Seconding Thomas Richard: even if I have some doubts in the formulation of the text
(for clients in the EC (*)) it is almost evident that using Maple's DLLs is covered
by their copyrights - you still 'use Maple', even if you do not use it as a complete
installation. Just in case that might be a reason for your question.

(*) I was 'encouraged' by Alejandro in a different thread to post in German (?)
but still can resist :-)

I was asking, not 'demanding'. Sorry, if I was not clear enough.

And do not think that English is The Language, take it as a kind
of concurrent "Lingua Franca".

So let us return to the very content

May I suggest that discussions are in English, if possible?

I do not post in German, and Markiyan does not use Ukrainian.

It would make things more readable. I think.

@David Mazziotti 

I rarely have that, though I often have problems in Classic working with objects form LinearAlgebra (but nobody knows why it is on my machine ...)

What I really wish: actual backup, like it is common on most other SW (for example Microsoft Office), i.e. a backup which still exist, if the file is saved - just the version before. Unfortunately Maple deletes the backup which is generated while working in a sheet.

I appended a worksheet to my suggestion, it should make it clear.

For the other points you may wish to read Maple's help about the commands.

For the 'complex' options: it is almost certain, that 2 = numerator can not have a solution in the Reals. It is a kind of hope, that it will be close to a maximum (also due to the increasing denominator)

Please do not forget: it is a brute suggestion, so for a proof you may need paper and pencil for estimates to justify it.

Ok, by x=arcsin(y) it reduces to compute, the following (0 < b, b < 1)

Int(y/(y^3+1)^(1/2)/(1-y^2),y = 0 .. b);

                       b
                      /
                     |             y
                     |   --------------------- dy
                     |     3     1/2    2
                    /    (y  + 1)    (-y  + 1)
                      0

which Maple can do in terms of Elliptics.

But is there an answer without using that, some more simple (?) result.

Ok, by x=arcsin(y) it reduces to compute, the following (0 < b, b < 1)

Int(y/(y^3+1)^(1/2)/(1-y^2),y = 0 .. b);

                       b
                      /
                     |             y
                     |   --------------------- dy
                     |     3     1/2    2
                    /    (y  + 1)    (-y  + 1)
                      0

which Maple can do in terms of Elliptics.

But is there an answer without using that, some more simple (?) result.

Used Mupad, quite similar in in syntax and you do not have to break fingers on
a German keyboard like entering MMA commands. Occassionally got help at usenet
by Thomas Richard or Maple users. Finally got a good offer by kind sales.

While professionally I would decide a different route: for my needs it was good.
Hope that M will improve beyond "Oberflächlichkeiten". Sigh.

Come on - it's worth to improve that stuff, we all like it - do it, you can if
you want it and dare!

Adding to Alejandro's answer:

That also happens for rationals (instead of floats) and in my own code
meanwhile I prefer something like

  if 0 < signum(3/10*Pi) then
  1 else 0;
  end if;
                                  1

Adding to Alejandro's answer:

That also happens for rationals (instead of floats) and in my own code
meanwhile I prefer something like

  if 0 < signum(3/10*Pi) then
  1 else 0;
  end if;
                                  1

S_n = partial sum (which Maple finds due to the given decomposition), then

eval(S_n, x=-1);
simplify(%) assuming n::nonnegint;

                                  0

Or likewise using 2*N+1 instead of N to tell Maple that it is odd.

S_n = partial sum (which Maple finds due to the given decomposition), then

eval(S_n, x=-1);
simplify(%) assuming n::nonnegint;

                                  0

Or likewise using 2*N+1 instead of N to tell Maple that it is odd.

By shooting positive integer values and numerical evaluation.

f:= x -> Sum((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))),n=0..infinity);
g:= x -> piecewise(abs(x)<1, -1 + 1/(1-x), 1 < abs(x), 1/(1-x), x=-1, 0, NULL) ;

plot(f-g, -8 ..8);

Seems to confirm it, but that is not a proof, it is a guess.

For a proof I tried f(x)=Sum((a(n)+a(n)^2)/(1-a(n)^3),n = 0 .. infinity)
with a(n)=x^(3^n), but had no good idea to proceed.

It may be better to confirm the above guess in another way.

You mean to prove that it is 1 / (1-x) ?

This defines a complex curve (which is roughly a real surface) in C^2 = R^4,
real or imaginary part intersects it with a plane to give real curve, but it
will still depend on 4 inputs.

At least you want to fix one of your 4 real parameters: formally you missed
to assign y2.