Prove a reduction formula for: $\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}}$ 
For $n > 1\in \Bbb N$, prove that:
  $$
J_n = \int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = \\
{1\over na}\left(x^{n-1}\sqrt{ax^2 + bx + c} - {b\over 2}(2n-1)J_{n-1} - c(n-1)J_{n-2}\right)
$$

I've been working on this for a while without any success. I've first tried to use the fact that:
$$
\int {P_n(x)dx\over \sqrt{ax^2 + bx + c}} = Q_{n-1}(x)\sqrt{ax^2 + bx + c} + \int {\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
where $Q_{n-1}(x)$ is a polymonial of $n-1$ degree at max and coeeficients for $Q_{n-1}(x)$ and $\lambda$ are yet to be determined. Using polynomials from the problem statement we get:
$$
\int \frac{x^n dx}{\sqrt{ax^2 + bx + c}} = Ax^{n-1}\sqrt{ax^2 + bx + c} + \int{\lambda dx \over \sqrt{ax^2 + bx + c}}
$$
Differente both parts of the equality, after which we need to find the coefficients:
$$
\frac{x^n dx}{\sqrt{ax^2 + bx + c}} = {d(Ax^{n-1}\sqrt{ax^2 + bx + c})\over dx} + {\lambda \over \sqrt{ax^2 + bx + c}} = \\
A(n-1)x^{n-2}\sqrt{ax^2 + bx + c} + Ax^{n-1}\frac{2ax + b}{2\sqrt{ax^2 + bx + c}} + {\lambda dx \over \sqrt{ax^2 + bx + c}}$$
Multiplying both sides by $\sqrt{ax^2 + bx + c}$ and skipping some algebraic transformation I was indeed able to get that:
$$
A = {1\over na}
$$
However, the term $\lambda$ appears to be equal to $0$. Which yields:
$$
J_n = {x^{n-1}\over na}\sqrt{ax^2 + bx + c}
$$
And this approach doesn't seem to lead anywhere. I've then tried a different technique. Let $b = 2b_0$, then:
$$
aJ_{n+2} = \int \frac{ax^{n+2}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
2b_0J_{n+1} = \int \frac{2b_0x^{n+1}dx}{\sqrt{ax^2 + 2b_0x+c}}\\
cJ_n = \int \frac{cx^{n}dx}{\sqrt{ax^2 + 2b_0x+c}}
$$
Summing left and right parts we get:
$$
aJ_{n+2} + 2b_0J{n+1} + cJ_n = \\
\int \frac{(ax^2 + 2b_0x + c)x^n dx}{\sqrt{ax^2 + 2b_0x + c}}= \\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx
$$
Integration by parts yields:
$$
u = \sqrt{ax^2 + 2b_0x + c}\\
du = {2ax + 2b_0\over 2\sqrt{ax^2 + 2b_0x + c}}dx\\
dv = x^n\\
v = {x^{n+1}\over n+1}\\
\int {x^n\sqrt{ax^2 + 2b_0x + c}} dx = uv - \int vdu = \\
= {x^{n+1}\over n+1} \sqrt{ax^2 + 2b0x + c} - {J_{n+2}\over n+2} - {J_{n+1}\over 2(n+1)}
$$
Which seems to be "the other way round". The question is what technique do I use to prove what's stated in the question section? Hopefully, I didn't make typos in the body. Thank you!
 A: Hint:
$$\dfrac{d(x^m\sqrt{ax^2+bx+c})}{dx}$$
$$=mx^{m-1}\sqrt{ax^2+bx+c}+\dfrac{x^m(2ax+b)}{2\sqrt{ax^2+bx+c}}$$
Now let $x^3(2ax+b)=(2x^2+dx+e)(ax^2+bx+c)-ce$ 
Find $d,e$ by comparing the coefficients of $x,x^2,x^3$
Integrate both sides of the first relationship  wrt $x$
A: Late answer but there is an elegant way to derive the reduction formula as follows:
We have
$$
aJ_n + \frac{b}{2}J_{n-1} = \int \frac{x^{n-1}(2ax+b)dx}{2\sqrt{ax^2+bx+c}} = \int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right)
$$
We then proceed by integrating by parts
$$
\int x^{n-1} d\left(\sqrt{ax^2+bx+c}\right) = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\int x^{n-2}\sqrt{ax^2+bx+c} dx
$$
To calculate $\int x^{n-2}\sqrt{ax^2+bx+c}$, note that
$$
aJ_n + bJ_{n-1} + cJ_{n-2} = \int \frac{x^{n-2}(ax^2+bx+c)}{\sqrt{ax^2+bx+c}} = \int x^{n-2}\sqrt{ax^2+bx+c}
$$
Substituting back
$$
aJ_n + \frac{b}{2}J_{n-1} = x^{n-1}\sqrt{ax^2+bx+c} - (n-1)\left(aJ_n + bJ_{n-1} + cJ_{n-2}\right)
$$
By grouping terms, we easily get the reduction formula. QED.
