UPDATE:
The Wikipedia article on Ramanujan's Master Theorem now mentions the method of brackets.
In order to evaluate the general case, namely $$\int_{0}^{\infty} \frac{x^{2\ \nu}\sin(yx)}{ (x^{2}+a^{2})^{ \mu +1}} \, \mathrm dx \, , \quad (-1 < \nu < \mu+1, \ y >0, \ a >0),$$ I'm going to use a technique referred to as the method of brackets.
The method of brackets is a generalization of Ramanujan's Master Theorem.
The term bracket refers to the association of the symbol $\langle a \rangle$ with the divergent integral $\int_{0}^{\infty} x^{a-1} \mathrm dx$.
I think some of the theory behind this technique is still a work in progress.
You can read about the method in the following papers:
Definite integrals by the method of brackets. Part 1
The method of brackets. Part 2: Examples and applications
Evaluation of entries in Gradshteyn and Ryzhik employing the method of brackets
On the Method of Brackets: Rules, Examples, Interpretations and Modifications
The hypergeometric representation of $\sin(yx)$ is $$yx \, _{0}F_{1} \left(; \frac{3}{2}; - \frac{-(yx)^{2}}{4} \right) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)} \frac{(yx)^{2n+1}}{4^{n}} = \sum_{n=0}^{\infty} \phi_{n} \, \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)} \frac{(yx)^{2n+1}}{4^{n}}.$$
And according to Rule 3.1 in the first paper, the function $(x^{2}+a^{2})^{-\mu -1}$ is assigned the bracket series $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \phi_{k,m} \, \frac{x^{2k} a^{2m}}{\Gamma(\mu+1)} \langle k + m+ \mu+1 \rangle.$$
Therefore, the integral $ \int_{0}^{\infty} x^{2 \nu}\sin(yx) (x^{2}+a^{2})^{- \mu -1} \, \mathrm dx$ corresponds to the bracket series $$\frac{1}{\Gamma(\mu+1)}\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \phi_{k,m,n} \, \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)} \frac{y^{2n+1}}{4^{n}} a^{2m} \langle k + m+ \mu+1 \rangle\langle2k+2n+2\nu+2 \rangle. $$
And the matrix equation corresponding to the vanishing of the brackets is
$$ \begin{pmatrix}
1 & 1 & 0 \\
2 & 0 & 2 \\
\end{pmatrix} \begin{pmatrix} k \\ m \\ n \end{pmatrix} = \begin{pmatrix} -\mu -1 \\ -2\nu -2 \end{pmatrix}.$$
Since the matrix equation is underdetermined, we will need to divide the evaluation of the bracket series into several cases.
Let's first let $m$ be a free parameter.
Then the brackets vanish if $k= -m-\mu-1$ and $n = m+ \mu-\nu$, with corresponding determinant $2$.
So according to Rule 3.3 in the first paper, the contribution to the integral is $$\begin{align} S_{1} &= \small \frac{1}{2\Gamma(\mu+1)} \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} + \mu - \nu + m \right)} \frac{y^{2m+ 2\mu - 2\nu+1}}{4^{m + \mu - \nu}} a^{2m} \, \Gamma \left(m+ \mu + 1 \right) \Gamma \left(\nu -\mu - m\right) \\ &= \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1} } \frac{1}{\Gamma(\mu - \nu + \frac{3}{2})} \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{\Gamma (\mu - \nu + \frac{3}{2}) }{\Gamma \left(\mu - \nu + \frac{3}{2} + m \right)} \, \frac{\Gamma( \mu + 1+ m)}{\Gamma(\mu+1)} \Gamma \left(\nu -\mu - m\right)\left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &= \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{1}{\Gamma(\mu - \nu + \frac{3}{2})}\sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{(\mu+1)_{m} }{(\mu-\nu + \frac{3}{2})_{m}} \, \frac{(-1)^{m}\Gamma(\mu - \nu +1) \Gamma(\nu - \mu)}{\Gamma(\mu - \nu +1 + m)}\left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &= \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{\Gamma(\nu - \mu)}{\Gamma(\mu - \nu + \frac{3}{2})}\sum_{m=0}^{\infty} \frac{1}{m!} \frac{(\mu+1)_{m} }{(\mu-\nu + \frac{3}{2})_{m}(\mu - \nu+1)_{m}} \left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &= \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{\Gamma(\nu - \mu)}{\Gamma(\mu - \nu + \frac{3}{2})} \, _{1}F_{2} \left(\mu+ 1; \mu - \nu + \frac{3}{2}, \mu - \nu+1; \frac{a^{2}y^{2}}{4} \right). \end{align}$$
Next let $n $ be a free parameter.
Then the brackets vanish if $k = - n - \nu- 1 $ and $m = n + \nu - \mu$, with corresponding determinant $-2$.
So according again to Rule 3.3, the contribution to the integral is $$\begin{align} S_{2} &= \small \frac{1}{|-2| \Gamma(\mu+1)} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{3}{2} \right) }{\Gamma \left(\frac{3}{2}+n \right)} \frac{y^{2n+1}}{4^{n}} a^{2n+2\nu - 2 \mu} \, \Gamma(n + \nu +1) \Gamma(\mu - \nu-n) \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{\Gamma \left(\frac{3}{2} \right) }{\Gamma \left(\frac{3}{2}+n \right)} \frac{\Gamma(\nu+1+n)}{\Gamma(\nu+1)} \Gamma(\mu - \nu-n) \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{(\nu+1)_{n}}{\left(\frac{3}{2} \right)_{n}} \frac{(-1)^{n}\Gamma(\nu- \mu +1) \Gamma(\mu - \nu)}{\Gamma(\nu - \mu+1+n)} \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \Gamma(\mu - \nu) \sum_{n=0}^{\infty} \frac{1}{n!}\frac{(\nu+1)_{n}}{\left(\frac{3}{2} \right)_{n} (\nu-\mu+1)_{n}} \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \Gamma(\mu - \nu) \, _{1}F_{2} \left(\nu+1; \nu-\mu+1, \frac{3}{2}; \frac{a^{2}y^{2}}{4} \right). \end{align}$$
Finally, if we let $k$ be a free parameter, we end up with a $_{3}F_{0}$ series. These series never converge. So according to Rule 3.4 in the first paper, we should discard this contribution.
EDIT: A $_{3}F_{0}$ series does reduce to a finite polynomial if at least one parameter is a nonpositive integer, but that's not a possibility here.
According to Rule 3.4, the value of the integral should be $S_{1}+S_{2}$.
Indeed, if we add $S_{1}$ and $S_{2}$ we miraculously get the result stated in Harry Bateman's Tables of Integral Transforms.
However, I'm not sure how to interpret the result when $\nu - \mu$ is zero or a negative integer.
