To expand, in the interval $(0 < x < c)$
$$f(x) = x^{2}$$
in a Fourier-Bessel function, I used:
$$ f(x) = \sum\limits_{n = 1}^{\infty}a_{n}J_{0}(\dfrac{\alpha_{n}x}{c})$$
Where
$$a_{n} = \dfrac{2}{c^{2}}\int_{0}^{c}x^{2} \cdot J_{0}(\dfrac{\alpha_{n}x}{c})xdx$$
To start developing I integrated the differential equation:
$$ x^{2}J''_{0}(\dfrac{\alpha_{n}x}{c}) + xJ'_{0}(\dfrac{\alpha_{n}x}{c}) + x^{2}J_{0}(\dfrac{\alpha_{n}x}{c}) = 0 \rightarrow \int_{0}^{c} x^{2}\cdot x J_{0}dx = -\int_{0}^{c} x^{2}\cdot x J''_{0}dx -\int_{0}^{c} x \cdot x J'_{0}dx $$
Since the Bessel functions have a weight of x and through integration by parts:
$$ \int_{0}^{c}x^{3}J''_{0}dx = \dfrac{c}{\alpha_{n}} J'_{0} - 3\dfrac{c}{\alpha_{n}}\int_{0}^{c}x^{2}J'_{0}dx = \dfrac{c}{\alpha_{n}} J'_{0} - 3(\dfrac{c}{\alpha_{n}})^{2}x^{2}J_{0} + 6(\dfrac{c}{\alpha_{n}})^{2}\int_{0}^{c}xJ_{0}dx$$
and
$$ \int_{0}^{0}x^{2}J'_{0}dx = \dfrac{c}{\alpha_{n}}x^{2}J_{0} - 2\dfrac{c}{\alpha_{n}}\int_{0}^{c}xJ_{0}dx$$
after applying the limits: $$ \int_{0}^{c} x^{3}J_{0}dx = -\dfrac{c^{4}}{\alpha_{n}}J'_{0}(\alpha_{n}) + (J_{0}(\alpha_{n}))^{2}c^{2}(\dfrac{3c^{2}}{\alpha_{n}^{2}} - \dfrac{c}{\alpha_{n}}) + (\dfrac{2c}{\alpha_{n}} - \dfrac{6c^{2}}{\alpha_{n}^{2}}) \int_{0}^{c}xJ_{0}dx$$
Since $$sJ_{0}(s) = \dfrac{d}{ds}(sJ_{1}(s))$$:
$$ \int_{0}^{c} x^{3}J_{0}dx = -\dfrac{c^{4}}{\alpha_{n}}J'_{0}(\alpha_{n}) + (J_{0}(\alpha_{n}))^{2}c^{2}(\dfrac{3c^{2}}{\alpha_{n}^{2}} - \dfrac{c}{\alpha_{n}}) + (\dfrac{2c}{\alpha_{n}} - \dfrac{6c^{2}}{\alpha_{n}^{2}}) cJ_{1}(\alpha_{n})$$
But I can't develop further!
Back to my old cookbook, we have $$\int x^n \,J_0(x)\,dx=\frac {x^{n+1}}{n+1}\,\,\, _1F_2\left(\frac{n+1}{2};1,\frac{n+3}{2};-\frac{x^2}{4}\right)$$ but, if $n$ is odd, after using bounds, fortunately Bessel functions are back.
So, if $$a_{n} = \dfrac{2}{c^{2}}\int_{0}^{c}x^{3} \, J_0\left(\frac{\alpha_n x}{c}\right)\,dx=\frac{2 c^2}{\alpha_n ^2}\,(2 J_2(\alpha_n )-\alpha J_3(\alpha_n )) $$ which is not your last result.