Find the ratio of integrals $\int_0^1 (1\pm x^4)^{-1/2}\,dx$ How to find this ratio

$$\frac{\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x^{4}}}\mathrm{d}x}{\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1-x^{4}}}\mathrm{d}x}$$

without evaluating each integral? The integrals themselves can be expressed as elliptic integrals, but the ratio may be simpler.

For the numerator, change variable to $t = \tan\frac{\theta}{2}$, we get
$$\int_0^1 \frac{dt}{\sqrt{1+t^4}} =
\int_0^1 \frac{dt}{\sqrt{(1+t^2)^4 - 2t^2}} =
\frac12\int_0^1 \frac{1}{\sqrt{1 - \frac12\left(\frac{2t}{1+t^2}\right)}}\frac{2dt}{1+t^2}\\
= \frac12 \int_0^{\pi/2}\frac{d\theta}{\sqrt{1-\frac12\sin^2\theta}}
= \frac12 K\left(\frac{1}{\sqrt{2}}\right)
$$
For the denominator, change variable to $t = \cos\theta$, we have
$$\int_0^1 \frac{dt}{\sqrt{1-t^4}}
= \int_{\pi/2}^0\frac{d\cos\theta}{\sqrt{(1-\cos^2\theta)(1+\cos^2\theta)}} = \int_0^{\pi/2}\frac{d\theta}{\sqrt{1+\cos^2\theta}}\\
= \int_0^{\pi/2}\frac{d\theta}{\sqrt{2-\sin^2\theta}}
= \frac{1}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}}\right)
$$
where $$K(k) = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}} = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}$$ 
is the complete elliptic integral of the first kind. 
Compare the two expressions, it is clear the desired ratio is $\displaystyle\;\frac{1}{\sqrt{2}}$.
Update
Thinking more about this, we can completely get rid of the use of elliptic integral.  
For the denominator, if we change variable to $t = \frac{1-s^2}{1+s^2}$, 
it becomes
$$
\int_1^0 \frac{d\left(\frac{1-s^2}{1+s^2}\right)}{\sqrt{1 - \left(\frac{1-s^2}{1+s^2}\right)^4}}
= 4\int_0^1 \frac{sds}{\sqrt{(1+s^2)^4 - (1-s^2)^4}}
= 4\int_0^1 \frac{sds}{\sqrt{8s^2 + 8s^6}}
= \sqrt{2}\int_0^1 \frac{ds}{\sqrt{1+s^4}}$$
This is nothing but $\sqrt{2}$ times the numerator!
