To evaluate
\begin{equation}
I=\frac{2\Gamma(m-\frac{1}{2})}{m\pi\Gamma(m)}\int\limits_{0}^{\infty}t^{\frac{1}{2}}{}_1F_1\left(\frac{1}{2},\frac{3}{2}-m,-t
\right)\,dt
\end{equation}
we replace the hypergeometric function by its representation in terms of
Laguerre polynomials (see
here)
\begin{equation}
{}_1F_1(a, a - n, z)=\frac{(-1)^nn!}{(1-a)_n}e^zL_n^{a-n-1}(-z)
\end{equation}
with $a=1/2,n=m-1,z=-t$, to express, after several simplifications,
\begin{equation}
I=\frac{2(-1)^{m-1}}{m\sqrt{\pi}}\int\limits_{0}^{\infty}t^{1/2}e^{-t}L_{m-1}^{1/2-m}(t)\,dt
\end{equation}
From the Rodrigues-type expression
\begin{equation}
L_n^\lambda(z)=\frac{e^zz^{-\lambda}}{n!}\frac{\partial^n}{\partial z^n}\left( z^{n+\lambda} e^{-z}\right)
\end{equation}
with $n=m-1,\lambda=1/2-m$,
\begin{equation}
L_{m-1}^{1/2-m}(z)=\frac{e^zz^{m-1/2}}{(m-1)!}\frac{\partial^{m-1}}{\partial z^{m-1}}\left( z^{-1/2} e^{-z}\right)
\end{equation}
Thus
\begin{equation}
I=\frac{2(-1)^{m-1}}{m!\sqrt{\pi}}\int\limits_{0}^{\infty}t^{m}\frac{\partial^{m-1}}{\partial t^{m-1}}\left( t^{-1/2} e^{-t}\right)\,dt
\end{equation}
By performing integrations by parts $m-2$ times,
\begin{equation}
\int\limits_{0}^{\infty}t^{m}\frac{\partial^{m-1}}{\partial t^{m-1}}\left( t^{-1/2} e^{-t}\right)\,dt=(-1)^{m-1}m!\int_0^\infty t^{1/2} e^{-t}\,dt
\end{equation}
Finally
\begin{equation}
I=1
\end{equation}
Effectively, the antiderivative express in terms of the gaussian hypergeometric function
$$I=\int\Big[1-(1-q)x^2)\Big]^{\frac{1}{1 - q}}\,dx=x \, _2F_1\left(\frac{1}{2},\frac{1}{q-1};\frac{3}{2};(1-q) x^2\right)$$ and the definite integral
$$J=\int_{-\infty}^{+\infty}\Big[1-(1-q)x^2)\Big]^{\frac{1}{1 - q}}\,dx=\sqrt{\frac{\pi}{q-1}}\,\,\frac{\Gamma \left(\frac{3-q}{2 (q-1)}\right)}{\Gamma
\left(\frac{1}{q-1}\right)}$$ provided that
$$2 \Re\left(\frac{q-2}{q-1}\right)<1\land (\Re(q)\geq 1\lor q\notin \mathbb{R})$$ which is your case with $1<q<3$.
For the second integral, I do not see any possible solution except a series expansion around $c=0$.
For simplicity, let $a=q-1$ and b=$\frac 1 {1-q}$ to make
$$K=\int_{-\infty}^{+\infty} \left(1+a x^2\right)^b \left(1+a (x+c)^2\right)^b\,dx$$
using
$$\left(1+a (x+c)^2\right)^{b}=\sum_{n=0}^\infty Q_n\, c^n$$
$$K=\sum_{n=0}^\infty c^n\,K_n \quad \text{where}\quad K_n=\int_{-\infty}^{+\infty} \left(1+a x^2\right)^b \,Q_n\,dx$$
We have $K_{2n+1}=0$ and
$$K_{2n}=(-1)^{\frac n2} a^{\frac {n-1}2}\sqrt \pi\ \frac{\Gamma \left(\frac{n-1}{2}-2 b\right)}{\Gamma \left(\frac{n}{2}+1\right)
\Gamma (n-2 b)}\, \prod_{k=0}^{\frac n2 -1}(b-k)^2$$
This means that $K$ is again an hypergoemetric function. I shall not write it since @Maxim gave the complete result (much more simpler than mine).
Best Answer
$\newcommand{\d}{\mathrm{d}}\newcommand{\B}{\mathfrak{B}}$We don't need hypergeometric functions here. $$\begin{align}I_n:&=\int_{\Bbb R}\frac{1}{(x^2+b^2)^{2n}}\,\d x\\&=\frac{2}{b^{4n}}\int_0^\infty\frac{1}{(x^2/b^2+1)^{2n}}\,\d x\\&=\frac{2}{b^{4n-1}}\int_0^\infty\frac{1}{(x^2+1)^{2n}}\,\d x\\&\overset{u=x^2+1}{=}\frac{1}{b^{4n-1}}\int_1^\infty u^{-2n}(u-1)^{-1/2}\,\d u\\&\overset{t=1/u}{=}\frac{1}{b^{4n-1}}\int_0^1 t^{2n-2}\left(\frac{1}{t}-1\right)^{-1/2}\,\d t\\&=\frac{1}{b^{4n-1}}\int_0^1t^{(2n-1/2)-1}(1-t)^{(1/2)-1}\,\d t\\&=\frac{1}{b^{4n-1}}\B\left(2n-\frac{1}{2},\,\frac{1}{2}\right)\\&=\frac{1}{b^{4n-1}}\cdot\frac{\Gamma\left(2n-\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{(2n-1)!}\\&=\frac{\sqrt{\pi}}{b^{4n-1}}\cdot\frac{1}{(2n-1)!}\cdot(2n-1-1/2)(2n-2-1/2)\cdots(1-1/2)\Gamma(1/2)\\&=\frac{\pi}{b^{4n-1}\cdot 2^{2n-1}}\cdot\frac{(4n-3)!!}{(2n-1)!}\end{align}$$
Where appeared $\B$ the Beta function and $\Gamma$ the gamma function, with which you clearly already are familiar. I also used $\Gamma(1/2)=\sqrt{\pi}$.