How to indefinitely integrate $\ln \ln \ln x$ in terms of special functions or a series of special functions? It's well known that
$$ \int \ln(x) dx = x \ln(x) - x + C $$
The next one can't be done in elementary functions but can still be expressed as
$$ \int \ln \ln (x) dx   = x \ln \ln x - \text{Li}(x) + C $$
Where "Li" is the non elementary Logarithmic integral. The next natural question then is:
$$ \int \ln \ln \ln (x)  dx = ...?$$
My playing around:
So my guess is that the answer should look something like
$$ x \ln \ln \ln(x) -...$$
Assuming that form i'm trying to then use special functions to crack open
$$ \int \frac{1}{\ln(x) \ln \ln(x) } dx$$
The expression $g(x) = \text{Li}(\ln(x))$ seems an interesting thing to look at, it differentiates to
$$ g'(x)  = \frac{1}{x \ln \ln(x)} $$
But I haven't been able to create series out of this quite yet.
 A: Unfortunately you cannot expect to have anything sufficiently simple. Suppose you have
$$\int f(x) dx = F(x)$$
Now we would like to find the connection $T$
$$\int \ln(f(x)) dx = T(F(x))$$
But look:
$$\ln(f(x)) = T'(F(x))f(x)$$
However, $T'(F(x)) \cdot f(x)$ is multiplicative to $f(x)$ and you have $\ln(f(x))$ to the left. We know no such expression that would be simple to state. On top of that you need some operator over $F(x)$ which is an integral(!). That simply will not do it.
To the rescue, you can simply reverse integration and keep on as far as you can:
$$\int \ln(f(x)) dx = x\ln(f(x)) + g(x)$$
$$\ln(f(x)) = \frac{x}{f(x)}f'(x) + \ln(f(x)) + g'(x)$$
This is making:
$$g'(x) = -\frac{x}{f(x)}f'(x)$$
$$g(x) = -\int x \, d\ln(f(x))$$
Now comes the funny part. I will show it for a triple and we can easily derive the formula:
$$d\ln\ln\ln(x)=\frac1{x\ln(x)\ln(\ln(x))}$$
It is simply
$$d\ln^{(n)}(x)=\frac1{x\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)}$$
or
$$\int \ln^{(n)}(x) dx = x\ln^{(n)}(x) - \int\frac1{\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)} \, dx$$
I know it sounds disappointing, and nobody can say if some theory can actually do something about this, but I very much doubt that anything better can be done in general.
However, notice, that in many of these cases you can use the telescoping rule, for this situation suitably adjusted to:
$$ \left ( x\sum_{n=0}^{+\infty}{f_n(x)} \right )' = x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=0}^{+\infty}{f_n(x)} $$
Now take
$$f_{n+1}(x) = - xf_n'(x), f_0(x)=\ln^{(k)}(x)$$
and you would telescopically cancel it all:
$$x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=0}^{+\infty}{f_n(x)} = x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=1}^{+\infty}{f_n(x)} + f_0(x) = $$ $$x\sum_{n=0}^{+\infty}{f_n'(x)} - x\sum_{n=0}^{+\infty}{f_n'(x)} + f_0(x) = \ln^{(k)}(x)$$
So it makes
$$ \left ( x\sum_{n=0}^{+\infty}{f_n(x)} \right )' = \ln^{(k)}(x) $$
or
$$ \int \ln^{(k)}(x) \, dx = x \left (\ln^{(k)}(x) - \frac1{\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)} + \sum_{n=2}^{+\infty}{f_n(x)} \right )$$
repeating the recursion
$$f_0(x)=\ln^{(k)}(x), f_{n+1}(x) = - xf_n'(x)$$
By this, at least you got rid of the integral sign. Important here is that the recursion is giving all slower and slower function, which is the case here.
This is adjusted from a more general approach (of the same author, so it is still all attempted contribution of mhe [my hamble existence]):
https://math.stackexchange.com/a/4186998/944732
In general you can find that
$$\int \ln^{(k)}(x) \, dx = x \left (\ln^{(k)}(x)  + \sum_{n=1}^{k-1}{\frac{(-1)^n}{\ln^{n}(x)}}\sum_{n \geq m_1 \geq m_2... \geq m_{k-1} \geq 1}{\frac{c_{n,m_1,...,m_{k-1}}}{\prod\limits_{j=1}^{k-1}(\ln^{(j+1)}(x))^{m_j}}} \right )$$
where constants $c_{n,m_1,...,m_{k-1}}$ are recursively defined, and combinatorial in nature:
$$c_{k_1,k_2,k_3,...,k_m}=\sum_{n=1}^{m}c_{k_1-\theta(1,n),k_2-\theta(2,n),k_3-\theta(3,n),...,k_m-\theta(m,n)} (k_n-1)$$
$$\theta(i,j)=0 \text{, except  } \theta(i,j)=1 \text{ if } i \leq j$$
where additionally
$$c_{k_1,k_2,k_3,...,k_m}=0 \text{  if any  } k_i < k_j \text{ when } i<j \text{ or } k_i \leq 0$$
and
$$c_{1,...,1}=1$$
A: A series answer.
For the related exponential form this is  Problem 4.30 in my paper
Edgar, G. A., Transseries for beginners, Real Anal. Exch. 35(2009-2010), No. 2, 253-310 (2010). ZBL1218.41019.
$$
\int e^{e^{e^y}} dy = e^{e^{e^y}} \sum_{j=1}^\infty e^{-je^y}
\left(\sum_{k=1}^j e^{-ky} c_{j,k}\right) + C
$$
The coefficients $c_{j,k}$ (beginning $1;1,1;2,3,2;6,11,12,6;\cdots$)
are related to Stirling numbers of the first kind:
$$
c_{j,k} = k! \left[j \atop k\right]
$$
Then
$$
\int \log\log\log x\;dx = x \log\log\log x -
x \sum_{j=1}^\infty (\log x)^{-j}
\left(\sum_{k=1}^j (\log\log x)^{-k} c_{j,k}\right) + C
$$
A: There is no common special function for this.
Substitute $y = \log \log \log x$ to get
$$
\int \log \log \log x\;dx = \int y e^y e^{e^y} e^{e^{e^y}} dy
\tag1$$
Noting that
$$
e^y e^{e^y} e^{e^{e^y}} = \frac{d}{dy} e^{e^{e^y}}
$$
integrate $(1)$ by parts to get
$$
y e^{e^{e^y}} - \int e^{e^{e^y}} dy
$$
The form
$$
\int e^{e^{e^y}} dy
$$
is the one most likely to be in integral tables.  But it is not.
A: Notice that besides all the answers above you can derive a recursion that is closing the case, albeit circumstantially.
First by definition
$$\ln^{(n)}(x)=\ln^{(n-1)}(\ln(x))$$
Next
$$\int f(\ln(x)) \, dx = \int f(u) \, dе^u = е^u f(u) - \int е^u f'(u) \, du $$
and then by induction
$$\int f(\ln(x)) \, dx = e^u \sum_{k=0}^{\infty} (-1)^k \frac{d^k f(u)}{du^k}  = x \sum_{k=0}^{\infty} (-1)^k \left (\frac{d^k f(u)}{du^k} \right )(\ln(x))$$
$$\int \ln^{(n)}(x) \, dx = x \sum_{k=0}^{\infty} (-1)^k \left (\frac{d^k}{du^k}\ln^{(n-1)}(u) \right )(\ln(x))$$
