Approximating $\int \frac{1}{1 + a p^4 + b p^6}$ I'm attempting to calculate an approximate "closed form" of the integral $$\int \frac{dp}{1 + a p^4 + b p^6}$$ 
as a function of $a$ and $b$, two small parameters (of the order of $10^{-2}$). I'm really not sure how to go about doing this.
First attempt at a solution:
I naively considered performing a series expansion with respect to $a$ and $b$, and then performing the integral, but when one of the limits is infinity the integral diverges.
I'm not really used to approximating integrals so I'm completely at a loss!
EDIT:
The title earlier showed a definite integral instead of an indefinite one. 
I also forgot to mention that $a>0$ and $b>0$.
 A: For an exact solution, what I would try is first to change variable $$B p^6=x^3\implies p=\frac{\sqrt{x}}{\sqrt[6]{B}}\implies dp=\frac{dx}{2 \sqrt[6]{B} \sqrt{x}}$$ which gives $$I=\int \frac{dp}{1 + A p^4 + B p^6}=\frac{1}{2 \sqrt[6]{B}}\int\frac{dx}{ \sqrt{x} \left(x^3+{A }{B^{-2/3}}x^2+1\right)}$$ Let us call $a,b,c$ the roots of the cubic equation ( two of them will be non-real complex conjugate since $\Delta=-{4 A^3}{B^{-2}}-27<0$) and use partial fraction decomposition to get $$\frac{1}{(x-a) (x-b) (x-c)}=\frac{1}{(a-b) (a-c)
   (x-a)}+\frac{1}{(b-a) (b-c) (x-b)}+\frac{1}{(c-a) (c-b) (x-c)}$$ So, we are left with the problem of $$J(d)=\int \frac{dx}{\sqrt{x} (x-d)}=-\frac{2}{\sqrt{d}}\tanh ^{-1}\left(\frac{\sqrt{x}}{\sqrt{d}}\right)$$ which seems to be manageable.
A: I'll assume $a > 0$ and $b > 0$.
By symmetry, the integral from $0$ to $\infty$ is half the integral from $-\infty$ to $\infty$.  You can use the Residue Theorem with a semicircular contour.  The result is that your integral is
$\pi i$ times the sum of the  residues of $1/(1 + a p^4 + b p^6)$ at the roots of $1+a p^4 + b p^6$ in the upper half plane.  
The residue of $1/f(p)$ at a simple root $p=r$ is $1/f'(r)$.
EDIT:
The indefinite integral can be written as
$$ \sum_r \dfrac{\ln(p-r)}{6 b r^4 + 4 a r^3}$$
where the sum is over the roots of $f(p) = 1 + a p^4 + b p^6$, assuming these are distinct: that is in fact the case when $a > 0$ and $b > 0$, since the discriminant of that polynomial is $-64 b (4 a^3 + 27 b^2)^2$.
A series expansion in powers of $a$ and $b$, valid for fixed $p$, is easy: just write
$$ \dfrac{1}{f(p)} = \sum_{k=0}^\infty (-a p^4 - b p^6)^k $$ 
and integrate term by term.
Thus the first few terms are
$$ \dfrac{1}{f(p)} = 1 - a p^4 - b p^6 + a^2 p^8 + 2 a b p^{10} + b^2 p^{12} + \ldots $$
$$ \int \dfrac{dp}{f(p)} = C + p - \frac{a}{5} p^5 - \frac{b}{7} p^7 + \frac{a^2}{9} p^9 + \frac{2ab}{11} p^{11} + \frac{b^2}{13} p^{13} + \ldots$$
