HP Prime integration issue

[LCieParagon]

(02-12-2021 10:53 PM)Albert Chan Wrote:  Hi, LCieParagon

I asked this before, and you are not supposed to see rootof(...) in Prime.
see https://www.hpmuseum.org/forum/thread-11...#pid105773

For your example, we can do a simple substitution, x = 2^(1/3)*y, dx = 2^(1/3)*dy
(y integral result copy/pasted from Hp Prime emulator)

I = ∫(1/(x^3-2), x)
  = 2^(1/3)/2 * ∫(1/(y^3-1), y)
  = 2^(1/3)/2 * (-atan((y+1/2)/(1/2*√3))/√3 - 1/6*ln(y^2+y+1) + 1/3*ln(abs(y-1)))

Back substitute y = x/2^(1/3), and we are done.

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The reason for substitution is y^3-1 = (y-1)*(y-w)*(y-w²), where w = exp(i*2*pi/3)

3/(y^3-1) = 1/(y-1) + w/(y-w) + w²/(y-w²)

I = 2^(1/3)/6 · ∫(3/(y^3-1), y)
  = 2^(1/3)/6 · (ln(y-1) + w·ln(y-w) + w²·ln(y-w²))

“I’m not supposed to see this in Prime”, but I do see it, and it doesn’t make sense to me. I can use simple algebra to do this integral by hand and while it looks pretty ugly, it’s actually not difficult to do. I can’t even simplify it because one matrix has a 7 parameter argument, and one 6. I can’t replace the empty spot with a 0, else I receive a wrong answer. There is no literature on rootof function on the HP Prime. I consider this a bug and should be fixed. I’m a math professor and this result is less than useless to me especially when a competitor calculator computes it in seconds.