\(\int \frac {-5-6 x^2-2 x^3-x^4+(5-5 x-x^2-x^3) \log (\frac {x+x \log (\log (5))}{\log (\log (5))})}{25 x^2+5 x^4+(25 x+5 x^3) \log (\frac {x+x \log (\log (5))}{\log (\log (5))})} \, dx\) [2174]

Optimal result

Integrand size = 85, antiderivative size = 34 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=\frac {1}{5} \left (-x+\log \left (\frac {20}{\left (\frac {5}{x}+x\right ) \left (x+\log \left (x+\frac {x}{\log (\log (5))}\right )\right )}\right )\right ) \] Output:

1/5*ln(20/(x+5/x)/(ln(x/ln(ln(5))+x)+x))-1/5*x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=\frac {1}{5} \left (-x+\log (x)-\log \left (5+x^2\right )-\log \left (x+\log \left (x \left (1+\frac {1}{\log (\log (5))}\right )\right )\right )\right ) \] Input:

Integrate[(-5 - 6*x^2 - 2*x^3 - x^4 + (5 - 5*x - x^2 - x^3)*Log[(x + x*Log 
[Log[5]])/Log[Log[5]]])/(25*x^2 + 5*x^4 + (25*x + 5*x^3)*Log[(x + x*Log[Lo 
g[5]])/Log[Log[5]]]),x]
 

Output:

(-x + Log[x] - Log[5 + x^2] - Log[x + Log[x*(1 + Log[Log[5]]^(-1))]])/5
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {7292, 27, 25, 7276, 2009}

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.

\(\displaystyle \int \frac {-x^4-2 x^3-6 x^2+\left (-x^3-x^2-5 x+5\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )-5}{5 x^4+\left (5 x^3+25 x\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )+25 x^2} \, dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-5 - 6*x^2 - 2*x^3 - x^4 + (5 - 5*x - x^2 - x^3)*Log[(x + x*Log[Log[5 
]])/Log[Log[5]]])/(25*x^2 + 5*x^4 + (25*x + 5*x^3)*Log[(x + x*Log[Log[5]]) 
/Log[Log[5]]]),x]
 

Output:

(-x + Log[x] - Log[5 + x^2] - Log[x + Log[x*(1 + Log[Log[5]]^(-1))]])/5
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {x}{5}+\frac {\ln \left (x \right )}{5}-\frac {\ln \left (x^{2}+5\right )}{5}-\frac {\ln \left (x +\ln \left (\frac {x \ln \left (\ln \left (5\right )\right )+x}{\ln \left (\ln \left (5\right )\right )}\right )\right )}{5}\)	\(36\)
parallelrisch	\(-\frac {x}{5}+\frac {\ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )}{5}-\frac {\ln \left (x +\ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )\right )}{5}-\frac {\ln \left (x^{2}+5\right )}{5}\)	\(46\)
norman	\(\frac {\ln \left (\frac {x \ln \left (\ln \left (5\right )\right )+x}{\ln \left (\ln \left (5\right )\right )}\right )}{5}-\frac {x}{5}-\frac {\ln \left (x +\ln \left (\frac {x \ln \left (\ln \left (5\right )\right )+x}{\ln \left (\ln \left (5\right )\right )}\right )\right )}{5}-\frac {\ln \left (x^{2}+5\right )}{5}\)	\(48\)
derivativedivides	\(-\frac {\left (-\ln \left (\ln \left (5\right )\right )-1\right ) \ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )+x \left (\ln \left (\ln \left (5\right )\right )+1\right )+\left (\ln \left (\ln \left (5\right )\right )+1\right ) \ln \left (\ln \left (\ln \left (5\right )\right ) \ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )+x \left (\ln \left (\ln \left (5\right )\right )+1\right )+\ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )\right )+\left (\ln \left (\ln \left (5\right )\right )+1\right ) \ln \left (x^{2} \left (\ln \left (\ln \left (5\right )\right )+1\right )^{2}+5 \ln \left (\ln \left (5\right )\right )^{2}+10 \ln \left (\ln \left (5\right )\right )+5\right )}{5 \left (\ln \left (\ln \left (5\right )\right )+1\right )}\)	\(116\)
default	\(-\frac {\left (-\ln \left (\ln \left (5\right )\right )-1\right ) \ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )+x \left (\ln \left (\ln \left (5\right )\right )+1\right )+\left (\ln \left (\ln \left (5\right )\right )+1\right ) \ln \left (\ln \left (\ln \left (5\right )\right ) \ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )+x \left (\ln \left (\ln \left (5\right )\right )+1\right )+\ln \left (\frac {x \left (\ln \left (\ln \left (5\right )\right )+1\right )}{\ln \left (\ln \left (5\right )\right )}\right )\right )+\left (\ln \left (\ln \left (5\right )\right )+1\right ) \ln \left (x^{2} \left (\ln \left (\ln \left (5\right )\right )+1\right )^{2}+5 \ln \left (\ln \left (5\right )\right )^{2}+10 \ln \left (\ln \left (5\right )\right )+5\right )}{5 \left (\ln \left (\ln \left (5\right )\right )+1\right )}\)	\(116\)

Input:

int(((-x^3-x^2-5*x+5)*ln((x*ln(ln(5))+x)/ln(ln(5)))-x^4-2*x^3-6*x^2-5)/((5 
*x^3+25*x)*ln((x*ln(ln(5))+x)/ln(ln(5)))+5*x^4+25*x^2),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=-\frac {1}{5} \, x - \frac {1}{5} \, \log \left (x^{2} + 5\right ) - \frac {1}{5} \, \log \left (x + \log \left (\frac {x \log \left (\log \left (5\right )\right ) + x}{\log \left (\log \left (5\right )\right )}\right )\right ) + \frac {1}{5} \, \log \left (x\right ) \] Input:

integrate(((-x^3-x^2-5*x+5)*log((x*log(log(5))+x)/log(log(5)))-x^4-2*x^3-6 
*x^2-5)/((5*x^3+25*x)*log((x*log(log(5))+x)/log(log(5)))+5*x^4+25*x^2),x, 
algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=- \frac {x}{5} + \frac {\log {\left (x \right )}}{5} - \frac {\log {\left (x + \log {\left (\frac {x \log {\left (\log {\left (5 \right )} \right )} + x}{\log {\left (\log {\left (5 \right )} \right )}} \right )} \right )}}{5} - \frac {\log {\left (x^{2} + 5 \right )}}{5} \] Input:

integrate(((-x**3-x**2-5*x+5)*ln((x*ln(ln(5))+x)/ln(ln(5)))-x**4-2*x**3-6* 
x**2-5)/((5*x**3+25*x)*ln((x*ln(ln(5))+x)/ln(ln(5)))+5*x**4+25*x**2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=-\frac {1}{5} \, x - \frac {1}{5} \, \log \left (x^{2} + 5\right ) - \frac {1}{5} \, \log \left (x + \log \left (x\right ) + \log \left (\log \left (\log \left (5\right )\right ) + 1\right ) - \log \left (\log \left (\log \left (5\right )\right )\right )\right ) + \frac {1}{5} \, \log \left (x\right ) \] Input:

integrate(((-x^3-x^2-5*x+5)*log((x*log(log(5))+x)/log(log(5)))-x^4-2*x^3-6 
*x^2-5)/((5*x^3+25*x)*log((x*log(log(5))+x)/log(log(5)))+5*x^4+25*x^2),x, 
algorithm="maxima")
 

Output:

-1/5*x - 1/5*log(x^2 + 5) - 1/5*log(x + log(x) + log(log(log(5)) + 1) - lo 
g(log(log(5)))) + 1/5*log(x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=-\frac {1}{5} \, x - \frac {1}{5} \, \log \left (x^{2} + 5\right ) - \frac {1}{5} \, \log \left (x + \log \left (x \log \left (\log \left (5\right )\right ) + x\right ) - \log \left (\log \left (\log \left (5\right )\right )\right )\right ) + \frac {1}{5} \, \log \left (x\right ) \] Input:

integrate(((-x^3-x^2-5*x+5)*log((x*log(log(5))+x)/log(log(5)))-x^4-2*x^3-6 
*x^2-5)/((5*x^3+25*x)*log((x*log(log(5))+x)/log(log(5)))+5*x^4+25*x^2),x, 
algorithm="giac")
 

Output:

-1/5*x - 1/5*log(x^2 + 5) - 1/5*log(x + log(x*log(log(5)) + x) - log(log(l 
og(5)))) + 1/5*log(x)
 

Mupad [B] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=\frac {\ln \left (x\right )}{5}-\frac {\ln \left (x-\ln \left (\ln \left (\ln \left (5\right )\right )\right )+\ln \left (\ln \left (\ln \left (5\right )\right )+1\right )+\ln \left (x\right )\right )}{5}-\frac {\ln \left (x^2+5\right )}{5}-\frac {x}{5} \] Input:

int(-(log((x + x*log(log(5)))/log(log(5)))*(5*x + x^2 + x^3 - 5) + 6*x^2 + 
 2*x^3 + x^4 + 5)/(log((x + x*log(log(5)))/log(log(5)))*(25*x + 5*x^3) + 2 
5*x^2 + 5*x^4),x)
 

Output:

log(x)/5 - log(x - log(log(log(5))) + log(log(log(5)) + 1) + log(x))/5 - l 
og(x^2 + 5)/5 - x/5
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38 \[ \int \frac {-5-6 x^2-2 x^3-x^4+\left (5-5 x-x^2-x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )}{25 x^2+5 x^4+\left (25 x+5 x^3\right ) \log \left (\frac {x+x \log (\log (5))}{\log (\log (5))}\right )} \, dx=-\frac {\mathrm {log}\left (x^{2}+5\right )}{5}-\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x +x}{\mathrm {log}\left (\mathrm {log}\left (5\right )\right )}\right )+x \right )}{5}+\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x +x}{\mathrm {log}\left (\mathrm {log}\left (5\right )\right )}\right )}{5}-\frac {x}{5} \] Input:

int(((-x^3-x^2-5*x+5)*log((x*log(log(5))+x)/log(log(5)))-x^4-2*x^3-6*x^2-5 
)/((5*x^3+25*x)*log((x*log(log(5))+x)/log(log(5)))+5*x^4+25*x^2),x)
 

Output:

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