@vv 

At least a weakness  and that is (probably) the reason for acer's version (division by zero)

It is not the integrand per se, using the series for MeijerG it can be seen that the integrand is

(-3*polylog(4,x^Pi)+Pi*ln(x)*polylog(3,x^Pi))/Pi^4/(-1+x)^2

@acer , it was vv, not me ...

yours is better, indeed

@Markiyan Hirnyk : yes

@Markiyan Hirnyk 

It is (a kind of) averaging over the singularity s. More formally split in s and use a change of variable to bring the left side to the right, cancelling the singularity. While for CPV one can not use change of variables in general this one is linear. I do not have a precise Lemma at hand.

f:= t -> (abs(t-(1/6)*Pi)+1)*exp(-sec(t))*cos(t)/(-1/4+sin(t)^2)+1/sin(t-(1/6)*Pi)^5;

'Int(f(t),t = 0 .. Pi/2, CPV)';
``='Int(f(t),t = 0 .. 1/3*Pi, CPV)+Int(f(t),t = 1/3*Pi .. 1/2*Pi)';
``='Int(f(1/6*Pi+t)+f(1/6*Pi-t),t = 0 .. 1/6*Pi)+Int(f(t),t = 1/3*Pi .. 1/2*Pi)';
evalf[30](%); # increase because of numerical singularity, now in t=0
                                        Pi
                    Int(f(t), t = 0 .. ----, CPV)
                                        2

                   = 3.80259236620486251646441207770

@Markiyan Hirnyk so what the numerical "CPV" values for the integral over Pi/6 - 1/2 ... Pi/6 + 1/2?

(the 2nd example would not work in Pi/2, but that's no so important)

@vv 

It reminds me of needing 15+3 decimals to represent a double precision

Increase precision. For example begin with restart; Digits:=40; and finally use SK0_SUM; SKA_SUM; %-%%; which gives 0.

Which means: the first 40 decimals are the same.

f(x) = f(y) instead (for me it is more clear)

 ... I will never understand why users go this route - opening the door for detailled tracking, see attached

dropbox.zip

The code editor is somewhat sparse. While coding (long time ago) I used to write down the skeleton for a command first, using (manual) indentations, always, and fill it up then.

By an indefinite integral you mean something that after differentiating gives you the expression from which you started. And that what I gave you: one always can write it as definite integral as above.

The above just shows a more simple looking task: find an indefinite integral for sqrt( exp(t) - t ), though you might read what Mariusz Iwaniuk said: probably no solution in terms of knwon functions.

Int((ln(xi)/xi^2-1/xi)^(1/2),xi=1..x);
simplify(%) assuming 0<x;
IntegrationTools[Change](%, xi=exp(t), t) assuming 0<x;
map(expand,%): simplify(%) assuming 0<x, t::real;
simplify(%, symbolic);

# to be verified:

Int((exp(t)-t)^(1/2),t = 0 .. ln(x))*I;
(ln(x)/x^2-1/x)^(1/2) = diff(%, x); simplify(%) assuming 0<x;
is(%);
                                 true

Int((ln(xi)/xi^2-1/xi)^(1/2),xi=1..x) is an anti-derivative and for 0 < x it is more simple to write it as
Int((exp(t)-t)^(1/2),t = 0 .. ln(x)) * I, numerical this causes no problem