Solve Elliptic Integral: Tips & Ideas

[Robin04]

Homework Statement

$\int dx \frac{1}{\sqrt{a \sin^2{x}+b (1-\cos{x})}}$

Relevant Equations

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I need to solve this integral which I suppose is an elliptic integral but don't know what kind, I'm not that familiar with them.
Mathematica says that it can be expressed with elementary functions and gives the solution:
$-\frac{2\, tanh^{-1}(\frac{\sqrt{2a+b}\cos{\frac{x}{2}}}{\sqrt{a+b+a\cos{x}}})\sqrt{a+b+a\cos{x}}\sin{\frac{x}{2}}}{\sqrt{2a+b}\sqrt{b-b\cos{x}+a\sin^2{x}}}$,
but I believe there is a simpler form (I need to invert it and express $x$ after), as it happened before that Mathematica gave a more complex form without being able to simplify it. Can you help me with some ideas?

 

[fresh_42]

WolframAlpha has an even uglier solution. However, I fed it with $\sqrt{-a(\cos x + \alpha)^2+\beta}$ in the denominator. To simplify it, one has to get rid of the translation at the cosine term. The Weierstraß substitution $t:= \tan(x/2)$ often helps in such situations. It turns a trigonometric integral into a rational one.

 

[Robin04]

WolframAlpha has an even uglier solution. However, I fed it with $\sqrt{-a(\cos x + \alpha)^2+\beta}$ in the denominator. To simplify it, one has to get rid of the translation at the cosine term. The Weierstraß substitution $t:= \tan(x/2)$ often helps in such situations. It turns a trigonometric integral into a rational one.

with $\alpha = \frac{b}{2a}, \beta = 1+\frac{b}{a}+\frac{b^2}{4a^2}$
$\frac{1}{\sqrt{a}}\int dx\frac{1}{\sqrt{\beta-(\cos{x}+\alpha)^2}}=\frac{2}{\sqrt{a}}\int \frac{dt}{1+t^2}\frac{1}{\sqrt{\beta-(\frac{1-t^2}{1+t^2}+\alpha)^2}}$
This is how you meant it? I don't see why this is simpler. Or getting rid of $\alpha$ is the key?

 

[fresh_42]

Well, simpler in the sense that there is only one trig function left, not two. I would continue and simplify the rational function obtained by the tangent substitution. The integral lists of rational polynomials, with or without root, are considerably longer than those for rational functions in trig terms. Now it looks like $\dfrac{p(t)}{\sqrt{q(t)}}$.

 

[Robin04]

Well, simpler in the sense that there is only one trig function left, not two. I would continue and simplify the rational function obtained by the tangent substitution. The integral lists of rational polynomials, with or without root, are considerably longer than those for rational functions in trig terms. Now it looks like $\dfrac{p(t)}{\sqrt{q(t)}}$.

$\frac{2}{\sqrt{a}}\int \frac{dt}{1+t^2}\frac{1}{\sqrt{\beta-(\frac{1-t^2}{1+t^2}+\alpha)^2}} =$
The denominator of the integrand:
$(1+t^2)\sqrt{\beta-(\frac{1-t^2}{1+t^2}+\alpha)^2} = \sqrt{(1+t^2)^2\beta-(1+t^2)^2(\frac{(1-t^2)^2}{(1+t^2)^2}+2\alpha \frac{1-t^2}{1+t^2}+\alpha^2)} = \sqrt{(1+t^2)^2\beta-(1-t^2)^2-2\alpha(1-t^2)(1+t^2)-(1+t^2)^2\alpha^2}=\sqrt{(\beta-\alpha^2)(1+t^2)^2-(1-t^2)^2-2\alpha(1-t^4)}=\sqrt{t^4(\beta-\alpha^2-1+2\alpha)+t^2(2\beta-2\alpha^2+2)+\beta-\alpha^2-1-2\alpha}$
If i substitute for $t^2$ then I have a cubic polynomial for which I didn't find any formula, and if the last $2\alpha$ constant would have a + sign my life would be saved but I haven't found any sign errors.

 

[fresh_42]

So we have $\gamma\displaystyle{\int} \dfrac{ds}{\sqrt{s}\cdot \sqrt{As^2+Bs+C}}$ with $s=t^2$. An integration by parts could help. Cp.
https://en.wikipedia.org/wiki/List_...nctions#Integrals_involving_R_=_√ax2_+_bx_+_chttps://de.wikibooks.org/wiki/Forme...#Integrale_die_√R_=_√(ax2_+_bx+_c)_beinhalten
There is no easy solution, since otherwise not both, Mathematica and WolframAlpha would have produced monstrous closed forms.

 

[Robin04]

So we have $\gamma\displaystyle{\int} \dfrac{ds}{\sqrt{s}\cdot \sqrt{As^2+Bs+C}}$ with $s=t^2$. An integration by parts could help. Cp.
https://en.wikipedia.org/wiki/List_...nctions#Integrals_involving_R_=_√ax2_+_bx_+_chttps://de.wikibooks.org/wiki/Forme...#Integrale_die_√R_=_√(ax2_+_bx+_c)_beinhalten
There is no easy solution, since otherwise not both, Mathematica and WolframAlpha would have produced monstrous closed forms.

$\gamma\displaystyle{\int} \dfrac{ds}{\sqrt{s}\cdot \sqrt{As^2+Bs+C}}=\gamma\left[\frac{2\sqrt{s}}{\sqrt{As^2+Bs+C}}+\displaystyle{\int}ds\frac{\sqrt{s}(B+2As)}{(As^2+Bs+C)^{3/2}}\right]=\gamma\left[ \frac{2\sqrt{s}}{\sqrt{As^2+Bs+C}}+B\displaystyle{\int}ds\frac{\sqrt{s}}{(As^2+Bs+C)^{3/2}}+2A\displaystyle{\int}ds\frac{s^{3/2}}{(As^2+Bs+C)^{3/2}}\right]$

I haven't found any of these in the lists you sent me :/

 

[fresh_42]

Yeah, it's still nasty. WolframAlpha spit out:
https://www.wolframalpha.com/input/?i=int+(1/+root+(x^3+Ax^2+Bx))dx+=Seems there is no easy closed form, at least none of which could be converted into $x= \ldots$

 

[Robin04]

Yeah, it's still nasty. WolframAlpha spit out:
https://www.wolframalpha.com/input/?i=int+(1/+root+(x^3+Ax^2+Bx))dx+=Seems there is no easy closed form, at least none of which could be converted into $x= \ldots$

Mathematica gives it with elliptic functions. I wonder how it found the solution to the original form without them... Always a mystery :/

 

[fresh_42]

I recently solved an integral $I(a,b):=\displaystyle{\int} \dfrac{dx}{\sqrt{(a^2+x^2)(b^2+x^2)}}$ with a split into substitutions $t = x \pm f(x;a,b)$ but that only gave me a recursion for $I$.

 

[Robin04]

I recently solved an integral $I(a,b):=\displaystyle{\int} \dfrac{dx}{\sqrt{(a^2+x^2)(b^2+x^2)}}$ with a split into substitutions $t = x \pm f(x;a,b)$ but that only gave me a recursion for $I$.

How do you obtain a recursion for an integral? :O

 

[fresh_42]

By a formula $I(a,b)=I(f(a,b),g(a,b))$ and then you can iterate $f,g$.