Please for some help with this integration problem.
$\displaystyle
\int \frac{dx}{\sqrt{(a x^4 + b x^2 + c)}}
$
i.e.
integral ((a x^4 + b x^2 + c)^–½) dx
Thanks for the rationalize approach. I will look into that.
The problem is based on predicting the rotation of a solid body after any given time. As far as I know there is no yet found solution. The integration is done numerically on a computer. I am desparately trying to get a theoretical formula. Alas, I am not that great with integration; I am at GCE A-level.
The equation is actually correctly stated.
Ok, I will continue with the computer method.Originally Posted by ThePerfectHacker
I actually broke down the numerical method into this itegration and a matrix inverse. I expect that this approach have been attempted before, and hoped against.
Thanks for your expertise.
I have a broken down the original equation using partial fractions
$\displaystyle
\frac{1}{\sqrt{(a x^4 + b x^2 + c)}}
$
Using the roots of a quadratic equation, let
$\displaystyle m = \frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\displaystyle n = \frac{-b-\sqrt{b^2-4ac}}{2a}$
Therefore becomes
$\displaystyle
\frac{1}{\sqrt{(x^2 + m)(x^2 + n)}}
$
Trying pratial fractions,
$\displaystyle
\frac{1}{\sqrt{(x^2 + m)(x^2 + n)}} = \frac{A}{\sqrt{(x^2 + m)}} + \frac{B}{\sqrt{(x^2 + n)}}
$
Therefore
$\displaystyle
A\sqrt{(x^2 + n)} + B\sqrt{(x^2 + m)} = 1
$
When $\displaystyle x = i \sqrt{m}$then $\displaystyle 1 = A \sqrt{((i \sqrt{m})^2+n)}$and $\displaystyle A=\frac{1}{\sqrt{-m+n}}=\frac{1}{\sqrt{n-m}}$
Likewise
$\displaystyle B=\frac{1}{\sqrt{m-n}}$
So the original becomes
$\displaystyle
\frac
{1}
{
(\sqrt{n-m})(\sqrt{x^2+m})
} +
\frac
{1}
{
(\sqrt{m-n})(\sqrt{x^2+n})
}
$
Expanding to avoid imaginary values
$\displaystyle
\frac
{1}
{\sqrt{nx^2+nm-mx^2-m^2} } +
\frac
{1}
{\sqrt{mx^2+mn-nx^2-n^2} }
$
The new form is
$\displaystyle
\frac{1}
{\sqrt{(n-m)x^2+(nm-m^2)} } +
\frac{1}
{\sqrt{(m-n)x^2+(mn-n^2)} }
$
How can this proceed further? I noticed that it does not fall exactly in the arcsin value for the integral
The equation was not correct due to the ± issue of the square root. I will try to re-formulate the equation keeping the sign in mind. Consequently, the result I gave in post one would be rather unsolvable.
I also looked at Elliptic Integral and Jacobian Elliptic Functions. I could not follow Closed Form.
You are right: This integral form is done in numerical methods.
I don't know if your still working on it, but maybe this may help The Integral Calculator
I think that I finally understand the closed form and the ellipsoid concept
Firstly, the first equation is not of a 'function' therefore the area under the graph would not be closed, and to explain it more vividly: the eqution is that of an ellipse!
EDIT: I think that the arctan integral is only valid for a range of x<1
Well, I did the equation for this over, if any is interested in helping. I am putting it in the Advance Topics section.