@ecterrab 

 Selecting with the mouse and scrolling takes too long, but this way it works acceptably and(!!!) preserves the output:

S1: place the cursor where to stop

S2: Select all the above with Crtl+Shift+Home (on Windows)

S3: from the menu: Evaluate -> Selection

Thank you

Before I read your post I was quite happy with what I saw on the screen.

Now I see these "absurdly" long negations everywhere. Here a sample from a text passage

I clearly am in favour of a shorter version.

@Carl Love 

That's a good workaround!

Since 2d is not affected by tiny rendering (below: left 4k right 2k) and the rendering changes without Maples GUI and kernel dooing anything, I think that the issue is more openGL related.

Thats how the the output looks like when I change back to 4k (without execution of any commands. i.e. it is the same plot as above):

(A while ago I contacted support with that issue but they could not reproduce it because they had no 4k monitors to test.)

@Carl Love 

My example was misleading. Older Maple version did not allow to use implicitplot3d without ranges so I put something that did not match the OPs screenshot.

I can exactly reproduce the tiny fonds and how it looks differently if I change to 2k (with Maple 2023)

@acer 

Thank you. I looked at the wrong spot for the Airy question. For "_M" I can't find anything.

Is there more information available in the console window?

@ecterrab 

I can see that turning the output of options off by default is not a good idea. I was more interested to switch the display of for a particular output not to expose readers who are not familiar with Maple with that detail. 

In this case I would of course also hide Maple Input.

Did you forget to attach the code?

@acer too easy. Thanks

@Joe Riel 

I tested int warnings on another elliptic expression

https://www.mapleprimes.com/questions/233304-How-To-Find-The-Inverse-Function-Of

This time it behaved as I would have expected (i.e. no supression of warnings after repeated tries). It seems that it depends on the type of integrand. I will leave it like this. Thank you

@acer 

This comes from applying the chain rule to

and then integrating over varphi.

That's a formal step to avoid using differentials. The left hand side integrated is nothing else than the time (as a function of an angle of a pendulum in this case. If varphi_t=varphi_0 the pendulum has made a half period.) It is easy to do this with differentials but Maple does not provide such a caculus (for a good reason).

Alternativley to that I could change variables (in this case I could not apply and equation that relates the variables because this is function varphi(t) that I try do derive).

So far the left hand side does not evaluate to t(varphi_t)-t(-varphi_0) and I am obligied to replaced this confusing term by t manually.

@sursumCorda 
I found an exception from your golden ratio discovery that works as expected

solve(0 <= ln(a/2) + ln(1 + a));

In this case also an additional relation between 0 and 1 does not change the result (as in my last reply to dharr).

solve({0 <= ln(a/2) + ln(1 + a),a>1/2});

So these things seem to come together.

@Joe Riel 

Thank you for the workaround but for what reason warnings disapear (in case of varphi without t)?

Is it a (new?) mechanism that Maple does not issue the same warnings after two times?

Just for your information: Here are related posts where I was more concerned to make warnings apear:

https://www.mapleprimes.com/questions/235722-Why-A-Restart-Is-Needed-To-Get-A-Helpfull

https://www.mapleprimes.com/questions/235770-Why-No-Warning-Is-Returned-With-2D-Input

@dharr Interesting findings!

The logarithmic substitutions for the original statement with ln

solve({a > 0, ln(a) + ln(1 + a) >= 0}, a);

lead to the statement you have given that the semi-algebraic solver solves successfully (and correct IMO).

solve({a>0,a*(a+1)>=1},a);

For me it looks like that the output is not assembled correctly (i.e. < and <= are treated the same way).

Here is another output that looks incorrectly combined in an OR fashion instead of AND

solve({0 <= ln(a) + ln(1 + a), 1 <= a}, a)

 With -1<a we are back to