Jtdylktuy

I would like to know a general closed form for
$$J=F_k(a_1,b_1,c_1;a_2,b_2,c_2)=int_0^kfrac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}mathrm{d}x$$
Where $k>0$, $4a_1c_1-b_1^2<0$, and $4a_2c_2-b_2^2<0$. I encounter forms of the integral all the time and I thought it would be beneficial to finally put that to rest. I already know that
$$I_k(a,b,c)=int_0^kfrac{mathrm{d}x}{ax^2+bx+c}=frac1gbigg[arctanbigg(frac{2ak+b}gbigg)-arctanbigg(frac bgbigg)bigg]$$
where $g=sqrt{4ac-b^2}$

Hence we have that $$J=a_1int_0^kfrac{x^2}{a_2x^2+b_2x+c_2}mathrm{d}x+b_1int_0^kfrac{x}{a_2x^2+b_2x+c_2}mathrm{d}x+c_1I_k(a_2,b_2,c_2)$$
So really one must focus on the integrals
$$H=int_0^kfrac{x^2}{a_2x^2+b_2x+c_2}mathrm{d}x$$
$$W=int_0^kfrac{x}{a_2x^2+b_2x+c_2}mathrm{d}x$$
They don't seem especially hard, but I just keep getting lost in the algebra with the constants. Could I have a bit of help? Thanks.

While it's natural to compute the integrals $$int_0^k frac{x^n ,dx}{a_2 x^2 + b_2 x + c_2},$$
it's somewhat more efficient to decompose the integral in a different way. (Herein we suppose $a_2 neq 0$; the case $a_2 = 0$ is easier.)

Notice that we can write the numerator of the integrand as
$$frac{a_1}{a_2}(a_2 x^2 + b_2 x + c_2) + B x + C$$
for some constants $B, C$ (that we can write explicitly in terms of the $a_i, b_i, c_i$), so we can write the integral as
$$int_0^k frac{a_1 x^2 + b_1 x + c_1}{a_2 x^2 + b_2 x + c_2} dx
= k cdot frac{a_1}{a_2} + color{#df0000}{int_0^k frac{B x + C}{a_2 x^2 + b_2 x + c_2} dx} .$$
Now, with a view toward applying the substitution $$u = a_2 x^2 + b_2 x + c_2, qquad du = 2 a_2 x + b_2 ,$$
we can write the numerator of the second integrand as $$B x + C = frac{B}{2 a_2} (2 a_2 x + b_2) + D$$
for some constant $D$ (that again we can write explicitly). So, the integral on the right-hand side is
$$color{#df0000}{int_0^k frac{B x + C}{a_2 x^2 + b_2 x + c_2} dx} = frac{B}{2 a_2} int_0^k frac{2 a_2 x + b_2}{a_2 x^2 + b_2 x + c} dx + D int_0^k frac{dx}{a_2 x^2 + b_2 x + c_2} .$$ Rewriting the first term on the r.h.s. with the above substitution gives
$$frac{B}{2 a_2} int_{c_2}^{a_2 k^2 + b_2 k + c_2} frac{du}{u},$$ and the second integral on the r.h.s. is the quantity you denoted $I_k(a, b, c)$, for which you've already found an expression.

Here is a fairly
simple-minded first step.

$begin{array}\
H
&=int_0^kdfrac{x^2}{a_2x^2+b_2x+c_2}dx\
&=dfrac1{a_2}int_0^kdfrac{x^2}{x^2+b^*_2x+c_2^*}dx\
&=dfrac1{a_2}int_0^kdfrac{x^2+b^*_2x+c_2^*-(b^*_2x+c_2^*)}{x^2+b^*_2x+c_2^*}dx\
&=dfrac1{a_2}int_0^kleft(1-dfrac{b^*_2x+c_2^*}{x^2+b^*_2x+c_2^*}right)dx\
&=dfrac1{a_2}left(k-int_0^kleft(dfrac{b^*_2x}{x^2+b^*_2x+c_2^*}right)dx-int_0^kleft(dfrac{c_2^*}{x^2+b^*_2x+c_2^*}right)dxright)\
end{array}
$

So $H$ reduces to
the other two,
one of which you know.

The process should work the same way with the integral you call $I$. Pull out $a$ in the denominator, complete the square, substitute $u = x + b/2a$, pull out $(4ac-b^{2})/4a^{2}$ from denominator and substitute $y = 2au/sqrt{4ac-b^{2}}$), and then you should arrive at three elementary integrals

$$ intfrac{x^{2},mathrm{d}x}{x^{2}+1} quadquad intfrac{x,mathrm{d}x}{x^{2}+1} quadquad intfrac{mathrm{d}x}{x^{2}+1}.$$

As a result of the first substitution, you should have all three of the integrals above to evaluate for integral $H$, and two of them for integral $W$.