Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: Bronstein pseudoelliptic
Replies: 0

 Search Thread: Advanced Search

 clicliclic@freenet.de Posts: 1,090 Registered: 4/26/08
Bronstein pseudoelliptic
Posted: Jan 12, 2013 12:50 PM
 Plain Text Reply

I believe I read somewhere that Bronstein's pseudo-elliptic integral

INT(x/SQRT(x^4 + 10*x^2 - 96*x - 71), x)

= - 1/8*LN(- (x^6 + 15*x^4 - 80*x^3 + 27*x^2 - 528*x + 781)
*SQRT(x^4 + 10*x^2 - 96*x - 71)
+ x^8 + 20*x^6 - 128*x^5 + 54*x^4 - 1408*x^3 + 3124*x^2 + 10001)

<http://mathforum.org/kb/message.jspa?messageID=1562809>

could now be solved by Mathematica, but according to the Wolfram
Integrator site, this is still not the case (the integral is still done
in terms of incomplete elliptic F, incomplete elliptic Pi, and Root
objects).

<http://integrals.wolfram.com/index.jsp>

A problem with the above elementary antiderivative is a jump near x =
3.531 (where the radicand is negative). Can the logarithm argument be
factored perhaps?

Martin.

© The Math Forum at NCTM 1994-2016. All Rights Reserved.