How can I evaluate $\int_0^1\frac{dx}{\sqrt{ax^4-3x^2+1}}$? 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 $(1)$.
 A: 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:
$$\int_0^1\frac{dt}{\sqrt{t^2-1}\sqrt{t^2-a}}=-\text K\left(\frac 1a\right)$$
A: 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  $\forall a > \frac 94$
