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No computer algebra system -- at least to my knowledge -- managed to either compute the integral $\int \log(x+e^x)\space \mathbb{d}x$, in terms of any known functions, or even just prove that it is not elementary.

Although it is pretty much obvious that no classical trick would help with getting the antiderivative, it is much less obvious -- to me at least --, as to why is this case so special that modern CASes either freeze (Wolfram), honestly report something like "Implementation incomplete (constant residues)" (Axiom), or a weird answer that is plain wrong if you try differentiating it (Mathcad).

In fact, there is a big family of functions that all crash the modern CASes, in form of $\int f(x+g(x)) \mathbb{d}x$, eg. $\sqrt{x+\cos(x)}$, $\sqrt[3]{x+\sin(x)}$. Mathcad even gives the wrong answer for the integrals of the latter two radicals.

So, my two questions are:

  • Are any of those inetegrals expressable in terms of any known functions?
  • Why do the modern computer algebra systems fail to prove these are not elementary/...okay, liouvillian?

If anyone reading this happen to know how Risch's algorithm works, I'd love to hear from him or her as to how Risch's algorithm is supposed to tackle this problem. I mean, proving the non-elementarity, what would be the right field extension tower for that, what would follow from it, and how would the algorithm show that this integral is not elementary/liouvillian/... -- or is it?

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  • $\begingroup$ CAS fail, because Math fail.$$\int \log (x+\exp (x)) \, dx=-\frac{x^2}{2}+x \left(\log \left(e^x+x\right)-\log \left(1+e^{-x} x\right)\right)+\int_1^{\infty } -x \log \left(1+e^{-t x} t x\right) \, dt+C$$.No hope for closed form. $\endgroup$
    – Mariusz Iwaniuk
    Commented Aug 25, 2019 at 11:23
  • $\begingroup$ Thanks for the input, but if you don't mind, 2 questions on what you wrote: (a) what is the motivation behind representing the integrand in such form? Never seen an approach like this before... (b) How does the latter integral help us conclude that there is no hope for a closed form? Is there maybe a standard technique I am missing? $\endgroup$
    – Thehx
    Commented Aug 25, 2019 at 11:42
  • $\begingroup$ (a)With define integral, You are more likely to find the answer,than with indefine integral,(b) $\int_1^{\infty } -x \log \left(1+e^{-t x} t x\right) \, dt=\sum _{n=0}^{\infty } \frac{(-1)^{1-n} x^{2+n} E_{-1-n}((1+n) x)}{1+n}$ if I put for example x=1 and simplify then I have:$\sum _{n=0}^{\infty } (-1)^{1-n} (1+n)^{-3-n} \Gamma (2+n,1+n)$,for this sum no hope for answer. $\endgroup$
    – Mariusz Iwaniuk
    Commented Aug 25, 2019 at 14:04
  • $\begingroup$ I'm not sure (b) can really be taken as a proof, since even though we can express some term as an infinite sum some parts of which are non-elementary, this does not prove that the sum itself is non-elementary too. I mean, when we break something into a sum of parts, the formula for a part could easily become more complicated than the sum formula... even if all summands in that infinite sum are non-elementary (which they likely are since I see special functions in your example), we can't guarantee that the sum expression itself is therefore non-elementary -- or I just don't know how we can. $\endgroup$
    – Thehx
    Commented Aug 26, 2019 at 0:41

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I was mistaken. Only algebraic examples like $\sqrt{x+e^x}$ cause Axiom to fail. As for the [ $\log(x+e^x)$ ] case, even sympy's risch_integrate proves the non-elementariness successfully. Wolfram still gets confused over this one though.

Axiom successfully proves non-elementariness of $\int\log(x+e^x)\mathbb{d}x$ with basic integrate command.

Sympy fails using general integrate function, but succeeds if explicitly asked to use risch_integrate. Right during the process of Hermite reduction, to be precise. Many thanks to Aaron Meurer (the author of risch_integrate) who explained what went on under the hood as he traced how the example was processed.

With the algebraic example, $\sqrt{x+e^x}$, Axiom honestly admits that "implementation incomplete: constant residues". Well... that algebraic case is the hardest one, is a known fact anyway.

Unfortunately, my math knowledge is nowhere near what is required to try and trace what happened with the algebraic example.

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