|Re: Symbolic integration|
Message #6 Posted by Karl Schneider on 10 June 2009, 1:29 a.m.,
in response to message #1 by Valentin Albillo
Welcome back, Valentin!
For integrands of the form
those who remember their calculus should be able to easily solve the integral for n = 1 or 2. For n = 3 or n = 4, looking up the closed-form solutions of cubic and quartic equations, substituting the roots into the denominator, expressing the integrand as a sum of partial fractions, and performing the elementary integrations should provide the answer. However, this gets tedious right away.
For the symbolic solutions for n > 4, I suppose that Mathematica is doing something like the following procedure found at the following link:
With n as an integer power of 2, there might possibly be simplifications, but maybe not. The sin, cos, sec, and csc terms are present for every solution with n >= 7, as far as I looked.
The number of terms in the integral generally equals the order of the polynomial, as might be expected from the basic approach described above. Exception: When one term is a scalar multiple of tan-1(x), as with n = 2, 6, 10, ...(?), there are only n-1 terms in the solution.
Edited: 10 June 2009, 3:36 a.m.