x)\right )}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle -\log (4) \log (\log (2)) \int \left (\frac {\left (3 x^3-15 x^2-\log (4) \log (5-x) x-\log (4) x+5 \log (4) \log (5-x)\right ) \log (x)}{(x-5) x^2 \left (x^2-\log (4) \log (5-x)\right )^2}-\frac {1}{x^2 \left (x^2-\log (4) \log (5-x)\right )}\right )dx\)

Input:

Int[(((-5*x^2 + x^3)*Log[4] + (5 - x)*Log[4]^2*Log[5 - x] + ((15*x^2 - 3*x 
^3)*Log[4] + x*Log[4]^2 + (-5 + x)*Log[4]^2*Log[5 - x])*Log[x])*Log[Log[2] 
])/(-5*x^6 + x^7 + (10*x^4 - 2*x^5)*Log[4]*Log[5 - x] + (-5*x^2 + x^3)*Log 
[4]^2*Log[5 - x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {2 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right ) \ln \left (x \right )}{x \left (2 \ln \left (2\right ) \ln \left (5-x \right )-x^{2}\right )}\)	\(31\)
parallelrisch	\(-\frac {2 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right ) \ln \left (x \right )}{x \left (2 \ln \left (2\right ) \ln \left (5-x \right )-x^{2}\right )}\)	\(31\)

Input:

int(((4*(-5+x)*ln(2)^2*ln(5-x)+4*x*ln(2)^2+2*(-3*x^3+15*x^2)*ln(2))*ln(x)+ 
4*(5-x)*ln(2)^2*ln(5-x)+2*(x^3-5*x^2)*ln(2))*ln(ln(2))/(4*(x^3-5*x^2)*ln(2 
)^2*ln(5-x)^2+2*(-2*x^5+10*x^4)*ln(2)*ln(5-x)+x^7-5*x^6),x,method=_RETURNV 
ERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\frac {2 \, \log \left (2\right ) \log \left (x\right ) \log \left (\log \left (2\right )\right )}{x^{3} - 2 \, x \log \left (2\right ) \log \left (-x + 5\right )} \] Input:

integrate(((4*(-5+x)*log(2)^2*log(5-x)+4*x*log(2)^2+2*(-3*x^3+15*x^2)*log( 
2))*log(x)+4*(5-x)*log(2)^2*log(5-x)+2*(x^3-5*x^2)*log(2))*log(log(2))/(4* 
(x^3-5*x^2)*log(2)^2*log(5-x)^2+2*(-2*x^5+10*x^4)*log(2)*log(5-x)+x^7-5*x^ 
6),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=- \frac {2 \log {\left (2 \right )} \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )}}{- x^{3} + 2 x \log {\left (2 \right )} \log {\left (5 - x \right )}} \] Input:

integrate(((4*(-5+x)*ln(2)**2*ln(5-x)+4*x*ln(2)**2+2*(-3*x**3+15*x**2)*ln( 
2))*ln(x)+4*(5-x)*ln(2)**2*ln(5-x)+2*(x**3-5*x**2)*ln(2))*ln(ln(2))/(4*(x* 
*3-5*x**2)*ln(2)**2*ln(5-x)**2+2*(-2*x**5+10*x**4)*ln(2)*ln(5-x)+x**7-5*x* 
*6),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\frac {2 \, \log \left (2\right ) \log \left (x\right ) \log \left (\log \left (2\right )\right )}{x^{3} - 2 \, x \log \left (2\right ) \log \left (-x + 5\right )} \] Input:

integrate(((4*(-5+x)*log(2)^2*log(5-x)+4*x*log(2)^2+2*(-3*x^3+15*x^2)*log( 
2))*log(x)+4*(5-x)*log(2)^2*log(5-x)+2*(x^3-5*x^2)*log(2))*log(log(2))/(4* 
(x^3-5*x^2)*log(2)^2*log(5-x)^2+2*(-2*x^5+10*x^4)*log(2)*log(5-x)+x^7-5*x^ 
6),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\frac {2 \, \log \left (2\right ) \log \left (x\right ) \log \left (\log \left (2\right )\right )}{x^{3} - 2 \, x \log \left (2\right ) \log \left (-x + 5\right )} \] Input:

integrate(((4*(-5+x)*log(2)^2*log(5-x)+4*x*log(2)^2+2*(-3*x^3+15*x^2)*log( 
2))*log(x)+4*(5-x)*log(2)^2*log(5-x)+2*(x^3-5*x^2)*log(2))*log(log(2))/(4* 
(x^3-5*x^2)*log(2)^2*log(5-x)^2+2*(-2*x^5+10*x^4)*log(2)*log(5-x)+x^7-5*x^ 
6),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=\int \frac {\ln \left (\ln \left (2\right )\right )\,\left (2\,\ln \left (2\right )\,\left (5\,x^2-x^3\right )-\ln \left (x\right )\,\left (2\,\ln \left (2\right )\,\left (15\,x^2-3\,x^3\right )+4\,x\,{\ln \left (2\right )}^2+4\,{\ln \left (2\right )}^2\,\ln \left (5-x\right )\,\left (x-5\right )\right )+4\,{\ln \left (2\right )}^2\,\ln \left (5-x\right )\,\left (x-5\right )\right )}{5\,x^6-x^7-2\,\ln \left (2\right )\,\ln \left (5-x\right )\,\left (10\,x^4-2\,x^5\right )+4\,{\ln \left (2\right )}^2\,{\ln \left (5-x\right )}^2\,\left (5\,x^2-x^3\right )} \,d x \] Input:

int((log(log(2))*(2*log(2)*(5*x^2 - x^3) - log(x)*(2*log(2)*(15*x^2 - 3*x^ 
3) + 4*x*log(2)^2 + 4*log(2)^2*log(5 - x)*(x - 5)) + 4*log(2)^2*log(5 - x) 
*(x - 5)))/(5*x^6 - x^7 - 2*log(2)*log(5 - x)*(10*x^4 - 2*x^5) + 4*log(2)^ 
2*log(5 - x)^2*(5*x^2 - x^3)),x)
 

Output:

int((log(log(2))*(2*log(2)*(5*x^2 - x^3) - log(x)*(2*log(2)*(15*x^2 - 3*x^ 
3) + 4*x*log(2)^2 + 4*log(2)^2*log(5 - x)*(x - 5)) + 4*log(2)^2*log(5 - x) 
*(x - 5)))/(5*x^6 - x^7 - 2*log(2)*log(5 - x)*(10*x^4 - 2*x^5) + 4*log(2)^ 
2*log(5 - x)^2*(5*x^2 - x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {\left (\left (-5 x^2+x^3\right ) \log (4)+(5-x) \log ^2(4) \log (5-x)+\left (\left (15 x^2-3 x^3\right ) \log (4)+x \log ^2(4)+(-5+x) \log ^2(4) \log (5-x)\right ) \log (x)\right ) \log (\log (2))}{-5 x^6+x^7+\left (10 x^4-2 x^5\right ) \log (4) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(4) \log ^2(5-x)} \, dx=-\frac {2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )}{x \left (2 \,\mathrm {log}\left (-x +5\right ) \mathrm {log}\left (2\right )-x^{2}\right )} \] Input:

int(((4*(-5+x)*log(2)^2*log(5-x)+4*x*log(2)^2+2*(-3*x^3+15*x^2)*log(2))*lo 
g(x)+4*(5-x)*log(2)^2*log(5-x)+2*(x^3-5*x^2)*log(2))*log(log(2))/(4*(x^3-5 
*x^2)*log(2)^2*log(5-x)^2+2*(-2*x^5+10*x^4)*log(2)*log(5-x)+x^7-5*x^6),x)
 

Output:

( - 2*log(log(2))*log(x)*log(2))/(x*(2*log( - x + 5)*log(2) - x**2))