\(\int \frac {(-2916 x-1458 x^2) \log (4+4 x+x^2)-108 x \log (\log (4+4 x+x^2))+(-54-27 x) \log (4+4 x+x^2) \log ^2(\log (4+4 x+x^2))}{(1458 x^4+729 x^5) \log (4+4 x+x^2)+(108 x^3+54 x^4) \log (4+4 x+x^2) \log ^2(\log (4+4 x+x^2))+(2 x^2+x^3) \log (4+4 x+x^2) \log ^4(\log (4+4 x+x^2))} \, dx\) [781]

Optimal result

Integrand size = 149, antiderivative size = 24 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {3}{x \left (3 x+\frac {1}{9} \log ^2\left (\log \left ((2+x)^2\right )\right )\right )} \] Output:

3/x/(3*x+1/9*ln(ln((2+x)^2))^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \left (27 x+\log ^2\left (\log \left ((2+x)^2\right )\right )\right )} \] Input:

Integrate[((-2916*x - 1458*x^2)*Log[4 + 4*x + x^2] - 108*x*Log[Log[4 + 4*x 
 + x^2]] + (-54 - 27*x)*Log[4 + 4*x + x^2]*Log[Log[4 + 4*x + x^2]]^2)/((14 
58*x^4 + 729*x^5)*Log[4 + 4*x + x^2] + (108*x^3 + 54*x^4)*Log[4 + 4*x + x^ 
2]*Log[Log[4 + 4*x + x^2]]^2 + (2*x^2 + x^3)*Log[4 + 4*x + x^2]*Log[Log[4 
+ 4*x + x^2]]^4),x]
 

Output:

27/(x*(27*x + Log[Log[(2 + x)^2]]^2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7239, 27, 25, 7238}

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.

\(\displaystyle \int \frac {(-27 x-54) \log \left (x^2+4 x+4\right ) \log ^2\left (\log \left (x^2+4 x+4\right )\right )-108 x \log \left (\log \left (x^2+4 x+4\right )\right )+\left (-1458 x^2-2916 x\right ) \log \left (x^2+4 x+4\right )}{\left (x^3+2 x^2\right ) \log \left (x^2+4 x+4\right ) \log ^4\left (\log \left (x^2+4 x+4\right )\right )+\left (729 x^5+1458 x^4\right ) \log \left (x^2+4 x+4\right )+\left (54 x^4+108 x^3\right ) \log \left (x^2+4 x+4\right ) \log ^2\left (\log \left (x^2+4 x+4\right )\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {27 \left (-(x+2) \log \left ((x+2)^2\right ) \left (54 x+\log ^2\left (\log \left ((x+2)^2\right )\right )\right )-4 x \log \left (\log \left ((x+2)^2\right )\right )\right )}{x^2 (x+2) \log \left ((x+2)^2\right ) \left (27 x+\log ^2\left (\log \left ((x+2)^2\right )\right )\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 27 \int -\frac {4 x \log \left (\log \left ((x+2)^2\right )\right )+(x+2) \log \left ((x+2)^2\right ) \left (\log ^2\left (\log \left ((x+2)^2\right )\right )+54 x\right )}{x^2 (x+2) \log \left ((x+2)^2\right ) \left (\log ^2\left (\log \left ((x+2)^2\right )\right )+27 x\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -27 \int \frac {4 x \log \left (\log \left ((x+2)^2\right )\right )+(x+2) \log \left ((x+2)^2\right ) \left (\log ^2\left (\log \left ((x+2)^2\right )\right )+54 x\right )}{x^2 (x+2) \log \left ((x+2)^2\right ) \left (\log ^2\left (\log \left ((x+2)^2\right )\right )+27 x\right )^2}dx\)

\(\Big \downarrow \) 7238

\(\displaystyle \frac {27}{x \left (27 x+\log ^2\left (\log \left ((x+2)^2\right )\right )\right )}\)

Input:

Int[((-2916*x - 1458*x^2)*Log[4 + 4*x + x^2] - 108*x*Log[Log[4 + 4*x + x^2 
]] + (-54 - 27*x)*Log[4 + 4*x + x^2]*Log[Log[4 + 4*x + x^2]]^2)/((1458*x^4 
 + 729*x^5)*Log[4 + 4*x + x^2] + (108*x^3 + 54*x^4)*Log[4 + 4*x + x^2]*Log 
[Log[4 + 4*x + x^2]]^2 + (2*x^2 + x^3)*Log[4 + 4*x + x^2]*Log[Log[4 + 4*x 
+ x^2]]^4),x]
 

Output:

27/(x*(27*x + Log[Log[(2 + x)^2]]^2))
 

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 7238

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

rule 7239

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

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {27}{\left ({\ln \left (\ln \left (x^{2}+4 x +4\right )\right )}^{2}+27 x \right ) x}\)	\(24\)

Input:

int(((-27*x-54)*ln(x^2+4*x+4)*ln(ln(x^2+4*x+4))^2-108*x*ln(ln(x^2+4*x+4))+ 
(-1458*x^2-2916*x)*ln(x^2+4*x+4))/((x^3+2*x^2)*ln(x^2+4*x+4)*ln(ln(x^2+4*x 
+4))^4+(54*x^4+108*x^3)*ln(x^2+4*x+4)*ln(ln(x^2+4*x+4))^2+(729*x^5+1458*x^ 
4)*ln(x^2+4*x+4)),x,method=_RETURNVERBOSE)
 

Output:

27/(ln(ln(x^2+4*x+4))^2+27*x)/x
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (\log \left (x^{2} + 4 \, x + 4\right )\right )^{2} + 27 \, x^{2}} \] Input:

integrate(((-27*x-54)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^2-108*x*log(log(x 
^2+4*x+4))+(-1458*x^2-2916*x)*log(x^2+4*x+4))/((x^3+2*x^2)*log(x^2+4*x+4)* 
log(log(x^2+4*x+4))^4+(54*x^4+108*x^3)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^ 
2+(729*x^5+1458*x^4)*log(x^2+4*x+4)),x, algorithm="fricas")
 

Output:

27/(x*log(log(x^2 + 4*x + 4))^2 + 27*x^2)
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{27 x^{2} + x \log {\left (\log {\left (x^{2} + 4 x + 4 \right )} \right )}^{2}} \] Input:

integrate(((-27*x-54)*ln(x**2+4*x+4)*ln(ln(x**2+4*x+4))**2-108*x*ln(ln(x** 
2+4*x+4))+(-1458*x**2-2916*x)*ln(x**2+4*x+4))/((x**3+2*x**2)*ln(x**2+4*x+4 
)*ln(ln(x**2+4*x+4))**4+(54*x**4+108*x**3)*ln(x**2+4*x+4)*ln(ln(x**2+4*x+4 
))**2+(729*x**5+1458*x**4)*ln(x**2+4*x+4)),x)
 

Output:

27/(27*x**2 + x*log(log(x**2 + 4*x + 4))**2)
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (2\right )^{2} + 2 \, x \log \left (2\right ) \log \left (\log \left (x + 2\right )\right ) + x \log \left (\log \left (x + 2\right )\right )^{2} + 27 \, x^{2}} \] Input:

integrate(((-27*x-54)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^2-108*x*log(log(x 
^2+4*x+4))+(-1458*x^2-2916*x)*log(x^2+4*x+4))/((x^3+2*x^2)*log(x^2+4*x+4)* 
log(log(x^2+4*x+4))^4+(54*x^4+108*x^3)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^ 
2+(729*x^5+1458*x^4)*log(x^2+4*x+4)),x, algorithm="maxima")
 

Output:

27/(x*log(2)^2 + 2*x*log(2)*log(log(x + 2)) + x*log(log(x + 2))^2 + 27*x^2 
)
 

Giac [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \log \left (\log \left (x^{2} + 4 \, x + 4\right )\right )^{2} + 27 \, x^{2}} \] Input:

integrate(((-27*x-54)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^2-108*x*log(log(x 
^2+4*x+4))+(-1458*x^2-2916*x)*log(x^2+4*x+4))/((x^3+2*x^2)*log(x^2+4*x+4)* 
log(log(x^2+4*x+4))^4+(54*x^4+108*x^3)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^ 
2+(729*x^5+1458*x^4)*log(x^2+4*x+4)),x, algorithm="giac")
 

Output:

27/(x*log(log(x^2 + 4*x + 4))^2 + 27*x^2)
 

Mupad [B] (verification not implemented)

Time = 3.35 (sec) , antiderivative size = 199, normalized size of antiderivative = 8.29 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27\,{\left (2\,\ln \left (x^2+4\,x+4\right )+x\,\ln \left (x^2+4\,x+4\right )\right )}^2\,\left (27\,x^2\,{\ln \left (x^2+4\,x+4\right )}^2+108\,x\,{\ln \left (x^2+4\,x+4\right )}^2+16\,x+108\,{\ln \left (x^2+4\,x+4\right )}^2\right )}{x\,\ln \left (x^2+4\,x+4\right )\,\left ({\ln \left (\ln \left (x^2+4\,x+4\right )\right )}^2+27\,x\right )\,\left (x+2\right )\,\left (27\,x^3\,{\ln \left (x^2+4\,x+4\right )}^3+162\,x^2\,{\ln \left (x^2+4\,x+4\right )}^3+16\,x^2\,\ln \left (x^2+4\,x+4\right )+324\,x\,{\ln \left (x^2+4\,x+4\right )}^3+32\,x\,\ln \left (x^2+4\,x+4\right )+216\,{\ln \left (x^2+4\,x+4\right )}^3\right )} \] Input:

int(-(log(4*x + x^2 + 4)*(2916*x + 1458*x^2) + 108*x*log(log(4*x + x^2 + 4 
)) + log(4*x + x^2 + 4)*log(log(4*x + x^2 + 4))^2*(27*x + 54))/(log(4*x + 
x^2 + 4)*(1458*x^4 + 729*x^5) + log(4*x + x^2 + 4)*log(log(4*x + x^2 + 4)) 
^4*(2*x^2 + x^3) + log(4*x + x^2 + 4)*log(log(4*x + x^2 + 4))^2*(108*x^3 + 
 54*x^4)),x)
 

Output:

(27*(2*log(4*x + x^2 + 4) + x*log(4*x + x^2 + 4))^2*(16*x + 27*x^2*log(4*x 
 + x^2 + 4)^2 + 108*x*log(4*x + x^2 + 4)^2 + 108*log(4*x + x^2 + 4)^2))/(x 
*log(4*x + x^2 + 4)*(27*x + log(log(4*x + x^2 + 4))^2)*(x + 2)*(162*x^2*lo 
g(4*x + x^2 + 4)^3 + 27*x^3*log(4*x + x^2 + 4)^3 + 32*x*log(4*x + x^2 + 4) 
 + 16*x^2*log(4*x + x^2 + 4) + 324*x*log(4*x + x^2 + 4)^3 + 216*log(4*x + 
x^2 + 4)^3))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-2916 x-1458 x^2\right ) \log \left (4+4 x+x^2\right )-108 x \log \left (\log \left (4+4 x+x^2\right )\right )+(-54-27 x) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )}{\left (1458 x^4+729 x^5\right ) \log \left (4+4 x+x^2\right )+\left (108 x^3+54 x^4\right ) \log \left (4+4 x+x^2\right ) \log ^2\left (\log \left (4+4 x+x^2\right )\right )+\left (2 x^2+x^3\right ) \log \left (4+4 x+x^2\right ) \log ^4\left (\log \left (4+4 x+x^2\right )\right )} \, dx=\frac {27}{x \left ({\mathrm {log}\left (\mathrm {log}\left (x^{2}+4 x +4\right )\right )}^{2}+27 x \right )} \] Input:

int(((-27*x-54)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^2-108*x*log(log(x^2+4*x 
+4))+(-1458*x^2-2916*x)*log(x^2+4*x+4))/((x^3+2*x^2)*log(x^2+4*x+4)*log(lo 
g(x^2+4*x+4))^4+(54*x^4+108*x^3)*log(x^2+4*x+4)*log(log(x^2+4*x+4))^2+(729 
*x^5+1458*x^4)*log(x^2+4*x+4)),x)
 

Output:

27/(x*(log(log(x**2 + 4*x + 4))**2 + 27*x))