\(\int \frac {x-x \log (x)+(-4 x^6 \log (x)+4 x^5 \log ^2(x)) \log (\frac {-5 x+5 \log (x)}{\log (x)})+(x \log (x)-\log ^2(x)) \log (\frac {-5 x+5 \log (x)}{\log (x)}) \log (\log (\frac {-5 x+5 \log (x)}{\log (x)}))}{(-x^3 \log (x)+x^2 \log ^2(x)) \log (\frac {-5 x+5 \log (x)}{\log (x)})} \, dx\) [828]

Optimal result

Integrand size = 115, antiderivative size = 21 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=x^4+\frac {\log \left (\log \left (5 \left (1-\frac {x}{\log (x)}\right )\right )\right )}{x} \] Output:

x^4+ln(ln(5-5*x/ln(x)))/x
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=x^4+\frac {\log \left (\log \left (5-\frac {5 x}{\log (x)}\right )\right )}{x} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 15.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(\frac {2 x^{5}+2 \ln \left (\ln \left (\frac {5 \ln \left (x \right )-5 x}{\ln \left (x \right )}\right )\right )}{2 x}\)	\(28\)
risch	\(\frac {\ln \left (\ln \left (5\right )+i \pi -\ln \left (\ln \left (x \right )\right )+\ln \left (x -\ln \left (x \right )\right )+\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )-x \right )}{\ln \left (x \right )}\right ) \left (\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )-x \right )}{\ln \left (x \right )}\right )+\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )\right ) \left (\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )-x \right )}{\ln \left (x \right )}\right )-\operatorname {csgn}\left (i \left (\ln \left (x \right )-x \right )\right )\right )}{2}+i \pi \operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )-x \right )}{\ln \left (x \right )}\right )^{2} \left (-\operatorname {csgn}\left (\frac {i \left (\ln \left (x \right )-x \right )}{\ln \left (x \right )}\right )-1\right )\right )}{x}+x^{4}\)	\(135\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=\frac {x^{5} + \log \left (\log \left (-\frac {5 \, {\left (x - \log \left (x\right )\right )}}{\log \left (x\right )}\right )\right )}{x} \] Input:

integrate(((-log(x)^2+x*log(x))*log((5*log(x)-5*x)/log(x))*log(log((5*log( 
x)-5*x)/log(x)))+(4*x^5*log(x)^2-4*x^6*log(x))*log((5*log(x)-5*x)/log(x))+ 
x-x*log(x))/(x^2*log(x)^2-x^3*log(x))/log((5*log(x)-5*x)/log(x)),x, algori 
thm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=x^{4} + \frac {\log {\left (\log {\left (\frac {- 5 x + 5 \log {\left (x \right )}}{\log {\left (x \right )}} \right )} \right )}}{x} \] Input:

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

Output:

x**4 + log(log((-5*x + 5*log(x))/log(x)))/x
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=\frac {x^{5} + \log \left (\log \left (5\right ) + \log \left (-x + \log \left (x\right )\right ) - \log \left (\log \left (x\right )\right )\right )}{x} \] Input:

integrate(((-log(x)^2+x*log(x))*log((5*log(x)-5*x)/log(x))*log(log((5*log( 
x)-5*x)/log(x)))+(4*x^5*log(x)^2-4*x^6*log(x))*log((5*log(x)-5*x)/log(x))+ 
x-x*log(x))/(x^2*log(x)^2-x^3*log(x))/log((5*log(x)-5*x)/log(x)),x, algori 
thm="maxima")
 

Output:

(x^5 + log(log(5) + log(-x + log(x)) - log(log(x))))/x
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=x^{4} + \frac {\log \left (\log \left (-5 \, x + 5 \, \log \left (x\right )\right ) - \log \left (\log \left (x\right )\right )\right )}{x} \] Input:

integrate(((-log(x)^2+x*log(x))*log((5*log(x)-5*x)/log(x))*log(log((5*log( 
x)-5*x)/log(x)))+(4*x^5*log(x)^2-4*x^6*log(x))*log((5*log(x)-5*x)/log(x))+ 
x-x*log(x))/(x^2*log(x)^2-x^3*log(x))/log((5*log(x)-5*x)/log(x)),x, algori 
thm="giac")
 

Output:

x^4 + log(log(-5*x + 5*log(x)) - log(log(x)))/x
 

Mupad [B] (verification not implemented)

Time = 3.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=\frac {\ln \left (\ln \left (-\frac {5\,\left (x-\ln \left (x\right )\right )}{\ln \left (x\right )}\right )\right )}{x}+x^4 \] Input:

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

Output:

log(log(-(5*(x - log(x)))/log(x)))/x + x^4
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {x-x \log (x)+\left (-4 x^6 \log (x)+4 x^5 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )+\left (x \log (x)-\log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right ) \log \left (\log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )\right )}{\left (-x^3 \log (x)+x^2 \log ^2(x)\right ) \log \left (\frac {-5 x+5 \log (x)}{\log (x)}\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (x \right )-5 x}{\mathrm {log}\left (x \right )}\right )\right )+x^{5}}{x} \] Input:

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

Output:

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