Question: Why do I get this unevaluated expression for indefinite integration?

I have an expression equal to the sum of N terms of the form Int(fn=1..N(x), x) and I want to replace each fn(x) by its Taylor (or series) expansion.

When the integrals are definite, like J1 below, I can easily obtain a new expression (K1) where the integrand has been replaced by some expansion.
But when the integral is indefinite, like J2, I get an evaluated expression for K2.

It seems I have to do some gymnastic (J3 --> K3) to get what I want

restart

J1 := Int(sin(p*x), x=0..1);
K1 := eval(J1, Int = ((a, b) -> Int(mtaylor(a, x=0, 5), b)));

Int(sin(p*x), x = 0 .. 1)

 

Int(p*x-(1/6)*p^3*x^3, x = 0 .. 1)

(1)

# undefined integration

J2 := Int(sin(p*x), x);

`Expected result` = Int(p*x-(1/6)*p^3*x^3, x);

K2 := eval(J2, Int = ((a, b) -> Int(mtaylor(a, x=0, 5), b)));

Int(sin(p*x), x)

 

`Expected result` = Int(p*x-(1/6)*p^3*x^3, x)

 

eval(Int(sin(p*x), x), {Int = (proc (a, b) options operator, arrow; Int(mtaylor(a, x = 0, 5), b) end proc)})

(2)

# undefined integration using Intat

J3 := Intat(op(1, J2), op(2, J2)=y);
eval(%, Intat = ((a, b) -> Intat(mtaylor(a, x=0, 5), b))):

K3 := IntegrationTools:-Change(convert(%, Int), y=x, x)

Intat(sin(p*x), x = y)

 

Int(p*x-(1/6)*p^3*x^3, x)

(3)
 

 

Download Integration.mw

Why do I get this unevaluatedform for K2?
Do I have to use Intat to get K3?

Thanks in advance

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