1. ## Integration Help

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

2. Are you certain it is not,
$\displaystyle \int \frac{xdx}{\sqrt{(a x^4 + b x^2 + c)}}$??

3. I havn't done any intergration yet, but what I remember from calculus is to rationalise the denominator

$\displaystyle \int dx \frac{\sqrt {ax^4 + bx^2 + c}}{ax^4 + bx^2 + c}$

Don't know if that helps...

4. 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.

5. Maybe there reason why it is done numerically is because it cannot be precisely computed!

There are cases like that so do not be supprised.

6. Originally Posted by ThePerfectHacker
Maybe there reason why it is done numerically is because it cannot be precisely computed!

There are cases like that so do not be supprised.
Ok, I will continue with the computer method.

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.

7. ## new equation form

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

8. Don't quote me on this.

I don't have my book with me at the moment, but the integral in question (I think!) appears in my Mechanics book. If I remember correctly, the integral can be reduced to an elliptic integral. They have no closed form.

-Dan

9. Originally Posted by topsquark
...elliptic integral...
I had a good feeling somebody was going to mention that!
But as far as my post #5 is correct. That is why they used numerical methods.

10. 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.

11. I don't know if your still working on it, but maybe this may help The Integral Calculator

12. Originally Posted by chancey
I don't know if your still working on it, but maybe this may help The Integral Calculator
Wow, this is great.

Thanks.

13. 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.