@acer Not all, but Maple2019 advertised for a big overhaul of simplifications like this, didn't it?

@acer Weren't things like this supposed to be fixed with version 2019 ?

@vv Sorry you are right. I mixed it up with

`assuming`([simplify(expand((coth(x)^(1/3)-tanh(x)^(1/3))^2+4))], [x > 0])

@vv Maple2019? With 2015 this does not work at all..

@Christopher2222 Thanks, but generally this does not work. Try this one

`assuming`([simplify(combine((coth(x)^(1/3)-tanh(x)^(1/3))^2+4, trig))], [x > 0])

@acer Windows10 + Maple2019

@acer Thanks!

How do I get the inline answer as in ctrl+= ?

@acer Ok thanks. That's unfortunate.

@acer Thank you. Can you also try the second example "for general a"?

So is it working now with 2019?

@vv Nonono...The answer was not useless since you reminded me about algcurves:-puiseux

I just said that I also "solved" it "manually", and apart from asympt/series/MultiSeries I'm not very familiar with other methods of which there are algcurves:-puiseux , but also else?

@vv That's how I did also solve it, but I thought there must have been some build in function as well...

Btw: What do you need the [] in alias(alpha=indets(a,RootOf)[]):  for ?

If I have

restart;
f:=RootOf(x^2+1,x)+RootOf(x^2+2,x)*RootOf(x^2+x+3,x);
alias(alpha=indets(f,RootOf)[1]);
evalf(alpha)

I can use it to select an operand, but in your case there is only one, so it should be obsolete or?

@vv 

Actually thanks...I completely forgot about algcurves:-puiseux

But do you know why series/asympt has issues with the zeros in rootof?

@Christian Wolinski Not sure what G[1]('%', indets('%', t));  is supposed to do. It's not doing anything for me.

@Kitonum Thanks first, but what if

f:=ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln(2*(1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))

?

I see that I can first do

op(f)

and then proceed termwise: