I am trying to learn how do somethings without using pattern matching and it is a struggle so far for me.

For an example, now I want to ask Maple to tell me if C[1] (which is a constant) exist in expression, as long as C[1] does not occur as argument to exp(....)

I'll explain why I want to do this and show small example and show how I ended solving it and ask if there is better way.

Given this expression (this can be result of dsolve for example. Made up here)

I just need to determine if the expression has C[1] anywhere, which is NOT an argument to exp(). In the above example, there is one.

The reason I want to find this, is that this is the constant of integration for first order ODE, and I want to repalce exp(C[1]) by constant C[1], but as long as there is no C[1] allready anywhere in expression on its, otherwise I need to introduce new constant C[2], which I do not want to do (since first order, should have only one constant of integration) and in this case will leave the expression as is and not try to simplify it.

But indets tells one that C[1] is there, not where. This is what I tried

restart;
expr:=1/exp(z)*arcsinh(x*exp(C[1]))+x*C[1]*sin(exp(x))+3*exp(C[1]*y)*sqrt(sin(exp(3*C[1])));
indets(expr, specfunc(exp));

Now I can find if C[1] is argument to exp(....) from the above or not by more processing. (using applyrule or subsindets, etc...)

But first I still need to to determine if there is C[1] that exist anywhere else, as long it is not inside exp(.....).  If I do 

indets(expr, C[1]);

But this does not tell me the information I want. It just says expression has C[1] somewhere.

I looked at evalindets and looked at applyrule and do not see how to use these to do what I want.

Basically I want to tell Maple this

     indets(expr,C[1], conditional( C[1] is not anywhere inside exp(.....) ))

I can solve this using pattern matching. But I fnd Maple pattern matching awkaward to use sometimes and trying to learn how to do things without it.

So this is how I ended solving it. I replace all the C[1] inside exp(....) with ZZZ. Then use indets again to check if C[1] still there or not. This tells me what I want. 

restart;
expr:=1/exp(z)*arcsinh(x*exp(C[1]))+x*C[1]*sin(exp(x))+3*exp(C[1]*y)*sqrt(sin(exp(3*C[1])));
new_expr:=subsindets(expr, specfunc(exp), ()-> ZZZ);
if indets(new_expr, C[1])<>{} then
   print("C[1] exist outside exp()");
else
   print("C[1] does not exist outside exp()");
fi;

"C[1] exist outside exp()"

And

restart;
expr:=1/exp(z)*arcsinh(x*exp(C[1]))+x*sin(exp(x))+3*exp(C[1]*y)*sqrt(sin(exp(3*C[1])));
new_expr:=subsindets(expr, specfunc(exp), ()-> ZZZ);
if indets(new_expr, C[1])<>{} then
   print("C[1] exist outside exp()");
else
   print("C[1] does not exist outside exp()");
fi;

"C[1] does not exist outside exp()"

is there a better way to do this?

Given some expression, I want to replace each exp(.....) in it with something else.

This is what I currently do. First I find all the exp() terms, then using a loop and use subs  like this

restart;
expr:=1/exp(z)*arcsinh(x*exp(C[1]))+x*sin(exp(x))+3*exp(C[1]*y)*sqrt(sin(exp(3*h)));

#find all exp terms
s:=select(x->has(x,exp),indets(expr));
s:=select(x->op(0,x)='exp',s)

Now, lets says I want to replace each with Z

I tried to use ~ but that did not work.

So now I am doing this

for item in s do
    expr:=subs(item=Z,expr);
od:
expr

is it possible do this without a loop? could not do it using ~

With map, I can do this map(x->subs(x=Z,expr),s) but this does not replace it inside expr where I want. 

Assignment is not allowed inside the above so I can not do  this map(x->expr:=subs(x=Z,expr),s)

And if I do expr:=map(x->subs(x=Z,expr),s)  it does not work either. it gives

How to do the replacement without using a do loop? 

This is the Maple code

restart;
result:=int(1/tanh(u),u)

simplify(result,size) ;
simplify(result,symbolic);
simplify(result,ln);
simplify(result,trig);

Any suggestions?

I can't figure out how Maple obtained this solution and looking for some ideas to try.

It is first order non-linear ode in y(x), which is separable.

ode:=diff(y(x),x)=x*ln(y(x));
dsolve([ode,y(1)=1],y(x))

But the general solution is

sol:=dsolve(ode)

Setting up manually an equation using the given condition in order to solve for _C1, produces no solution. 

eq:=subs([y(x)=1,x=1],sol);
solve(eq,_C1)

Warning, solutions may have been lost
 

Also 

coulditbe(exp(RootOf(1 + 2*Ei(1, -_Z) + 2*_C1))=1)

   FAIL

So how did Maple solve for the constant of integration which results in particular solution y(x)=1 that is supposed to satisfy the condition y(1)=1?  

It is clear that y(x)=1 satisfies the ODE itself. But I am asking about how it also satisfies y(1)=1

(odetst says it does satisfy the ODE and condition as well. So Maple must have done something very smart under the cover)

Next I tried

ode:=diff(y(x),x)=x*ln(y(x));
sol:=dsolve(ode,y(x));
sol:=DEtools:-remove_RootOf(sol);
sol:=subs([y(x)=1,x=1],sol)

And now

solve(sol,_C1)

Error, (in Ei) numeric exception: division by zero
 

Just wondering how did Maple decide that y(x)=1 satisfies y(1)=1? I do not see it.

Using Maple 2020.1. But same result on Maple 2019

From help, it says

coulditbe routine returns true if there is a possible value of x1 that satisfies prop1

my question is, how to find out this condition/possible values that Maple found?  This infomration is very useful, but now I do not see how to obtain it. All what coulditbe retuirn is true or false.

Context of why I am asking:  Sometimes odetest do not verify its own solutions. And coulditbe can help in finding under what conditions the solution can satisfy the ode. Here is an example

restart;
ode:=diff(y(x),x) = abs(y(x))+1;
solExplicit:=dsolve(ode);
offset := odetest~([solExplicit],ode)

gives

[exp(-x)/_C1 - abs((-exp(-x) + _C1)/_C1) - 1, exp(x)*_C1 - abs(exp(x)*_C1 - 1) - 1]

Both solution fail odetest. 

coulditbe~(offset,0)

gives true

So there are assumptions/conditions which makes the solution satisfy the ODE. In this case, by inspection one can see what these conditions are. They are, for one solution:

(-exp(-x) + _C1)/_C1  >0

and for the other, the condition is

exp(x)*_C1 - 1 >0

Under these assumptions, odetest would have given 0 for each odetest.

And it is this information I wanted to obtain automatically from coulditbe.

In Mathematica, Reduce is used for this. Reduce gives conditions under which something is satisfied. For example, 

Reduce[ C[1] Exp[x] - Abs[C[1] Exp[x] - 1] - 1 == 0, {x, C[1]}, Reals]

Gives

C[1] >= Exp[-x]

While the above in  Maple

coulditbe( C[1]*exp(x)- abs( C[1]*exp(x)-1)-1 = 0)

gives true  only, but without the important information, true under what conditions.

Is there a different command in Maple which could give this information?