reduction of elliptic integrals to standard form

Any integral of the form $\\int R(x,\\sqrt{P(x)})\\,dx$ , where $R$ is a rationalfunction and $P$ is a polynomial of degree 3 or 4 can be expressed as a linearcombination of elementary functions and elliptic integrals of the first, second,and third kinds.

To begin, we will assume that $P$ has no repeated roots. Were this not the case,we could simply pull the repeated factor out of the radical and be left with apolynomial of degree of 1 or 2 inside the square root and express the integral interms of inverse trigonometric functions.

Make a change of variables $z=(ax+b)/(cx+d)$ . By choosing the coefficients $a, b, c, d$ suitably, one can cast P into either Jacobi’s normal form $P(z)=(1-z^{2})(1-k^{2}z^{2})$ or Weierstrass’ normal form $P(z)=4z^{3}-g_{2}z-g_{3}$ .

Note that

R(z,\\sqrt{P(z)})=\\frac{A(z)+B(z)\\sqrt{P(z)}}{C(z)+D(z)\\sqrt{P(z)}}

for suitable polynomials $A, B, C, D$ . We can rationalize the denominator like so:

\\frac{A(z)+B(z)\\sqrt{P(z)}}{C(z)+D(z)\\sqrt{P(z)}}\\times\\frac{C(z)-D(z)\\sqrt{P(%z)}}{C(z)-D(z)\\sqrt{P(z)}}=F(z)+G(z)\\sqrt{P(z)}

The rational functions $F$ and $G$ appearing in the foregoing equation are defined like so:

	$\\displaystyle F(z)$	$\\displaystyle=$	$\\displaystyle\\frac{A(z)C(z)-B(z)D(z)P(z)}{C^{2}(z)-D^{2}(z)P(z)}$
	$\\displaystyle G(z)$	$\\displaystyle=$	$\\displaystyle 2\\frac{B(z)C(z)-A(z)D(z)}{C^{2}(z)-D^{2}(z)P(z)}$

Since $\\int F(z)\\,dz$ may be expressed in terms of elementary functions,we shall focus our attention on the remaining piece, $\\int G(z)\\sqrt{P(z)}\\,dz$ ,which we shall write as $\\int H(z)/\\sqrt{P(z)}\\,dz$ , where $H=PG$ ..Because we may decompose $H$ into partial fractions, it suffices to considerthe following cases, which we shall all $A_{n}$ and $B_{n}$ :

A_{n}(z)=\\int\\frac{z^{n}}{\\sqrt{P(z)}}\\,dz

B_{n}(z,r)=\\int\\frac{1}{(z-r)^{n}\\sqrt{P(z)}}\\,dz

Here, $n$ is a non-negative integer and $r$ is a complex number.

We will reduce thes further using integration by parts.Taking antiderivatives, we have:

\\int\\frac{z^{n-1}(zP^{\\prime}(z)+2nP(z))}{2\\sqrt{P(z)}}\\,dz=z^{n}\\sqrt{P(z)}+C

\\int\\frac{(z-r)P^{\\prime}(z)-2nP(z)}{2(z-r)^{n+1}\\sqrt{P(z)}}\\,dz=\\frac{\\sqrt{%P(z)}}{(z-r)^{n}}+C

These identities will allow us to express $A_{n}$ ’s and $B_{n}$ ’s with large $n$ in terms of ones with smaller $n$ ’s.

At this point, it is convenient to employ the specific form of the polynominal $P$ .We will first conside the Weierstrass normal form and then the Jacobi normal form.

Substituting into our identities and collecting terms, we find

4(2n+3)A_{n+2}=(2n+1)g_{2}A_{n}+2ng_{3}A_{n-1}+z^{n}\\sqrt{4z^{3}-g_{2}x-g_{3}}+C

2n(4r^{3}-g_{2}r-g_{3})B_{n+1}+(2n-1)(12r^{2}-g_{2})B_{n}+24(n-1)rB_{n-1}+4(2n%-3)B_{n-2}+\\frac{\\sqrt{4z^{3}-g_{2}x-g_{3}}}{(z-r)^{n}}+C=0

Note that there are some cases which can be integrated in elementary terms. Namely, suppose that the power is odd:

\\int z^{2m+1}\\sqrt{(1-z^{2})(1-k^{2}z^{2})}\\,dz

Then we may make a change of variables $y=z^{2}$ to obtain

\\frac{1}{2}\\int y^{2m}\\sqrt{(1-y)(1-k^{2}y)}\\,dy,

which may be integrated using elementary functions.

Next, we derive some identities using integration by parts. Since

d\\left((1-z^{2})(1-k^{2}z^{2})\\sqrt{(1-z^{2})(1-k^{2}z^{2})}\\right)=\\left(%\\frac{9}{2}k^{2}z^{3}-3(1+k^{2})z\\right)\\sqrt{(1-z^{2})(1-k^{2}z^{2})}\\,dz,

we have

	$\\displaystyle(2m+1)$		$\\displaystyle\\int z^{2m}(1-z^{2})(1-k^{2}z^{2})\\sqrt{(1-z^{2})(1-k^{2}z^{2})}%\\,dz$
	$\\displaystyle+$		$\\displaystyle\\int z^{2m+1}\\left(\\frac{9}{2}k^{2}z^{3}-3(1+k^{2})z\\right)\\sqrt{%(1-z^{2})(1-k^{2}z^{2})}\\,dz$
	$\\displaystyle=$		$\\displaystyle z^{2m+1}(1-z^{2})(1-k^{2}z^{2})\\sqrt{(1-z^{2})(1-k^{2}z^{2})}+C$

By colecting terms, this identity may be rewritten as follows:

	$\\displaystyle\\left(1+2m+\\frac{9}{2}k^{2}\\right)$		$\\displaystyle\\int z^{2m+4}\\sqrt{(1-z^{2})(1-k^{2}z^{2})}\\,dz-$
	$\\displaystyle(4+2m)(1+k^{2})$		$\\displaystyle\\int z^{2m+2}\\sqrt{(1-z^{2})(1-k^{2}z^{2})}\\,dz+$
			$\\displaystyle\\int z^{2m}\\sqrt{(1-z^{2})(1-k^{2}z^{2})}=$
			$\\displaystyle x^{2k+1}(1-z^{2})(1-k^{2}z^{2})\\sqrt{(1-z^{2})(1-k^{2}z^{2})}+C$

By repeated use of this identity, we may express any integral of the form $\\int z^{2m}\\sqrt{P(z)}\\,dz$ as the sum of a linear combination of $\\int z^{2}\\sqrt{P(z)}\\,dz$ and $\\int\\sqrt{P(z)}\\,dz$ and the product of a polyomial and $\\sqrt{P(z)}$ .

Likewise, we can use integration by parts to simplify integrals of the form

\\int\\frac{\\sqrt{P(z)}}{(z-r)^{n}}\\,dz

Will finish later — saving in case of computer crash.