\(\int \frac {3 x+3 x^2+(9+12 x+3 x^2) \log (3+x)+(-6 x^2-2 x^3) \log (3+x) \log (x \log (3+x))+(-9+6 x+3 x^2) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{(-12 x^2-16 x^3-4 x^4) \log (3+x) \log (x \log (3+x))+(36 x+48 x^2+12 x^3) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx\) [6779]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 6860, 78, 6816} \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=-\frac {\log (x)}{4}+\frac {1}{2} \log (x+1)+\frac {1}{4} \log (x-3 \log (\log (x \log (x+3)))) \]

[In]

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]]*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]]]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

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

Rule 6860

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] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 x (1+x)+(3+x) \log (3+x) \left (-3 (1+x)+\log (x \log (3+x)) \left (2 x^2-3 (-1+x) \log (\log (x \log (3+x)))\right )\right )}{4 x \left (3+4 x+x^2\right ) \log (3+x) \log (x \log (3+x)) (x-3 \log (\log (x \log (3+x))))} \, dx \\ & = \frac {1}{4} \int \frac {-3 x (1+x)+(3+x) \log (3+x) \left (-3 (1+x)+\log (x \log (3+x)) \left (2 x^2-3 (-1+x) \log (\log (x \log (3+x)))\right )\right )}{x \left (3+4 x+x^2\right ) \log (3+x) \log (x \log (3+x)) (x-3 \log (\log (x \log (3+x))))} \, dx \\ & = \frac {1}{4} \int \left (\frac {-1+x}{x (1+x)}+\frac {-3 x-9 \log (3+x)-3 x \log (3+x)+3 x \log (3+x) \log (x \log (3+x))+x^2 \log (3+x) \log (x \log (3+x))}{x (3+x) \log (3+x) \log (x \log (3+x)) (x-3 \log (\log (x \log (3+x))))}\right ) \, dx \\ & = \frac {1}{4} \int \frac {-1+x}{x (1+x)} \, dx+\frac {1}{4} \int \frac {-3 x-9 \log (3+x)-3 x \log (3+x)+3 x \log (3+x) \log (x \log (3+x))+x^2 \log (3+x) \log (x \log (3+x))}{x (3+x) \log (3+x) \log (x \log (3+x)) (x-3 \log (\log (x \log (3+x))))} \, dx \\ & = \frac {1}{4} \log (x-3 \log (\log (x \log (3+x))))+\frac {1}{4} \int \left (-\frac {1}{x}+\frac {2}{1+x}\right ) \, dx \\ & = -\frac {\log (x)}{4}+\frac {1}{2} \log (1+x)+\frac {1}{4} \log (x-3 \log (\log (x \log (3+x)))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{4} (-\log (x)+2 \log (1+x)+\log (x-3 \log (\log (x \log (3+x))))) \]

[In]

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[Log[x*Log[3 + x]]]),x]

[Out]

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

Maple [A] (verified)

Time = 32.61 (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	\(-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (1+x \right )}{2}+\frac {\ln \left (-\frac {x}{3}+\ln \left (\ln \left (x \right )+\ln \left (\ln \left (3+x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \ln \left (3+x \right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left (3+x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left (3+x \right )\right )+\operatorname {csgn}\left (i \ln \left (3+x \right )\right )\right )}{2}\right )\right )}{4}\)	\(78\)

[In]

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)

[Out]

-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)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{4} \, \log \left (-x + 3 \, \log \left (\log \left (x \log \left (x + 3\right )\right )\right )\right ) \]

[In]

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")

[Out]

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 = 0.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=- \frac {\log {\left (x \right )}}{4} + \frac {\log {\left (- \frac {x}{3} + \log {\left (\log {\left (x \log {\left (x + 3 \right )} \right )} \right )} \right )}}{4} + \frac {\log {\left (x + 1 \right )}}{2} \]

[In]

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)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{4} \, \log \left (-\frac {1}{3} \, x + \log \left (\log \left (x\right ) + \log \left (\log \left (x + 3\right )\right )\right )\right ) \]

[In]

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")

[Out]

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)

none

Time = 1.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {1}{4} \, \log \left (x - 3 \, \log \left (\log \left (x \log \left (x + 3\right )\right )\right )\right ) + \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x\right ) \]

[In]

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")

[Out]

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 = 14.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+3 x^2+\left (9+12 x+3 x^2\right ) \log (3+x)+\left (-6 x^2-2 x^3\right ) \log (3+x) \log (x \log (3+x))+\left (-9+6 x+3 x^2\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))}{\left (-12 x^2-16 x^3-4 x^4\right ) \log (3+x) \log (x \log (3+x))+\left (36 x+48 x^2+12 x^3\right ) \log (3+x) \log (x \log (3+x)) \log (\log (x \log (3+x)))} \, dx=\frac {\ln \left (x+1\right )}{2}+\frac {\ln \left (\ln \left (\ln \left (x\,\ln \left (x+3\right )\right )\right )-\frac {x}{3}\right )}{4}-\frac {\ln \left (x\right )}{4} \]

[In]

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)

[Out]

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