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In this post, I reached an integral in the form

$$\int_0^1\frac{dx}{\sqrt{ax^4+bx^2+c}}\tag{1}$$ where $b=-3$, $c=1$. I am stack here.

WolframAlpha did the indefinite integral. But, I couldn't get the result for the definite integral.

Thanks for any hint, answer or reference suggestion for solving integrals in the form $(1)$.

Bob Dobbs
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2 Answers2

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Using the integral definition of the inverse Jacobi sn function,

$$\frac d{dz}\text{sn}^{-1}(z,m)=\frac1{\sqrt{1-z^2}\sqrt{1-mz^2}}$$

we get:

$$\frac d{dz}\text{sn}^{-1}(a z,m)=\frac a{\sqrt{1-a^2z^2}\sqrt{1-ma^2z^2}}=\frac1{a\sqrt m\sqrt{\frac1{a^2}-z^2}\sqrt{\frac1{a^2m}-z^2}}= -\frac 1{a\sqrt m\sqrt{z^2-\frac1{a^2}}\sqrt{z^2-\frac1{m a^2}}} $$

Therefore:

$$\frac1{\sqrt m}\frac d{dz}\text{sn}^{-1}\left(\frac z{\sqrt a},\frac 1m\right)=-\frac{\sqrt a}{\sqrt{z^2-a}\sqrt{z^2-am}}$$

and finally,

$$\int_0^1\frac{dx}{\sqrt{(x^2-a)(x^2-b)}}=-\frac{\text{sn}^{-1}\left(\frac1{\sqrt a},\frac ab\right)}{\sqrt a}= -\frac{\text{sn}^{-1}\left(\frac1{\sqrt b},\frac ba\right)}{\sqrt b} $$

Shown here. Also, using inverse Jacobi NS $\text{ns}^{-1}(z,m)$, and $z\in\Bbb R,m<1$ simplifies to:

$$\int_0^1\frac{dx}{\sqrt{(x^2-a)(x^2-b)}}= -\frac{\text{ns}^{-1}\left(\sqrt a,\frac ab\right)}{\sqrt a}= -\frac{\text{ns}^{-1}\left(\sqrt b,\frac ba\right)}{\sqrt b} $$

Using special case formulas, one finds Elliptic K:

$$\int_0^1\frac{dt}{\sqrt{t^2-1}\sqrt{t^2-a}}=-\text K\left(\frac 1a\right)$$

Тyma Gaidash
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1

The problem is that Wolfram Alpha does not know anything about $a$.

Without any assumptions, the result write

$$-i \sqrt{\frac{\sqrt{9-4 a}-3}{2 a}}\times $$ $$F\left(i \sinh ^{-1}\left( \sqrt{\frac{2a}{\sqrt{9-4 a}-3}}\right)|\frac{(9-2 a)-3 \sqrt{9-4 a}}{2 a}\right)$$ which will simplify depending on the value of $a$.

It is a real $\forall a > \frac 94$

Claude Leibovici
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