I am working on an integral and I need some help. From the book Irresistible Integrals, I am given the problem of obtaining a closed form for:
$$N(j,n;a)=\int_0^\infty \frac{x^{2j}\mathrm dx}{(x^4+2ax^2+1)^{n+1}},\quad n,j\in\Bbb N,\, 0\leq j\leq 2n+1,\,a>-1$$
I have been able to make significant progress, but there is an integral that I am stuck on. Here's what I've done. Using the substitution $x=1/u$, it is easily shown that $$N(j,n;a)=N(2n-j+1,n;a)\tag{1}$$ With $x=\tan\frac{t}2$, we have that $$N(j,n;a)=\frac12\int_0^\pi \frac{\tan(t/2)^{2j}\sec(t/2)^2\mathrm dt}{\left[\tan(t/2)^4+2a\tan(t/2)^2+1\right]^{n+1}}$$ Which boils down to $$N(j,n;a)=\frac1{2(2-2a)^{n+1}}\int_0^\pi\frac{(1+\cos x)^{2n-j+1}(1-\cos x)^j\mathrm dx}{\left[\frac{1+a}{1-a}+\cos(x)^2\right]^{n+1}}\tag{2}$$ And from $(1)$, we have that $$N(j,n;a)=\frac1{2(2-2a)^{n+1}}\int_0^\pi\frac{(1+\cos x)^j(1-\cos x)^{2n-j+1}\mathrm dx}{\left[\frac{1+a}{1-a}+\cos(x)^2\right]^{n+1}}\tag{3}$$ So we average $(2)$ and $(3)$ to get $$\frac1{4(2-2a)^{n+1}}\int_0^\pi\frac{(1+\cos x)^j(1-\cos x)^{2n-j+1}+(1+\cos x)^{2n-j+1}(1-\cos x)^j}{\left[\frac{1+a}{1-a}+\cos(x)^2\right]^{n+1}}\mathrm dx$$ To which we may apply the binomial theorem twice to get $$N(j,n;a)=\frac1{4(2-2a)^{n+1}}\sum_{r=0}^{j}\sum_{k=0}^{2n-j+1}R_{r,k}^{j,n}T_{n}^{r,k}(a)$$ Where $$R_{r,k}^{j,n}=\left[(-1)^r+(-1)^k\right]{2n-j+1\choose k}{j\choose r}$$ And $$T_{n}^{r,k}(a)=\int_0^\pi \frac{\cos(x)^{r+k}\mathrm dx}{\left[\frac{1+a}{1-a}+\cos(x)^2\right]^{n+1}}$$ It is probably easier to consider the integral $$V_n^{w}(a)=\int_0^\pi \frac{\cos(x)^w\mathrm dx}{[a+\cos(x)^2]^{n+1}}\tag{4}$$ So we have that $$T_{n}^{r,k}(a)=V_n^{r+k}\left(\frac{1+a}{1-a}\right)$$ Other than inventing $V_n^w(a)$, I have been unable to make any progress on the remaining integral. I tried making a tangent-half angle substitution, but that didn't give me any new ideas. I also tried $u=2x$ to get rid of the $\cos(x)^2$ in the denominator of $(4)$, and while it worked, it gave me a nasty $(1+\cos u)^{w/2}$ in the numerator which I could not apply the binomial theorem to unless $w$ was even. In the case of $w=2m$ for $m\in\Bbb N_0$ we would have $$V_n^{2m}(a)=2^{n-m+1}\int_0^\pi \frac{(1+\cos u)^m\mathrm du}{(2a+1+\cos u)^{n+1}}$$ Which is, with the binomial theorem, $$V_n^{2m}(a)=2^{n-m+1}\sum_{i=0}^{m}{m\choose i}\int_0^\pi \frac{\cos(u)^i\mathrm du}{(2a+1+\cos u)^{n+1}}$$ $$V_n^{2m}(a)=2^{n-m+1}\sum_{i=0}^{m}{m\choose i}V_n^{i}(2a+1)$$ Although this does not help much either. Could I have some help, preferably with $(4)$? Thanks.
Update
As we showed, $$V_n^{2m}(a)=2^{n-m+1}\sum_{i=0}^{m}{m\choose i}V_n^{i}(2a+1)$$ This is actually very significant, because I have reason to believe that the definition $$R_{r,k}^{j,n}=\left[(-1)^r+(-1)^k\right]{2n-j+1\choose k}{j\choose r}$$ actually ensures that $r+k$ (that matter) are always going to be even. This is because $$(-1)^r+(-1)^k=\begin{cases} 2 &: r,k\,\text{even}\\ -2 &: r,k\,\text{odd}\\ 0 &: \text{otherwise} \end{cases}$$ And since $\text{even}+\text{even}=\text{even}$ and $\text{odd}+\text{odd}=\text{even}$, we have that $$R_{r,k}^{j,n}=0$$ for all odd $r+k$. Hence we have $$N(j,n;a)=\frac1{4(2-2a)^{n+1}}\sum_{0\leq r\leq j\\0\leq k\leq 2n-j+1\\ r+k\,\text{even} }C_{r,k}^{j,n}T_{n}^{r,k}(a)$$ With $$C_{r,k}^{j,n}=(-1)^r{2n-j+1\choose k}{j\choose r}$$
So indeed we need only consider $$V_n^{2m}(a)=2^{n-m+1}\sum_{i=0}^{m}{m\choose i}V_n^{i}(2a+1)$$ Although I am unsure as to how one can find an explicit form of this integral. Thanks in advance.