integration - Integral $int_1^inftyfrac{dx}{1+2^x+3^x}$

Can the integral $$\int_1^\infty\dfrac{dx}{1+2^x+3^x}$$ be given in closed form?

This question arises naturally when I considered doing integrals. What makes an integral hard? Well, the integrand, of course. So why is it hard to integrate some integrands? For starters, I need to say that there are definite and indefinite integrals and that might affect the difficulty of the integral, but in both cases the situation seems the same.

Let's say $f(x)$ is the (elementary) integrand. For starters, consider short expressions for $f(x)$. That makes sense, because logically long expressions are harder to integrate on average. So, we consider elementary functions $f(x)$ with a short expression.

If $f(x)$ is mainly a product of simple functions, in other words, if $f(x)$ contains more products then sums then by using integration by parts it is clear that the integral of $f(x)$ is more likely to be 'solvable'. Similarly, if $f(x)$ contains more products than compositions, it is easier in general. So, in order of difficulty:

$$ \text{products}<\text{compositions}<\text{sums} $$

The argument: there is only one formula for sums when it comes to integrals or derivatives and that is the trivial $\int a(x) + b(x) dx = \int a(x) dx + \int b(x) dx$
or $(a(x)+b(x))' = (a(x))' + (b(x))'$. However, the formula for products and compositions are more powerful and lead more to success.

As an example, I will give three integrals of about the same notational length.
They are similar but the argument above seems to make a point. Which of the functions below do you consider easier to integrate?

$$\color{Red}{\sin(x) (e^x + \cos(x))},\,\, \color{Blue}{e^{\sin(x)} \cos(x)}\, \text{or}\color{Green}{\dfrac{1+\cos(x)}{e^x+\sin(x)}}?$$

Or maybe some examples from MSE itself? Consider the list of integrals below.

$$\int_1^\infty\dfrac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\,dx$$

This value is known to agree with a closed form for the first $9000$ digits, yet no satisfactory proof has been given. All the following integrals, though, have closed forms:

$$\int_{-1}^1\frac1x\sqrt{\dfrac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \, dx$$

$$\int_0^1\dfrac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\,dx$$

$$\int_0^1\log\log\left(\dfrac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\, dx$$

$$\int_0^\infty\dfrac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}\,dx$$

$$\int_0^1\dfrac{dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}$$

$$\int_0^{\frac{\ln^22}4}\,\dfrac{\arccos\frac{\exp\sqrt x}{\sqrt2}}{1-\exp\sqrt{4\,x}}\,dx$$

So, why is the first one so much more difficult or impossible?

Note that the first integral contains more additions while the others have more compositions and products. Those compositions and products lead to possible ways to attack the problem with substitutions, pattern recognitions, integration by parts and rewritting them as (not too complicated) infinite sums.

Many deceivingly simple looking integrals are of the form $\displaystyle\int f\left(\frac{1}{a(x)+b(x)}\right)g(x)\,dx$, where $a$ and $b$ are not both polynomials and $f$ is not the exponential, sine or cosine. It is also hard to see how to use contour integrals to deal with integrands that contain a lot of sums, in particular when $\int_1^{\infty}f(x)\,dx$ does not equal an integer.

There have been complains about a lack of motivation for posting integrals on MSE, so hereby I did show my motivation.

As for showing how far I got, I must admit I am nowhere. I do not know how to start with this integral since all the methods I know well do not seem to help or at least I do not see it.