\(\int \frac {4+4 x-x^2+(4+4 x-x^2) \log (\frac {4+4 x-x^2}{x})+\log (x) (6 x^2-x^3+(-4+5 x^2-x^3) \log (\frac {4+4 x-x^2}{x})) \log (\log (x))}{\log (x) (-4 x-4 x^2+x^3+(-4 x-4 x^2+x^3) \log (\frac {4+4 x-x^2}{x})) \log (\log (x))} \, dx\) [6839]

Optimal result

Integrand size = 133, antiderivative size = 27 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x+\log \left (\frac {5 \left (x+x \log \left (4+\frac {4}{x}-x\right )\right )}{\log (\log (x))}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 2.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6820, 6860, 45, 6816, 2339, 29} \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x+\log (x)+\log \left (\log \left (-x+\frac {4}{x}+4\right )+1\right )-\log (\log (\log (x))) \]

[In]

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

[Out]

-x + Log[x] + Log[1 + Log[4 + 4/x - x]] - Log[Log[Log[x]]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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 {-4-4 x+x^2+(-6+x) x^2 \log (x) \log (\log (x))+\left (-4-4 x+x^2\right ) \log \left (4+\frac {4}{x}-x\right ) (1+(-1+x) \log (x) \log (\log (x)))}{x \left (4+4 x-x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right ) \log (x) \log (\log (x))} \, dx \\ & = \int \left (\frac {6 x^2-x^3-4 \log \left (4+\frac {4}{x}-x\right )+5 x^2 \log \left (4+\frac {4}{x}-x\right )-x^3 \log \left (4+\frac {4}{x}-x\right )}{x \left (-4-4 x+x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right )}-\frac {1}{x \log (x) \log (\log (x))}\right ) \, dx \\ & = \int \frac {6 x^2-x^3-4 \log \left (4+\frac {4}{x}-x\right )+5 x^2 \log \left (4+\frac {4}{x}-x\right )-x^3 \log \left (4+\frac {4}{x}-x\right )}{x \left (-4-4 x+x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right )} \, dx-\int \frac {1}{x \log (x) \log (\log (x))} \, dx \\ & = \int \frac {(-6+x) x^2+\left (4-5 x^2+x^3\right ) \log \left (4+\frac {4}{x}-x\right )}{x \left (4+4 x-x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right )} \, dx-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\log (x)\right ) \\ & = \int \left (\frac {1-x}{x}+\frac {4+x^2}{x \left (-4-4 x+x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right )}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (\log (x))\right ) \\ & = -\log (\log (\log (x)))+\int \frac {1-x}{x} \, dx+\int \frac {4+x^2}{x \left (-4-4 x+x^2\right ) \left (1+\log \left (4+\frac {4}{x}-x\right )\right )} \, dx \\ & = \log \left (1+\log \left (4+\frac {4}{x}-x\right )\right )-\log (\log (\log (x)))+\int \left (-1+\frac {1}{x}\right ) \, dx \\ & = -x+\log (x)+\log \left (1+\log \left (4+\frac {4}{x}-x\right )\right )-\log (\log (\log (x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x+\log (x)+\log \left (1+\log \left (4+\frac {4}{x}-x\right )\right )-\log (\log (\log (x))) \]

[In]

Integrate[(4 + 4*x - x^2 + (4 + 4*x - x^2)*Log[(4 + 4*x - x^2)/x] + Log[x]*(6*x^2 - x^3 + (-4 + 5*x^2 - x^3)*L
og[(4 + 4*x - x^2)/x])*Log[Log[x]])/(Log[x]*(-4*x - 4*x^2 + x^3 + (-4*x - 4*x^2 + x^3)*Log[(4 + 4*x - x^2)/x])
*Log[Log[x]]),x]

[Out]

-x + Log[x] + Log[1 + Log[4 + 4/x - x]] - Log[Log[Log[x]]]

Maple [A] (verified)

Time = 15.80 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11

method	result	size
default	\(-x +\ln \left (x \right )+\ln \left (\ln \left (-\frac {x^{2}-4 x -4}{x}\right )+1\right )-\ln \left (\ln \left (\ln \left (x \right )\right )\right )\)	\(30\)
parallelrisch	\(-1+\ln \left (x \right )-\ln \left (\ln \left (\ln \left (x \right )\right )\right )+\ln \left (\ln \left (-\frac {x^{2}-4 x -4}{x}\right )+1\right )-x\)	\(31\)
parts	\(-x +\ln \left (x \right )+\ln \left (\ln \left (\frac {-x^{2}+4 x +4}{x}\right )+1\right )-\ln \left (\ln \left (\ln \left (x \right )\right )\right )\)	\(31\)
risch	\(\ln \left (x \right )-x +\ln \left (\ln \left (x^{2}-4 x -4\right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i \left (x^{2}-4 x -4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4 x -4\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{2}-4 x -4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4 x -4\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4 x -4\right )}{x}\right )}^{3}-2 \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4 x -4\right )}{x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4 x -4\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \pi +2 i \ln \left (x \right )-2 i\right )}{2}\right )-\ln \left (\ln \left (\ln \left (x \right )\right )\right )\)	\(171\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x + \log \left (x\right ) + \log \left (\log \left (-\frac {x^{2} - 4 \, x - 4}{x}\right ) + 1\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=- x + \log {\left (x \right )} + \log {\left (\log {\left (\frac {- x^{2} + 4 x + 4}{x} \right )} + 1 \right )} - \log {\left (\log {\left (\log {\left (x \right )} \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x + \log \left (x\right ) + \log \left (\log \left (-x^{2} + 4 \, x + 4\right ) - \log \left (x\right ) + 1\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

-x + log(x) + log(log(-x^2 + 4*x + 4) - log(x) + 1) - log(log(log(x)))

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=-x + \log \left (x\right ) + \log \left (\log \left (-x^{2} + 4 \, x + 4\right ) - \log \left (x\right ) + 1\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

-x + log(x) + log(log(-x^2 + 4*x + 4) - log(x) + 1) - log(log(log(x)))

Mupad [B] (verification not implemented)

Time = 12.86 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {4+4 x-x^2+\left (4+4 x-x^2\right ) \log \left (\frac {4+4 x-x^2}{x}\right )+\log (x) \left (6 x^2-x^3+\left (-4+5 x^2-x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))}{\log (x) \left (-4 x-4 x^2+x^3+\left (-4 x-4 x^2+x^3\right ) \log \left (\frac {4+4 x-x^2}{x}\right )\right ) \log (\log (x))} \, dx=\ln \left (\ln \left (\frac {-x^2+4\,x+4}{x}\right )+1\right )-\ln \left (\ln \left (\ln \left (x\right )\right )\right )-x+\ln \left (x\right ) \]

[In]

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

[Out]

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