\(\int \frac {(5+(-2-x) \log (x)+(2+x) \log (\frac {x}{4+2 x})) \log (\log (5))}{(1250+525 x-48 x^2+x^3) \log (x)+(-1250-525 x+48 x^2-x^3) \log (\frac {x}{4+2 x})+((500+230 x-10 x^2) \log (x)+(-500-230 x+10 x^2) \log (\frac {x}{4+2 x})) \log (-\log (x)+\log (\frac {x}{4+2 x}))+((50+25 x) \log (x)+(-50-25 x) \log (\frac {x}{4+2 x})) \log ^2(-\log (x)+\log (\frac {x}{4+2 x}))} \, dx\) [383]

Optimal result

Integrand size = 169, antiderivative size = 28 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=\frac {\log (\log (5))}{x-5 \left (5+\log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )\right )} \] Output:

ln(ln(5))/(x-25-5*ln(-ln(x)+ln(x/(4+2*x))))
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=\frac {\log (\log (5))}{-25+x-5 \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \] Input:

Integrate[((5 + (-2 - x)*Log[x] + (2 + x)*Log[x/(4 + 2*x)])*Log[Log[5]])/( 
(1250 + 525*x - 48*x^2 + x^3)*Log[x] + (-1250 - 525*x + 48*x^2 - x^3)*Log[ 
x/(4 + 2*x)] + ((500 + 230*x - 10*x^2)*Log[x] + (-500 - 230*x + 10*x^2)*Lo 
g[x/(4 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]] + ((50 + 25*x)*Log[x] + (- 
50 - 25*x)*Log[x/(4 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]]^2),x]
 

Output:

Log[Log[5]]/(-25 + x - 5*Log[-Log[x] + Log[x/(4 + 2*x)]])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {27, 7239, 7237}

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.

\(\displaystyle \int \frac {\log (\log (5)) \left ((-x-2) \log (x)+(x+2) \log \left (\frac {x}{2 x+4}\right )+5\right )}{\left (\left (-10 x^2+230 x+500\right ) \log (x)+\left (10 x^2-230 x-500\right ) \log \left (\frac {x}{2 x+4}\right )\right ) \log \left (\log \left (\frac {x}{2 x+4}\right )-\log (x)\right )+\left (x^3-48 x^2+525 x+1250\right ) \log (x)+\left (-x^3+48 x^2-525 x-1250\right ) \log \left (\frac {x}{2 x+4}\right )+\left ((25 x+50) \log (x)+(-25 x-50) \log \left (\frac {x}{2 x+4}\right )\right ) \log ^2\left (\log \left (\frac {x}{2 x+4}\right )-\log (x)\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \log (\log (5)) \int \frac {-((x+2) \log (x))+(x+2) \log \left (\frac {x}{2 (x+2)}\right )+5}{25 \left ((x+2) \log (x)-(x+2) \log \left (\frac {x}{2 (x+2)}\right )\right ) \log ^2\left (\log \left (\frac {x}{2 (x+2)}\right )-\log (x)\right )+10 \left (\left (-x^2+23 x+50\right ) \log (x)-\left (-x^2+23 x+50\right ) \log \left (\frac {x}{2 (x+2)}\right )\right ) \log \left (\log \left (\frac {x}{2 (x+2)}\right )-\log (x)\right )+\left (x^3-48 x^2+525 x+1250\right ) \log (x)-\left (x^3-48 x^2+525 x+1250\right ) \log \left (\frac {x}{2 (x+2)}\right )}dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7237

\(\displaystyle -\frac {\log (\log (5))}{-x+5 \log \left (\log \left (\frac {x}{2 (x+2)}\right )-\log (x)\right )+25}\)

Input:

Int[((5 + (-2 - x)*Log[x] + (2 + x)*Log[x/(4 + 2*x)])*Log[Log[5]])/((1250 
+ 525*x - 48*x^2 + x^3)*Log[x] + (-1250 - 525*x + 48*x^2 - x^3)*Log[x/(4 + 
 2*x)] + ((500 + 230*x - 10*x^2)*Log[x] + (-500 - 230*x + 10*x^2)*Log[x/(4 
 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]] + ((50 + 25*x)*Log[x] + (-50 - 2 
5*x)*Log[x/(4 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]]^2),x]
 

Output:

-(Log[Log[5]]/(25 - x + 5*Log[-Log[x] + Log[x/(2*(2 + x))]]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7237

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si 
mp[q*(y^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 103.88 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.79

\[\frac {\ln \left (\ln \left (5\right )\right )}{x -5 \ln \left (-\ln \left (2\right )-\ln \left (2+x \right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x}{2+x}\right ) \left (-\operatorname {csgn}\left (\frac {i x}{2+x}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{2+x}\right )+\operatorname {csgn}\left (\frac {i}{2+x}\right )\right )}{2}\right )-25}\]

Input:

int(((-2-x)*ln(x)+ln(x/(4+2*x))*(2+x)+5)*ln(ln(5))/(((25*x+50)*ln(x)+(-25* 
x-50)*ln(x/(4+2*x)))*ln(-ln(x)+ln(x/(4+2*x)))^2+((-10*x^2+230*x+500)*ln(x) 
+(10*x^2-230*x-500)*ln(x/(4+2*x)))*ln(-ln(x)+ln(x/(4+2*x)))+(x^3-48*x^2+52 
5*x+1250)*ln(x)+(-x^3+48*x^2-525*x-1250)*ln(x/(4+2*x))),x)
 

Output:

ln(ln(5))/(x-5*ln(-ln(2)-ln(2+x)-1/2*I*Pi*csgn(I*x/(2+x))*(-csgn(I*x/(2+x) 
)+csgn(I*x))*(-csgn(I*x/(2+x))+csgn(I/(2+x))))-25)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=\frac {\log \left (\log \left (5\right )\right )}{x - 5 \, \log \left (-\log \left (x\right ) + \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right )\right ) - 25} \] Input:

integrate(((-2-x)*log(x)+log(x/(4+2*x))*(2+x)+5)*log(log(5))/(((25*x+50)*l 
og(x)+(-25*x-50)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2*x)))^2+((-10*x^2+2 
30*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2 
*x)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(4+2*x 
))),x, algorithm="fricas")
 

Output:

log(log(5))/(x - 5*log(-log(x) + log(1/2*x/(x + 2))) - 25)
 

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=- \frac {\log {\left (\log {\left (5 \right )} \right )}}{- x + 5 \log {\left (- \log {\left (x \right )} + \log {\left (\frac {x}{2 x + 4} \right )} \right )} + 25} \] Input:

integrate(((-2-x)*ln(x)+ln(x/(4+2*x))*(2+x)+5)*ln(ln(5))/(((25*x+50)*ln(x) 
+(-25*x-50)*ln(x/(4+2*x)))*ln(-ln(x)+ln(x/(4+2*x)))**2+((-10*x**2+230*x+50 
0)*ln(x)+(10*x**2-230*x-500)*ln(x/(4+2*x)))*ln(-ln(x)+ln(x/(4+2*x)))+(x**3 
-48*x**2+525*x+1250)*ln(x)+(-x**3+48*x**2-525*x-1250)*ln(x/(4+2*x))),x)
                                                                                    
                                                                                    
 

Output:

-log(log(5))/(-x + 5*log(-log(x) + log(x/(2*x + 4))) + 25)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=\frac {\log \left (\log \left (5\right )\right )}{x - 5 \, \log \left (-\log \left (2\right ) - \log \left (x + 2\right )\right ) - 25} \] Input:

integrate(((-2-x)*log(x)+log(x/(4+2*x))*(2+x)+5)*log(log(5))/(((25*x+50)*l 
og(x)+(-25*x-50)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2*x)))^2+((-10*x^2+2 
30*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2 
*x)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(4+2*x 
))),x, algorithm="maxima")
 

Output:

log(log(5))/(x - 5*log(-log(2) - log(x + 2)) - 25)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=\frac {\log \left (\log \left (5\right )\right )}{x - 5 \, \log \left (-\log \left (2\right ) - \log \left (x + 2\right )\right ) - 25} \] Input:

integrate(((-2-x)*log(x)+log(x/(4+2*x))*(2+x)+5)*log(log(5))/(((25*x+50)*l 
og(x)+(-25*x-50)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2*x)))^2+((-10*x^2+2 
30*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2 
*x)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(4+2*x 
))),x, algorithm="giac")
 

Output:

log(log(5))/(x - 5*log(-log(2) - log(x + 2)) - 25)
 

Mupad [B] (verification not implemented)

Time = 2.96 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=-\frac {\ln \left (\ln \left (5\right )\right )}{5\,\ln \left (\ln \left (\frac {x}{2\,x+4}\right )-\ln \left (x\right )\right )-x+25} \] Input:

int(-(log(log(5))*(log(x/(2*x + 4))*(x + 2) - log(x)*(x + 2) + 5))/(log(lo 
g(x/(2*x + 4)) - log(x))*(log(x/(2*x + 4))*(230*x - 10*x^2 + 500) - log(x) 
*(230*x - 10*x^2 + 500)) + log(x/(2*x + 4))*(525*x - 48*x^2 + x^3 + 1250) 
- log(x)*(525*x - 48*x^2 + x^3 + 1250) + log(log(x/(2*x + 4)) - log(x))^2* 
(log(x/(2*x + 4))*(25*x + 50) - log(x)*(25*x + 50))),x)
 

Output:

-log(log(5))/(5*log(log(x/(2*x + 4)) - log(x)) - x + 25)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {\left (5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )\right ) \log (\log (5))}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx=-\frac {\mathrm {log}\left (\mathrm {log}\left (5\right )\right )}{5 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{2 x +4}\right )-\mathrm {log}\left (x \right )\right )-x +25} \] Input:

int(((-2-x)*log(x)+log(x/(4+2*x))*(2+x)+5)*log(log(5))/(((25*x+50)*log(x)+ 
(-25*x-50)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2*x)))^2+((-10*x^2+230*x+5 
00)*log(x)+(10*x^2-230*x-500)*log(x/(4+2*x)))*log(-log(x)+log(x/(4+2*x)))+ 
(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(4+2*x))),x)
 

Output:

( - log(log(5)))/(5*log(log(x/(2*x + 4)) - log(x)) - x + 25)