Nice! 

Using val:=simplify(evala(%), size) reduces that a lot. 

Some guessing (deg = 12 as your first example) lets one try to look at 

  val/Pi*12; evalf(%); 
  identify(%, all); convert(%, sin);

and 'proofing' the identity

  'val' = Pi/12 / sin(5/24*Pi);
  convert(%, RootOf);
  simplify(%);
  evala(%);
  is(%);

                                 true

The expression is a sum and it seems that this gives troubles
due to cancellation, killing ~ 30 decimal places. Thus one needs
increased precision.

Find a sheet attached (for the original problem, have not checked
for subsequent modification).

MP_fsolve_complex_ca.mws

The expression is a sum and it seems that this gives troubles
due to cancellation, killing ~ 30 decimal places. Thus one needs
increased precision.

Find a sheet attached (for the original problem, have not checked
for subsequent modification).

MP_fsolve_complex_ca.mws

Maple knows some integrals, which are not directly know to 'int'
This is one - but Maple can do the suggested Mellin transform:

with(inttrans);
mellin(f(x),x,q+1): '%'= convert(%, Int);

                                       infinity
                                      /
                                     |                q
           mellin(f(x), x, q + 1) =  |          f(x) x  dx
                                     |
                                    /
                                      0

'mellin(1/(exp(x)-1),x,1+q)': '%'=%;
                  1
       mellin(----------, x, 1 + q) = GAMMA(1 + q) Zeta(1 + q)
              exp(x) - 1

Please upload your code. either in a *.txt file for copy & paste
or as Maple sheet. That would make it more clear, no?

I was pondering about that, it exhibits a 'weakness' of Maple,
if viewed a different way (but also leads to an answer):

I was pondering about that, it exhibits a 'weakness' of Maple,
if viewed in a different way (but also leads to an answer):

You transform to Int(ln((t+3)/(3-t))/(-t^2+1)^(1/2)/t,t = 0 .. 1)
and view at it as special situation of a=3 for the function (a <= 1)

  F:= a -> Int(ln((t+a)/(a-t))/(-t^2+1)^(1/2)/t,t = 0 .. 1)

Maple does not find value( F(3) ), but it finds value( F(a) ) and
from that the result follows as well:

  F(a); #eval(%, a=3); # does not work for 'value' ...
  value(%);
  simplify(%)  assuming 1<=a;

                            Pi arcsin(1/a)

  eval(%, a=3);

                            Pi arcsin(1/3)

PS: I converted that from a 'comment' to an 'answer' - hope you do not mind.

I was pondering about that, it exhibits a 'weakness' of Maple,
if viewed a different way (but also leads to an answer):

I was pondering about that, it exhibits a 'weakness' of Maple,
if viewed in a different way (but also leads to an answer):

You transform to Int(ln((t+3)/(3-t))/(-t^2+1)^(1/2)/t,t = 0 .. 1)
and view at it as special situation of a=3 for the function (a <= 1)

  F:= a -> Int(ln((t+a)/(a-t))/(-t^2+1)^(1/2)/t,t = 0 .. 1)

Maple does not find value( F(3) ), but it finds value( F(a) ) and
from that the result follows as well:

  F(a); #eval(%, a=3); # does not work for 'value' ...
  value(%);
  simplify(%)  assuming 1<=a;

                            Pi arcsin(1/a)

  eval(%, a=3);

                            Pi arcsin(1/3)

PS: I converted that from a 'comment' to an 'answer' - hope you do not mind.

The most compact form is Pi*arcsin(1/3), which follows from the above "PS":

1/2*Pi^2-arccos(1/3)*Pi; 
convert(%, arcsin): expand(%);

                  Pi arcsin(1/3)

Maple suggests that through applying 'identify' to result/Pi ...

Sorry for the late addendum

The most compact form is Pi*arcsin(1/3), which follows from the above "PS":

1/2*Pi^2-arccos(1/3)*Pi; 
convert(%, arcsin): expand(%);

                  Pi arcsin(1/3)

Maple suggests that through applying 'identify' to result/Pi ...

Sorry for the late addendum

Thank you - I just had some luck to finally find a route :-)

Thank you - I just had some luck to finally find a route :-)

By symmetry only half the interval is needed. Changing x = 1/2*arccos(3*xi/(3*xi+2))
gives an integral, that Maple solves in terms of the dilog function. Simplifying then
leads to the asserted result. For details please see the appended sheet.

hirnyk_ugly_int.mws

PS: arctan(sqrt(8)) = arccos(1/3), if I trust 'identify', that would have been smarter.
Indeed: cos(arctan(2*2^(1/2))) equals 1/3 after hitting the enter key ...

By symmetry only half the interval is needed. Changing x = 1/2*arccos(3*xi/(3*xi+2))
gives an integral, that Maple solves in terms of the dilog function. Simplifying then
leads to the asserted result. For details please see the appended sheet.

hirnyk_ugly_int.mws

PS: arctan(sqrt(8)) = arccos(1/3), if I trust 'identify', that would have been smarter.
Indeed: cos(arctan(2*2^(1/2))) equals 1/3 after hitting the enter key ...

Yes, Matrix size is roughly 32 MB

M:=LinearAlgebra:-RandomMatrix(10^6, 2, generator=0..0.5, datatype=float[8]);
in a fresh sheet and then saved.

Yes, Matrix size is roughly 32 MB

M:=LinearAlgebra:-RandomMatrix(10^6, 2, generator=0..0.5, datatype=float[8]);
in a fresh sheet and then saved.