\(\int \frac {-25+20 x^2+(5 x-4 x^3) \log (5)+(20 x^2-4 x^3 \log (5)) \log ^2(x)+(-5+x \log (5)) \log (\frac {5-x \log (5)}{\log (5)})+\log (x) (-25-20 x^2+(4 x+4 x^3) \log (5)+(-5+x \log (5)) \log (\frac {5-x \log (5)}{\log (5)}))}{(60 x^2-20 e^3 x^2+20 x^3+(-12 x^3+4 e^3 x^3-4 x^4) \log (5)) \log ^2(x)+\log (x) (25 x-20 x^3+(-5 x^2+4 x^4) \log (5)+(5 x-x^2 \log (5)) \log (\frac {5-x \log (5)}{\log (5)}))} \, dx\) [1014]

Optimal result

Integrand size = 200, antiderivative size = 36 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=\log \left (-3+e^3-x+\frac {x-\frac {5+\log \left (-x+\frac {5}{\log (5)}\right )}{4 x}}{\log (x)}\right ) \] Output:

ln(exp(3)+(x-1/4*(5+ln(5/ln(5)-x))/x)/ln(x)-3-x)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.36 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=-\log (x)-\log (\log (x))+\log \left (5-4 x^2+12 x \log (x)-4 e^3 x \log (x)+4 x^2 \log (x)+\log \left (-x+\frac {5}{\log (5)}\right )\right ) \] Input:

Integrate[(-25 + 20*x^2 + (5*x - 4*x^3)*Log[5] + (20*x^2 - 4*x^3*Log[5])*L 
og[x]^2 + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]] + Log[x]*(-25 - 20*x^ 
2 + (4*x + 4*x^3)*Log[5] + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]]))/(( 
60*x^2 - 20*E^3*x^2 + 20*x^3 + (-12*x^3 + 4*E^3*x^3 - 4*x^4)*Log[5])*Log[x 
]^2 + Log[x]*(25*x - 20*x^3 + (-5*x^2 + 4*x^4)*Log[5] + (5*x - x^2*Log[5]) 
*Log[(5 - x*Log[5])/Log[5]])),x]
 

Output:

-Log[x] - Log[Log[x]] + Log[5 - 4*x^2 + 12*x*Log[x] - 4*E^3*x*Log[x] + 4*x 
^2*Log[x] + Log[-x + 5/Log[5]]]
 

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.

\(\displaystyle \int \frac {\left (5 x-4 x^3\right ) \log (5)+20 x^2+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+\log (x) \left (\left (4 x^3+4 x\right ) \log (5)-20 x^2+(x \log (5)-5) \log \left (\frac {5-x \log (5)}{\log (5)}\right )-25\right )+(x \log (5)-5) \log \left (\frac {5-x \log (5)}{\log (5)}\right )-25}{\left (20 x^3-20 e^3 x^2+60 x^2+\left (-4 x^4+4 e^3 x^3-12 x^3\right ) \log (5)\right ) \log ^2(x)+\left (-20 x^3+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\left (4 x^4-5 x^2\right ) \log (5)+25 x\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (5 x-4 x^3\right ) \log (5)+20 x^2+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+\log (x) \left (\left (4 x^3+4 x\right ) \log (5)-20 x^2+(x \log (5)-5) \log \left (\frac {5-x \log (5)}{\log (5)}\right )-25\right )+(x \log (5)-5) \log \left (\frac {5-x \log (5)}{\log (5)}\right )-25}{x (5-x \log (5)) \log (x) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 x \log (x)}{-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5}+\frac {20 x}{(x \log (5)-5) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {20 x}{(5-x \log (5)) \log (x) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {4 \left (x^2+1\right ) \log (5)}{(5-x \log (5)) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {\left (5-4 x^2\right ) \log (5)}{(5-x \log (5)) \log (x) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {(-\log (x)-1) \log \left (\frac {5}{\log (5)}-x\right )}{x \log (x) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {25}{x (x \log (5)-5) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}+\frac {25}{x (x \log (5)-5) \log (x) \left (-4 x^2+4 x^2 \log (x)+12 \left (1-\frac {e^3}{3}\right ) x \log (x)+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\log (x)-\log (\log (x))+\frac {20 \int \frac {1}{-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5}dx}{\log (5)}+5 \int \frac {1}{x \left (-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5\right )}dx+4 \int \frac {x}{-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5}dx+\frac {20 \int \frac {1}{\log (x) \left (-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5\right )}dx}{\log (5)}+5 \int \frac {1}{x \log (x) \left (-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5\right )}dx+4 \int \frac {x}{\log (x) \left (-4 \log (x) x^2+4 x^2-12 \left (1-\frac {e^3}{3}\right ) \log (x) x-\log \left (\frac {5}{\log (5)}-x\right )-5\right )}dx+\frac {20 \int \frac {1}{4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5}dx}{\log (5)}+4 \left (3-e^3\right ) \int \frac {1}{4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5}dx+5 \int \frac {1}{x \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx+\frac {4 \left (25+\log ^2(5)\right ) \int \frac {1}{(5-x \log (5)) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx}{\log (5)}+5 \log (5) \int \frac {1}{(x \log (5)-5) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx+\frac {100 \int \frac {1}{(x \log (5)-5) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx}{\log (5)}+\frac {20 \int \frac {1}{\log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx}{\log (5)}+5 \int \frac {1}{x \log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx+4 \int \frac {x}{\log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx-\frac {5 \left (20-\log ^2(5)\right ) \int \frac {1}{(5-x \log (5)) \log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx}{\log (5)}+\frac {100 \int \frac {1}{(5-x \log (5)) \log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx}{\log (5)}+5 \log (5) \int \frac {1}{(x \log (5)-5) \log (x) \left (4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5\right )}dx+4 \left (3-e^3\right ) \int \frac {\log (x)}{4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5}dx+8 \int \frac {x \log (x)}{4 \log (x) x^2-4 x^2+12 \left (1-\frac {e^3}{3}\right ) \log (x) x+\log \left (\frac {5}{\log (5)}-x\right )+5}dx\)

Input:

Int[(-25 + 20*x^2 + (5*x - 4*x^3)*Log[5] + (20*x^2 - 4*x^3*Log[5])*Log[x]^ 
2 + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]] + Log[x]*(-25 - 20*x^2 + (4 
*x + 4*x^3)*Log[5] + (-5 + x*Log[5])*Log[(5 - x*Log[5])/Log[5]]))/((60*x^2 
 - 20*E^3*x^2 + 20*x^3 + (-12*x^3 + 4*E^3*x^3 - 4*x^4)*Log[5])*Log[x]^2 + 
Log[x]*(25*x - 20*x^3 + (-5*x^2 + 4*x^4)*Log[5] + (5*x - x^2*Log[5])*Log[( 
5 - x*Log[5])/Log[5]])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.47

\[-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (-4 \ln \left (x \right ) {\mathrm e}^{3+\ln \left (x \right )}+4 x^{2} \ln \left (x \right )+12 x \ln \left (x \right )-4 x^{2}-\ln \left (\ln \left (5\right )\right )+\ln \left (-x \ln \left (5\right )+5\right )+5\right )\]

Input:

int(((-4*x^3*ln(5)+20*x^2)*ln(x)^2+((x*ln(5)-5)*ln((-x*ln(5)+5)/ln(5))+(4* 
x^3+4*x)*ln(5)-20*x^2-25)*ln(x)+(x*ln(5)-5)*ln((-x*ln(5)+5)/ln(5))+(-4*x^3 
+5*x)*ln(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*ln(5)-20*x^2*exp(3)+2 
0*x^3+60*x^2)*ln(x)^2+((-x^2*ln(5)+5*x)*ln((-x*ln(5)+5)/ln(5))+(4*x^4-5*x^ 
2)*ln(5)-20*x^3+25*x)*ln(x)),x)
 

Output:

-ln(x)-ln(ln(x))+ln(-4*ln(x)*exp(3+ln(x))+4*x^2*ln(x)+12*x*ln(x)-4*x^2-ln( 
ln(5))+ln(-x*ln(5)+5)+5)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.89 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=\log \left (x - e^{3} + 3\right ) + \log \left (\frac {4 \, x^{2} - 4 \, {\left (x^{2} - x e^{3} + 3 \, x\right )} \log \left (x\right ) - \log \left (-\frac {x \log \left (5\right ) - 5}{\log \left (5\right )}\right ) - 5}{x^{2} - x e^{3} + 3 \, x}\right ) - \log \left (\log \left (x\right )\right ) \] Input:

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5) 
/log(5))+(4*x^3+4*x)*log(5)-20*x^2-25)*log(x)+(x*log(5)-5)*log((-x*log(5)+ 
5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*lo 
g(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log( 
5)+5)/log(5))+(4*x^4-5*x^2)*log(5)-20*x^3+25*x)*log(x)),x, algorithm="fric 
as")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.47 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=- \log {\left (x \right )} + \log {\left (4 x^{2} \log {\left (x \right )} - 4 x^{2} - 4 x e^{3} \log {\left (x \right )} + 12 x \log {\left (x \right )} + \log {\left (\frac {- x \log {\left (5 \right )} + 5}{\log {\left (5 \right )}} \right )} + 5 \right )} - \log {\left (\log {\left (x \right )} \right )} \] Input:

integrate(((-4*x**3*ln(5)+20*x**2)*ln(x)**2+((x*ln(5)-5)*ln((-x*ln(5)+5)/l 
n(5))+(4*x**3+4*x)*ln(5)-20*x**2-25)*ln(x)+(x*ln(5)-5)*ln((-x*ln(5)+5)/ln( 
5))+(-4*x**3+5*x)*ln(5)+20*x**2-25)/(((4*x**3*exp(3)-4*x**4-12*x**3)*ln(5) 
-20*x**2*exp(3)+20*x**3+60*x**2)*ln(x)**2+((-x**2*ln(5)+5*x)*ln((-x*ln(5)+ 
5)/ln(5))+(4*x**4-5*x**2)*ln(5)-20*x**3+25*x)*ln(x)),x)
 

Output:

-log(x) + log(4*x**2*log(x) - 4*x**2 - 4*x*exp(3)*log(x) + 12*x*log(x) + l 
og((-x*log(5) + 5)/log(5)) + 5) - log(log(x))
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=\log \left (-4 \, x^{2} + 4 \, {\left (x^{2} - x {\left (e^{3} - 3\right )}\right )} \log \left (x\right ) + \log \left (-x \log \left (5\right ) + 5\right ) - \log \left (\log \left (5\right )\right ) + 5\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \] Input:

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5) 
/log(5))+(4*x^3+4*x)*log(5)-20*x^2-25)*log(x)+(x*log(5)-5)*log((-x*log(5)+ 
5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*lo 
g(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log( 
5)+5)/log(5))+(4*x^4-5*x^2)*log(5)-20*x^3+25*x)*log(x)),x, algorithm="maxi 
ma")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.39 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=\log \left (4 \, x^{2} \log \left (x\right ) - 4 \, x e^{3} \log \left (x\right ) - 4 \, x^{2} + 12 \, x \log \left (x\right ) + \log \left (-x \log \left (5\right ) + 5\right ) - \log \left (\log \left (5\right )\right ) + 5\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \] Input:

integrate(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5) 
/log(5))+(4*x^3+4*x)*log(5)-20*x^2-25)*log(x)+(x*log(5)-5)*log((-x*log(5)+ 
5)/log(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*lo 
g(5)-20*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log( 
5)+5)/log(5))+(4*x^4-5*x^2)*log(5)-20*x^3+25*x)*log(x)),x, algorithm="giac 
")
 

Output:

log(4*x^2*log(x) - 4*x*e^3*log(x) - 4*x^2 + 12*x*log(x) + log(-x*log(5) + 
5) - log(log(5)) + 5) - log(x) - log(log(x))
 

Mupad [B] (verification not implemented)

Time = 3.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.39 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=\ln \left (\ln \left (5-x\,\ln \left (5\right )\right )-\ln \left (\ln \left (5\right )\right )+4\,x^2\,\ln \left (x\right )+12\,x\,\ln \left (x\right )-4\,x^2-4\,x\,{\mathrm {e}}^3\,\ln \left (x\right )+5\right )-\ln \left (\ln \left (x\right )\right )-\ln \left (x\right ) \] Input:

int(-(log(5)*(5*x - 4*x^3) + log(-(x*log(5) - 5)/log(5))*(x*log(5) - 5) + 
log(x)*(log(5)*(4*x + 4*x^3) + log(-(x*log(5) - 5)/log(5))*(x*log(5) - 5) 
- 20*x^2 - 25) + 20*x^2 - log(x)^2*(4*x^3*log(5) - 20*x^2) - 25)/(log(x)^2 
*(log(5)*(12*x^3 - 4*x^3*exp(3) + 4*x^4) + 20*x^2*exp(3) - 60*x^2 - 20*x^3 
) - log(x)*(25*x + log(-(x*log(5) - 5)/log(5))*(5*x - x^2*log(5)) - log(5) 
*(5*x^2 - 4*x^4) - 20*x^3)),x)
                                                                                    
                                                                                    
 

Output:

log(log(5 - x*log(5)) - log(log(5)) + 4*x^2*log(x) + 12*x*log(x) - 4*x^2 - 
 4*x*exp(3)*log(x) + 5) - log(log(x)) - log(x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {-25+20 x^2+\left (5 x-4 x^3\right ) \log (5)+\left (20 x^2-4 x^3 \log (5)\right ) \log ^2(x)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )+\log (x) \left (-25-20 x^2+\left (4 x+4 x^3\right ) \log (5)+(-5+x \log (5)) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )}{\left (60 x^2-20 e^3 x^2+20 x^3+\left (-12 x^3+4 e^3 x^3-4 x^4\right ) \log (5)\right ) \log ^2(x)+\log (x) \left (25 x-20 x^3+\left (-5 x^2+4 x^4\right ) \log (5)+\left (5 x-x^2 \log (5)\right ) \log \left (\frac {5-x \log (5)}{\log (5)}\right )\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (\mathrm {log}\left (\frac {-\mathrm {log}\left (5\right ) x +5}{\mathrm {log}\left (5\right )}\right )-4 \,\mathrm {log}\left (x \right ) e^{3} x +4 \,\mathrm {log}\left (x \right ) x^{2}+12 \,\mathrm {log}\left (x \right ) x -4 x^{2}+5\right )-\mathrm {log}\left (x \right ) \] Input:

int(((-4*x^3*log(5)+20*x^2)*log(x)^2+((x*log(5)-5)*log((-x*log(5)+5)/log(5 
))+(4*x^3+4*x)*log(5)-20*x^2-25)*log(x)+(x*log(5)-5)*log((-x*log(5)+5)/log 
(5))+(-4*x^3+5*x)*log(5)+20*x^2-25)/(((4*x^3*exp(3)-4*x^4-12*x^3)*log(5)-2 
0*x^2*exp(3)+20*x^3+60*x^2)*log(x)^2+((-x^2*log(5)+5*x)*log((-x*log(5)+5)/ 
log(5))+(4*x^4-5*x^2)*log(5)-20*x^3+25*x)*log(x)),x)
 

Output:

 - log(log(x)) + log(log(( - log(5)*x + 5)/log(5)) - 4*log(x)*e**3*x + 4*l 
og(x)*x**2 + 12*log(x)*x - 4*x**2 + 5) - log(x)