3.69.39 $\int \frac{4+4x-x^2+(4+4x-x^2)\log (\frac{4+4x-x^2}{x})+\log (x)(6x^2-x^3+(-4+5x^2-x^3)\log (\frac{4+4x-x^2}{x}))\log (\log (x))}{\log (x)(-4x-4x^2+x^3+(-4x-4x^2+x^3)\log (\frac{4+4x-x^2}{x}))\log (\log (x))}dx$

Optimal. Leaf size=27 $$-x+\log \left(\frac{5(x+x\log (4+\frac{4}{x}-x))}{\log (\log (x))}\right)$$

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Rubi [A]  time = 3.45, antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 6, integrand size = 133, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.045, Rules used = {6688, 6728, 43, 6684, 2302, 29} $$\begin{array}{c}-x+\log (x)+\log (\log (-x+\frac{4}{x}+4)+1)-\log (\log (\log (x)))\end{array}$$

Antiderivative was successfully verified.

[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 43

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 2302

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 6684

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

Rule 6688

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

Rule 6728

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-4-4x+x^2+(-6+x)x^2\log (x)\log (\log (x))+(-4-4x+x^2)\log (4+\frac{4}{x}-x)(1+(-1+x)\log (x)\log (\log (x)))}{x(4+4x-x^2)(1+\log (4+\frac{4}{x}-x))\log (x)\log (\log (x))}dx\\ & =\int (\frac{6x^2-x^3-4\log (4+\frac{4}{x}-x)+5x^2\log (4+\frac{4}{x}-x)-x^3\log (4+\frac{4}{x}-x)}{x(-4-4x+x^2)(1+\log (4+\frac{4}{x}-x))}-\frac{1}{x\log (x)\log (\log (x))})dx\\ & =\int \frac{6x^2-x^3-4\log (4+\frac{4}{x}-x)+5x^2\log (4+\frac{4}{x}-x)-x^3\log (4+\frac{4}{x}-x)}{x(-4-4x+x^2)(1+\log (4+\frac{4}{x}-x))}dx-\int \frac{1}{x\log (x)\log (\log (x))}dx\\ & =\int \frac{(-6+x)x^2+(4-5x^2+x^3)\log (4+\frac{4}{x}-x)}{x(4+4x-x^2)(1+\log (4+\frac{4}{x}-x))}dx-\mathrm{Subst}(\int \frac{1}{x\log (x)}dx,x,\log (x))\\ & =\int (\frac{1-x}{x}+\frac{4+x^2}{x(-4-4x+x^2)(1+\log (4+\frac{4}{x}-x))})dx-\mathrm{Subst}(\int \frac{1}{x}dx,x,\log (\log (x)))\\ & =-\log (\log (\log (x)))+\int \frac{1-x}{x}dx+\int \frac{4+x^2}{x(-4-4x+x^2)(1+\log (4+\frac{4}{x}-x))}dx\\ & =\log (1+\log (4+\frac{4}{x}-x))-\log (\log (\log (x)))+\int (-1+\frac{1}{x})dx\\ & =-x+\log (x)+\log (1+\log (4+\frac{4}{x}-x))-\log (\log (\log (x)))\end{array}\end{array}$$

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Mathematica [A]  time = 0.12, size = 26, normalized size = 0.96 $$\begin{array}{c}-x+\log (x)+\log (1+\log (4+\frac{4}{x}-x))-\log (\log (\log (x)))\end{array}$$

Antiderivative was successfully verified.

[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]]]

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fricas [A]  time = 0.52, size = 29, normalized size = 1.07 $$\begin{array}{c}-x+\log (x)+\log (\log (-\frac{x^2-4x-4}{x})+1)-\log (\log (\log (x)))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.32, size = 30, normalized size = 1.11 $$\begin{array}{c}-x+\log (x)+\log (\log (-x^2+4x+4)-\log (x)+1)-\log (\log (\log (x)))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.24, size = 53, normalized size = 1.96

method	result	size
default	$-\ln (\ln (\ln (x)))-\ln (-\frac{x^2-4x-4}{x})-x+\ln (x^2-4x-4)+\ln (\ln (-\frac{x^2-4x-4}{x})+1)$	$53$
risch	$\ln (x)-x+\ln (\ln (x^2-4x-4)-\frac{i(\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i(x^2-4x-4)\right)\mathrm{csgn}\left(\frac{i(x^2-4x-4)}{x}\right)-\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i(x^2-4x-4)}{x}\right)}^2+2\pi \mathrm{csgn}{\left(\frac{i(x^2-4x-4)}{x}\right)}^2-\pi \mathrm{csgn}\left(i(x^2-4x-4)\right)\mathrm{csgn}{\left(\frac{i(x^2-4x-4)}{x}\right)}^2-\pi \mathrm{csgn}{\left(\frac{i(x^2-4x-4)}{x}\right)}^3-2\pi -2i\ln (x)+2i)}{2})-\ln (\ln (\ln (x)))$	$173$

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [A]  time = 0.42, size = 30, normalized size = 1.11 $$\begin{array}{c}-x+\log (x)+\log (\log (-x^2+4x+4)-\log (x)+1)-\log (\log (\log (x)))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 4.39, size = 30, normalized size = 1.11 $$\begin{array}{c}\ln (\ln \left(\frac{-x^2+4x+4}{x}\right)+1)-\ln (\ln (\ln (x)))-x+\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.88, size = 26, normalized size = 0.96 $$\begin{array}{c}-x+\log (x)+\log (\log \left(\frac{-x^2+4x+4}{x}\right)+1)-\log (\log (\log (x)))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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