$\int\frac{a^4-x^4}{(x^4+a^2x^2+a^4)^{3/2}} dx$

Question

$$\int\frac{a^4-x^4}{(x^4+a^2x^2+a^4)^{3/2}} dx$$

I have tried rewriting $a^4-x^4$ as $x^4+a^2x^2+a^4-(2a^4+a^2x^2)$ and $x^4+a^2x^2+a^4$ as $(x^2+a^2+ax)^\frac{3}{2}(x^2+a^2-ax)^\frac{3}{2}$ which on breaking into partial fractions gives $$\int (x^4+a^2x^2+a^4)^{-1/2} +\int\frac{a^2(2a^2+x^2)}{(x^2+a^2+ax)^\frac{3}{2}(x^2+a^2-ax)^\frac{3}{2}}$$ Then I took $\frac{x}{a}$ as one variable $y$ but I didn't get much from this.

(Also I tried inputting this into Wolfram Alpha but for some reason it can't do it, no idea what I'm doing wrong)

Answer 1

Divide numerator and denominator by $x^3$

$$I = \int \frac{\frac{a^4}{x^3}-x}{\left(x^2+a^2+\frac{a^4}{x^2}\right)^{\frac{3}{2}}}\:dx = -\frac{1}{2}\int \frac{d\left(x^2+\frac{a^4}{x^2}\right)}{\left(x^2+\frac{a^4}{x^2}+a^2\right)^{\frac{3}{2}}} = \frac{1}{\sqrt{x^2+\frac{a^4}{x^2}+a^2}}+C$$

$$\implies \boxed{I = \frac{x}{\sqrt{x^4+a^2x^2+a^4}}+C}$$

Answer 2

We could even do it without integration. $$I=\int\frac{a^4-x^4}{(x^4+a^2x^2+a^4)^{3/2}} dx=\frac 1a \int \frac{1-y^4}{(y^4+y^2+1)^{3/2}} dy$$ Now, because of the power in denominator, assume that $$\int \frac{1-y^4}{(y^4+y^2+1)^{3/2}} dy=\frac {P_n(y)}{(y^4+y^2+1)^{1/2}}$$ Differentiate both sides and simplify to get $$1-y^4=\left(y^4+y^2+1\right) P_n'(y)-\left(2 y^3+y\right) P(y)$$ Comparing the degrees on both sides it is clear that $n=1$. So make $P_1(y)=\alpha+\beta y$ and replace $$1-y^4=\beta -\alpha y-2 \alpha y^3-\beta y^4$$ then $\alpha=0$ and $\beta=1$