How to evaluate these indefinite integrals with $\sqrt{1+x^4}$? These integrals are supposed to have an elementary closed form, but Mathematica only returns something in terms of elliptic integrals. I got them from the book Treatise on Integral Calculus by Edwards. How can we evaluate them?

$$
I = \int{\frac{\sqrt{1+x^4}}{1-x^4}dx}\\
J = \int{\frac{x^2}{(1-x^4)\sqrt{1+x^4}}} dx 
$$


The problem is on p. 319 in the 1921 edition of Volume I. Write 
$$
I = \int \frac{1+x^4}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,\qquad
J = \int \frac{x^2}{(1-x^4)} \frac{dx}{\sqrt{1+x^4}}
,
$$
and note that $I=\frac{1}{2}(A+B)$ and $J=\frac{1}{4}(A-B)$, where
$$
A = \int\frac{1+x^2}{1-x^2} \frac{dx}{\sqrt{1+x^4}}
,\qquad
B = \int\frac{1-x^2}{1+x^2} \frac{dx}{\sqrt{1+x^4}}
.
$$
These integrals $A$ and $B$ appear in an exercise on p. 103, and can be solved by setting $z=\frac{\sqrt{1+x^4}}{x}$ (for $x > 1$, say, so that the change of variables is invertible; the final result doesn't depend on this assumption, as can be checked by differentiating it). This gives
$$
\frac{dz}{dx} = \frac{(x^2+1)(x^2-1)}{x^2 \sqrt{1+x^4}}
,\quad
z^2 = x^2 + \frac{1}{x^2}
,\quad
z^2 \pm 4 = \left( x \pm \frac{1}{x} \right)^2 = \left( \frac{x^2 \pm 1}{x} \right)^2
,
$$
so that
$$
A = -\int \frac{dz}{z^2-2}
,\qquad
B = -\int \frac{dz}{z^2+2}
,
$$
and I think you can take it from there!
(Another option would be to do as Edwards suggests on p. 311, and evaluate $A$ and $B$ by letting $z=1/(x-\frac{1}{x})$ and $z=1/(x+\frac{1}{x})$, respectively.)
