\(\int \frac {3 x+3 x^2+(9+12 x+3 x^2) \log (3+x)+(-6 x^2-2 x^3) \log (3+x) \log (x \log (3+x))+(-9+6 x+3 x^2) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{(-12 x^2-16 x^3-4 x^4) \log (3+x) \log (x \log (3+x))+(36 x+48 x^2+12 x^3) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx\) [779]

Optimal result

Integrand size = 143, antiderivative size = 29 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{4} \log \left (\frac {5 (1+x)^2 (-x+3 \log (\log (x \log (3+x))))}{x}\right ) \] Output:

1/4*ln(5*(1+x)^2*(3*ln(ln(x*ln(3+x)))-x)/x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{4} (-\log (x)+2 \log (1+x)+\log (x-3 \log (\log (x \log (3+x))))) \] Input:

Integrate[(3*x + 3*x^2 + (9 + 12*x + 3*x^2)*Log[3 + x] + (-6*x^2 - 2*x^3)* 
Log[3 + x]*Log[x*Log[3 + x]] + (-9 + 6*x + 3*x^2)*Log[3 + x]*Log[x*Log[3 + 
 x]]*Log[Log[x*Log[3 + x]]])/((-12*x^2 - 16*x^3 - 4*x^4)*Log[3 + x]*Log[x* 
Log[3 + x]] + (36*x + 48*x^2 + 12*x^3)*Log[3 + x]*Log[x*Log[3 + x]]*Log[Lo 
g[x*Log[3 + x]]]),x]
 

Output:

(-Log[x] + 2*Log[1 + x] + Log[x - 3*Log[Log[x*Log[3 + x]]]])/4
 

Rubi [A] (verified)

Time = 3.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7292, 27, 25, 7279, 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.

\(\displaystyle \int \frac {3 x^2+\left (3 x^2+12 x+9\right ) \log (x+3)+\left (3 x^2+6 x-9\right ) \log (x+3) \log (x \log (x+3)) \log (\log (x \log (x+3)))+\left (-2 x^3-6 x^2\right ) \log (x+3) \log (x \log (x+3))+3 x}{\left (12 x^3+48 x^2+36 x\right ) \log (x+3) \log (\log (x \log (x+3))) \log (x \log (x+3))+\left (-4 x^4-16 x^3-12 x^2\right ) \log (x+3) \log (x \log (x+3))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-3 x^2-\left (3 x^2+12 x+9\right ) \log (x+3)-\left (3 x^2+6 x-9\right ) \log (x+3) \log (x \log (x+3)) \log (\log (x \log (x+3)))-\left (-2 x^3-6 x^2\right ) \log (x+3) \log (x \log (x+3))-3 x}{4 x \left (x^2+4 x+3\right ) \log (x+3) \log (x \log (x+3)) (x-3 \log (\log (x \log (x+3))))}dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7279

\(\displaystyle -\frac {1}{4} \int \left (\frac {1-x}{x (x+1)}+\frac {-\log (x+3) \log (x \log (x+3)) x^2+3 \log (x+3) x-3 \log (x+3) \log (x \log (x+3)) x+3 x+9 \log (x+3)}{x (x+3) \log (x+3) \log (x \log (x+3)) (x-3 \log (\log (x \log (x+3))))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} (-\log (x)+2 \log (x+1)+\log (x-3 \log (\log (x \log (x+3)))))\)

Input:

Int[(3*x + 3*x^2 + (9 + 12*x + 3*x^2)*Log[3 + x] + (-6*x^2 - 2*x^3)*Log[3 
+ x]*Log[x*Log[3 + x]] + (-9 + 6*x + 3*x^2)*Log[3 + x]*Log[x*Log[3 + x]]*L 
og[Log[x*Log[3 + x]]])/((-12*x^2 - 16*x^3 - 4*x^4)*Log[3 + x]*Log[x*Log[3 
+ x]] + (36*x + 48*x^2 + 12*x^3)*Log[3 + x]*Log[x*Log[3 + x]]*Log[Log[x*Lo 
g[3 + x]]]),x]
 

Output:

(-Log[x] + 2*Log[1 + x] + Log[x - 3*Log[Log[x*Log[3 + x]]]])/4
 

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 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

rule 7292

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

Maple [A] (verified)

Time = 60.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (1+x \right )}{2}+\frac {\ln \left (x -3 \ln \left (\ln \left (x \ln \left (3+x \right )\right )\right )\right )}{4}\)	\(27\)
risch	\(-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (1+x \right )}{2}+\frac {\ln \left (-\frac {x}{3}+\ln \left (\ln \left (x \right )+\ln \left (\ln \left (3+x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \ln \left (3+x \right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left (3+x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left (3+x \right )\right )+\operatorname {csgn}\left (i \ln \left (3+x \right )\right )\right )}{2}\right )\right )}{4}\)	\(78\)

Input:

int(((3*x^2+6*x-9)*ln(3+x)*ln(x*ln(3+x))*ln(ln(x*ln(3+x)))+(-2*x^3-6*x^2)* 
ln(3+x)*ln(x*ln(3+x))+(3*x^2+12*x+9)*ln(3+x)+3*x^2+3*x)/((12*x^3+48*x^2+36 
*x)*ln(3+x)*ln(x*ln(3+x))*ln(ln(x*ln(3+x)))+(-4*x^4-16*x^3-12*x^2)*ln(3+x) 
*ln(x*ln(3+x))),x,method=_RETURNVERBOSE)
 

Output:

-1/4*ln(x)+1/2*ln(1+x)+1/4*ln(x-3*ln(ln(x*ln(3+x))))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{4} \, \log \left (-x + 3 \, \log \left (\log \left (x \log \left (x + 3\right )\right )\right )\right ) \] Input:

integrate(((3*x^2+6*x-9)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-2 
*x^3-6*x^2)*log(3+x)*log(x*log(3+x))+(3*x^2+12*x+9)*log(3+x)+3*x^2+3*x)/(( 
12*x^3+48*x^2+36*x)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-4*x^4- 
16*x^3-12*x^2)*log(3+x)*log(x*log(3+x))),x, algorithm="fricas")
 

Output:

1/2*log(x + 1) - 1/4*log(x) + 1/4*log(-x + 3*log(log(x*log(x + 3))))
 

Sympy [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=- \frac {\log {\left (x \right )}}{4} + \frac {\log {\left (- \frac {x}{3} + \log {\left (\log {\left (x \log {\left (x + 3 \right )} \right )} \right )} \right )}}{4} + \frac {\log {\left (x + 1 \right )}}{2} \] Input:

integrate(((3*x**2+6*x-9)*ln(3+x)*ln(x*ln(3+x))*ln(ln(x*ln(3+x)))+(-2*x**3 
-6*x**2)*ln(3+x)*ln(x*ln(3+x))+(3*x**2+12*x+9)*ln(3+x)+3*x**2+3*x)/((12*x* 
*3+48*x**2+36*x)*ln(3+x)*ln(x*ln(3+x))*ln(ln(x*ln(3+x)))+(-4*x**4-16*x**3- 
12*x**2)*ln(3+x)*ln(x*ln(3+x))),x)
 

Output:

-log(x)/4 + log(-x/3 + log(log(x*log(x + 3))))/4 + log(x + 1)/2
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{4} \, \log \left (-\frac {1}{3} \, x + \log \left (\log \left (x\right ) + \log \left (\log \left (x + 3\right )\right )\right )\right ) \] Input:

integrate(((3*x^2+6*x-9)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-2 
*x^3-6*x^2)*log(3+x)*log(x*log(3+x))+(3*x^2+12*x+9)*log(3+x)+3*x^2+3*x)/(( 
12*x^3+48*x^2+36*x)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-4*x^4- 
16*x^3-12*x^2)*log(3+x)*log(x*log(3+x))),x, algorithm="maxima")
 

Output:

1/2*log(x + 1) - 1/4*log(x) + 1/4*log(-1/3*x + log(log(x) + log(log(x + 3) 
)))
 

Giac [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{4} \, \log \left (x - 3 \, \log \left (\log \left (x \log \left (x + 3\right )\right )\right )\right ) + \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) \] Input:

integrate(((3*x^2+6*x-9)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-2 
*x^3-6*x^2)*log(3+x)*log(x*log(3+x))+(3*x^2+12*x+9)*log(3+x)+3*x^2+3*x)/(( 
12*x^3+48*x^2+36*x)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-4*x^4- 
16*x^3-12*x^2)*log(3+x)*log(x*log(3+x))),x, algorithm="giac")
 

Output:

1/4*log(x - 3*log(log(x*log(x + 3)))) + 1/2*log(x + 1) - 1/4*log(x)
 

Mupad [B] (verification not implemented)

Time = 3.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {\ln \left (x+1\right )}{2}+\frac {\ln \left (\ln \left (\ln \left (x\,\ln \left (x+3\right )\right )\right )-\frac {x}{3}\right )}{4}-\frac {\ln \left (x\right )}{4} \] Input:

int(-(3*x + log(x + 3)*(12*x + 3*x^2 + 9) + 3*x^2 - log(x + 3)*log(x*log(x 
 + 3))*(6*x^2 + 2*x^3) + log(x + 3)*log(x*log(x + 3))*log(log(x*log(x + 3) 
))*(6*x + 3*x^2 - 9))/(log(x + 3)*log(x*log(x + 3))*(12*x^2 + 16*x^3 + 4*x 
^4) - log(x + 3)*log(x*log(x + 3))*log(log(x*log(x + 3)))*(36*x + 48*x^2 + 
 12*x^3)),x)
 

Output:

log(x + 1)/2 + log(log(log(x*log(x + 3))) - x/3)/4 - log(x)/4
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {\mathrm {log}\left (3 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x +3\right ) x \right )\right )-x \right )}{4}+\frac {\mathrm {log}\left (x +1\right )}{2}-\frac {\mathrm {log}\left (x \right )}{4} \] Input:

int(((3*x^2+6*x-9)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-2*x^3-6 
*x^2)*log(3+x)*log(x*log(3+x))+(3*x^2+12*x+9)*log(3+x)+3*x^2+3*x)/((12*x^3 
+48*x^2+36*x)*log(3+x)*log(x*log(3+x))*log(log(x*log(3+x)))+(-4*x^4-16*x^3 
-12*x^2)*log(3+x)*log(x*log(3+x))),x)
 

Output:

(log(3*log(log(log(x + 3)*x)) - x) + 2*log(x + 1) - log(x))/4