\(\int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log (\frac {5 \log (x)}{e^3})}{(4 x^2-x^3) \log (x)+(-3 x+x^2) \log (x) \log (\frac {5 \log (x)}{e^3})} \, dx\) [406]

Optimal result

Integrand size = 60, antiderivative size = 24 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\log \left (\frac {x-(-3+x) \left (x-\log \left (\frac {5 \log (x)}{e^3}\right )\right )}{x^2}\right ) \] Output:

ln((x-(-3+x)*(x-ln(5*ln(x)/exp(3))))/x^2)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx \] Input:

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

Output:

Integrate[(-3 + x - 4*x*Log[x] + (6 - x)*Log[x]*Log[(5*Log[x])/E^3])/((4*x 
^2 - x^3)*Log[x] + (-3*x + x^2)*Log[x]*Log[(5*Log[x])/E^3]), 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.

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

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )-3}{x \log (x) \left (-x^2+x \log (\log (x))+x (1+\log (5))-3 \log (5 \log (x))+9\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x-4 x \log (x)-3}{x \log (x) \left (-x^2+x \log (\log (x))+x (1+\log (5))-3 \log (5 \log (x))+9\right )}+\frac {(6-x) \left (\log (\log (x))-3 \left (1-\frac {\log (5)}{3}\right )\right )}{x \left (-x^2+x \log (\log (x))+x (1+\log (5))-3 \log (5 \log (x))+9\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 \int \frac {1}{-x^2+\log (\log (x)) x+(1+\log (5)) x-3 \log (5 \log (x))+9}dx-\frac {2}{3} (2-\log (5)) \int \frac {x}{-x^2+\log (\log (x)) x+(1+\log (5)) x-3 \log (5 \log (x))+9}dx+\frac {1}{3} \int \frac {x^2}{-x^2+\log (\log (x)) x+(1+\log (5)) x-3 \log (5 \log (x))+9}dx+\int \frac {1}{\log (x) \left (-x^2+\log (\log (x)) x+(1+\log (5)) x-3 \log (5 \log (x))+9\right )}dx+\frac {2}{3} \int \frac {x \log (\log (x))}{-x^2+\log (\log (x)) x+(1+\log (5)) x-3 \log (5 \log (x))+9}dx+4 \int \frac {1}{x^2-\log (\log (x)) x-(1+\log (5)) x+3 \log (5 \log (x))-9}dx+\frac {1}{3} (1+\log (5)) \int \frac {x}{x^2-\log (\log (x)) x-(1+\log (5)) x+3 \log (5 \log (x))-9}dx+\frac {2}{3} \int \frac {x^2}{x^2-\log (\log (x)) x-(1+\log (5)) x+3 \log (5 \log (x))-9}dx+3 \int \frac {1}{x \log (x) \left (x^2-\log (\log (x)) x-(1+\log (5)) x+3 \log (5 \log (x))-9\right )}dx+\frac {1}{3} \int \frac {x \log (\log (x))}{x^2-\log (\log (x)) x-(1+\log (5)) x+3 \log (5 \log (x))-9}dx-\frac {x}{3}-2 \log (x)\)

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
risch	\(\ln \left (-3+x \right )-2 \ln \left (x \right )+\ln \left (\ln \left (5 \ln \left (x \right ) {\mathrm e}^{-3}\right )-\frac {\left (x -4\right ) x}{-3+x}\right )\)	\(30\)
parallelrisch	\(\ln \left (x^{2}-\ln \left (5 \ln \left (x \right ) {\mathrm e}^{-3}\right ) x -4 x +3 \ln \left (5 \ln \left (x \right ) {\mathrm e}^{-3}\right )\right )-2 \ln \left (x \right )\)	\(37\)

Input:

int(((-x+6)*ln(x)*ln(5*ln(x)/exp(3))-4*x*ln(x)+x-3)/((x^2-3*x)*ln(x)*ln(5* 
ln(x)/exp(3))+(-x^3+4*x^2)*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

ln(-3+x)-2*ln(x)+ln(ln(5*ln(x)*exp(-3))-(x-4)*x/(-3+x))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\log \left (x - 3\right ) - 2 \, \log \left (x\right ) + \log \left (-\frac {x^{2} - {\left (x - 3\right )} \log \left (5 \, e^{\left (-3\right )} \log \left (x\right )\right ) - 4 \, x}{x - 3}\right ) \] Input:

integrate(((6-x)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*lo 
g(x)*log(5*log(x)/exp(3))+(-x^3+4*x^2)*log(x)),x, algorithm="fricas")
 

Output:

log(x - 3) - 2*log(x) + log(-(x^2 - (x - 3)*log(5*e^(-3)*log(x)) - 4*x)/(x 
 - 3))
 

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=- 2 \log {\left (x \right )} + \log {\left (x - 3 \right )} + \log {\left (\log {\left (\frac {5 \log {\left (x \right )}}{e^{3}} \right )} + \frac {- x^{2} + 4 x}{x - 3} \right )} \] Input:

integrate(((6-x)*ln(x)*ln(5*ln(x)/exp(3))-4*x*ln(x)+x-3)/((x**2-3*x)*ln(x) 
*ln(5*ln(x)/exp(3))+(-x**3+4*x**2)*ln(x)),x)
 

Output:

-2*log(x) + log(x - 3) + log(log(5*exp(-3)*log(x)) + (-x**2 + 4*x)/(x - 3) 
)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\log \left (x - 3\right ) - 2 \, \log \left (x\right ) + \log \left (-\frac {x^{2} - x {\left (\log \left (5\right ) + 1\right )} - {\left (x - 3\right )} \log \left (\log \left (x\right )\right ) + 3 \, \log \left (5\right ) - 9}{x - 3}\right ) \] Input:

integrate(((6-x)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*lo 
g(x)*log(5*log(x)/exp(3))+(-x^3+4*x^2)*log(x)),x, algorithm="maxima")
 

Output:

log(x - 3) - 2*log(x) + log(-(x^2 - x*(log(5) + 1) - (x - 3)*log(log(x)) + 
 3*log(5) - 9)/(x - 3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\log \left (-x^{2} + x \log \left (5 \, \log \left (x\right )\right ) + x - 3 \, \log \left (5 \, \log \left (x\right )\right ) + 9\right ) - 2 \, \log \left (x\right ) \] Input:

integrate(((6-x)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*lo 
g(x)*log(5*log(x)/exp(3))+(-x^3+4*x^2)*log(x)),x, algorithm="giac")
 

Output:

log(-x^2 + x*log(5*log(x)) + x - 3*log(5*log(x)) + 9) - 2*log(x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (x \right )}{e^{3}}\right ) x -3 \,\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (x \right )}{e^{3}}\right )-x^{2}+4 x \right )-2 \,\mathrm {log}\left (x \right ) \] Input:

int(((6-x)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*log(x)*l 
og(5*log(x)/exp(3))+(-x^3+4*x^2)*log(x)),x)
 

Output:

log(log((5*log(x))/e**3)*x - 3*log((5*log(x))/e**3) - x**2 + 4*x) - 2*log( 
x)