Given

restart;
eq:=A=(1/2+x+y)^(3);
the_rhs:=solve(eq,A);

I asked solve to solve the above equation for A, expecting to get back (1/2+x+y)^(3), but it returns instead this

I looked at solve options, and tried number of them, but no change. I was looking for option to tell Maple not to simplify as it solves.

I also tried to see if the solution returned can be put back to the original form, and could not so far find a way, tried simplify and some options. But I did not try every possible method as there are do many.

Compare to Mathematica, which keeps the solution the same, as what one would expect. I see no reason to change it

ClearAll[A,x,y]
eq=A==(1/2+x+y)^3
Solve[eq,A]

The reason I am asking, is that it now makes parsing a little harder as I am looking for something in the form (expression)^power   as the solution.  i.e. the type to be `^`.    Now the type shows up as `*` because Maple for some reason changed it. 

It will easier if Maple did not do that, or if there is a way to change the expression back to the way it was. If all this fails, I have to just make the parsing handle this extra case form if needed.

Any suggestions?

Maple 2022.1

Update

Found a way after lots of trials and errors

simplify(the_rhs,[power,symbolic]);

But it would have been better if Maple did not do the simplification in the first place. But I could not find an option to tell it to do that while solving.

Maple 2022.1 on windows 10

restart;
w:=y-x/(c-x):
p1:=plots:-implicitplot(eval(w,c=2),x=-6..9,y=-3..3):
p2:=plots:-implicitplot(eval(w,c=2),x=-6..10,y=-3..3):
p3:=plots:-implicitplot(eval(w,c=2),x=-6..11,y=-3..3):
[p1,p2,p3]

Gives

Notice p2. The vertical line is gone. This happens when the range x=-6..10 and it shows back again when x=-6..11 or x=-6..9

Why does this happen? It is the same equation.

I tried different things but can't figure how to make Maple give a simplified result from int command without these RootOf with sums in them. Here is an example.

restart;
anti:=int(1/(a+x^(1/5)),x);

The strange thing is that doing

eval(anti,a=1)

Keeps these rootsOf and Sums there. But doing

anti:=int(1/(1+x^(1/5)),x);

Gives simplified result without RootOf

I tried DEtools:-remove_RootOf and assumptions and value() but nothing worked. I tried assumptions on a, such as assuming a::integer,a>0 and other things. 

I need similar result as Mathematica:

ClearAll[x, a];
Integrate[1/(a + x^(1/5)), x]

I am sure there is an option in Maple to do that, i.e. give result without these RootOf and Sums. But so far, I could not find it. I looked at options to int() command also. 

Any one know what option to use or how to simplifies it the the above result by Mathematica? 

Maple 2022.1 on windows 10

Is it considered OK to give singular solution to an ode which does not satisfy the ode?

I thought may be because it is singular solution and not general solution, then that is allowed sometimes. But it seems strange to me to give a solution which cleary do not satisfy the ode, even if it is singular.  But may be because it is singular, it is not required to satisfy the ode? Here is an example

restart;
ode_1:=diff(y(x),x)=sqrt(1+x+y(x)) ;
sol_1:=DEtools:-dalembertsol(ode_1);

Gives

But the first given "solution" (singular) does not satisfy the ode. Plugging it into the ode gives  -1=1.  Also odetest(y(x)=-x,ode_1) agrees it does not satisfy the ode. It gives -2 and not zero.

What do others think about this?  

This works

restart;
ode := diff(y(x),x)=a0+a1*y(x)+a2*y(x)^2+a3*y(x)^3;
DEtools:-abelsol(ode);

But this does not

restart;
ode := diff(y(x),x)=a0+a1*y(x)+a2*y(x)^2+a3*y(x)^3;
name_of_solver:=abelsol;
DEtools:-name_of_solver(ode);

Then I found this made it work

eval(DEtools:-name_of_solver)(ode);

So I am guessing what happens is this: doing DEtools:-name_of_solver Maple first searched for a proc called name_of_solver inside DEtools package, and found none. So it does not work.

But adding eval first, then name_of_solver is evaluated and replaced by abelsol and after that the call is made, and it then finds this proc inside the DEtools package.

But my question is, why is eval needed here?

Is it not automatically happens that a variable is replaced by its value? So I expected DEtools:-name_of_solver to automatically become DEtools:-abelsol and only then the call is made.

Why the rule of evaluation is different in this case?