How can we integrate integral(s) of this type?

Question

So I was able to free $dx$ from the power. Now only wolfram can solve this Integral. How can I do this on my own?

$$r=\int_0^1\left(\frac{x^{12}}{(1-x^4)^3}+1\right)^{1/4}~dx$$

Answer 1

Hint:

$\int_0^1\left(\dfrac{x^{12}}{(1-x^4)^3}+1\right)^\frac{1}{4}~dx$

$=\int_0^1\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~d\left(x^\frac{1}{4}\right)$

$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~dx$

$=\dfrac{1}{4}\int_1^0(1-x)^{-\frac{3}{4}}\left(\dfrac{(1-x)^3}{x^3}+1\right)^\frac{1}{4}~d(1-x)$

$=\dfrac{1}{4}\int_0^1(1-x)^{-\frac{3}{4}}\left(\dfrac{3x^2-3x+1}{x^3}\right)^\frac{1}{4}~dx$

$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}(3x^2-3x+1)^\frac{1}{4}~dx$

Which relates to Appell Hypergeometric Function