The integration of $\frac1{[(1-x^2)^{1/4}+\sqrt x]^9 (1-x^2)^{1/4}}$

Question

How to solve the following integral?
$$ {\tt I} = \int_{0}^{1}\frac{{\rm d}x} {\left[\left(1 - x^{2}\right)^{1/4} + \sqrt{x}\right]^{9} \left(1 - x^{2}\right)^{1/4}} $$

My attempt: I substituted x = $\cos\theta$, then it reduces to $$ {\tt I} = \int_{0}^{\pi/2}\frac{{\rm (\sqrt{\sin\theta})}{\rm d}\theta} {\left[(\sqrt{\sin\theta}) + (\sqrt{\cos\theta})\right]^{9}} $$ after that on applying King's rule and adding the two integrals we get $$ {\tt I} = \frac{1}{2}\int_{0}^{\pi/2}\frac{{\rm d}\theta} {\left(\cos^4\theta\right) \left[1 + (\sqrt{\tan\theta})\right]^{8}} $$ Now I substituted $t = (1+\sqrt{\tan\theta})$, then I got $$ {\tt I} = \frac{1}{2}\int_{1}^{\infty}\frac{{\rm ((t-1)^5 + (t-1)) d}{\rm t}} {t^{8}} $$ Further expanding the numerator and integrating individual terms we get the answer ${\tt I} = \frac{1}{21}$.

This method is tedious and lengthy due to the big numerator. What would be a better approach(shorter) to this question?

Answer 1

I had the same solution in mind, up to the last substitution where I used $u=\sqrt{\tan\theta}$, followed by $u=\dfrac{1-v}{1+v}$:

$$\begin{align*} I &= \frac12 \int_0^\tfrac\pi2 \frac{\sec^4\theta}{\left(1+\sqrt{\tan \theta}\right)^8} \, d\theta \\ &= \int_0^\infty \frac{u \left(1+u^4\right)}{(1+u)^8} \, du \\ &= \frac1{64} \int_{-1}^1 \left(1+5v^2-5v^4-v^6\right) \, dv \end{align*}$$