Prove:
$$\mathscr{I}_{\space\text{n}}:=\int_0^\infty\text{n}\cdot x\cdot\exp\left(\text{n}\cdot x\cdot i\right)\cdot\mathscr{J}_0\left(x\cdot\sqrt{1-\text{n}^2}\right)\space\text{d}x=-\frac{1}{\text{n}}\cdot\frac{1}{\left(2-\frac{1}{\text{n}^2}\right)^\frac{3}{2}}\tag1$$
My work:
I used Laplace transform, in the following way:
$$\mathscr{L}_x\left[\mathscr{I}_{\space\text{n}}\right]_{\left(\text{s}\right)}:=\text{n}\cdot\int_0^\infty\frac{x\cdot\exp\left(\text{n}\cdot x\cdot i\right)\cdot\mathscr{J}_0\left(x\cdot\sqrt{1-\text{n}^2}\right)}{\exp\left(\text{s}\cdot x\right)}\space\text{d}x\tag2$$
Using the 'frequency-domain derivative' property of the Laplace transform:
$$\mathscr{L}_x\left[\mathscr{I}_{\space\text{n}}\right]_{\left(\text{s}\right)}=-\text{n}\cdot\frac{\partial}{\partial\space\text{s}}\left\{\mathscr{L}_x\left[\exp\left(\text{n}\cdot x\cdot i\right)\cdot\mathscr{J}_0\left(x\cdot\sqrt{1-\text{n}^2}\right)\right]_{\left(\text{s}\right)}\right\}\tag3$$
Now, using the 'frequency shifting' property of the Laplace transform:
$$\mathscr{L}_x\left[\mathscr{I}_{\space\text{n}}\right]_{\left(\text{s}\right)}=-\text{n}\cdot\frac{\partial}{\partial\space\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_0\left(x\cdot\sqrt{1-\text{n}^2}\right)\right]_{\left(\text{s}-\text{n}\cdot i\right)}\right\}\tag4$$
For the Bessel function we know:
$$\mathscr{J}_0\left(x\right):=\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{x^{2\text{p}}}{2^{2\text{p}}}\tag5$$
So, we get:
$$\mathscr{J}_0\left(x\cdot\sqrt{1-\text{n}^2}\right)=\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{\left(x\cdot\sqrt{1-\text{n}^2}\right)^{2\text{p}}}{2^{2\text{p}}}=$$ $$\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot x^{2\text{p}}\tag6$$
Combining:
$$\mathscr{L}_x\left[\mathscr{I}_{\space\text{n}}\right]_{\left(\text{s}\right)}=-\text{n}\cdot\frac{\partial}{\partial\space\text{s}}\left\{\mathscr{L}_x\left[\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot x^{2\text{p}}\right]_{\left(\text{s}-\text{n}\cdot i\right)}\right\}=$$ $$-\text{n}\cdot\frac{\partial}{\partial\space\text{s}}\left\{\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot\mathscr{L}_x\left[x^{2\text{p}}\right]_{\left(\text{s}-\text{n}\cdot i\right)}\right\}=$$ $$-\text{n}\cdot\frac{\partial}{\partial\space\text{s}}\left\{\sum_{\text{p}=0}^\infty\frac{\left(-1\right)^\text{p}}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot\frac{\left(2\text{p}\right)!}{\left(\text{s}-\text{n}\cdot i\right)^{1+2\text{p}}}\right\}=$$ $$-\text{n}\cdot\sum_{\text{p}=0}^\infty\left(-1\right)^\text{p}\cdot\frac{\left(2\text{p}\right)!}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot\frac{1+2\text{p}}{\left(\text{n}+\text{s}\cdot i\right)^2}\cdot\frac{1}{\left(\text{s}-\text{n}\cdot i\right)^{2\text{p}}}\tag7$$
Now, when $\text{s}=0$, we get:
$$\lim_{\text{s}\to0}\space\mathscr{L}_x\left[\mathscr{I}_{\space\text{n}}\right]_{\left(\text{s}\right)}=-\text{n}\cdot\sum_{\text{p}=0}^\infty\frac{\left(2\text{p}\right)!}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot\frac{1+2\text{p}}{\text{n}^{2\cdot\left(1+\text{p}\right)}}\cdot\frac{1}{\left(-1\right)^\text{p}\cdot i^{2\text{p}}}=$$ $$-\text{n}\cdot\left\{\frac{1}{\text{n}^2}+\sum_{\text{p}=1}^\infty\frac{\left(2\text{p}\right)!}{\left(\text{p}!\right)^2}\cdot\frac{\left(1-\text{n}^2\right)^\text{p}}{2^{2\text{p}}}\cdot\frac{1+2\text{p}}{\text{n}^{2\cdot\left(1+\text{p}\right)}}\cdot\frac{1}{\left(-1\right)^\text{p}\cdot i^{2\text{p}}}\right\}\tag8$$
How to continue?
There's a formula that one can use or prove:
$$\int_0^\infty e^{-ax}J_0(bx)dx=\frac{1}{\sqrt{a^2+b^2}}.$$
The proof uses the series expansion of the bessel function $J_n(x)$, which is given by
$$J_n(x)=\sum_{p=0}^\infty\frac{(-1)^p}{p!(p+n)!}\left(\frac{x}{2}\right)^{n+2p},$$
and then integration term by term and finally recognizing that the series is the Taylor expansion of the function $1/\sqrt{a^2+b^2}$. Given the integration formula, we can take derivative with respect to $-a$ to obtain
$$\int_0^\infty x\,e^{-ax}J_0(bx)dx=\frac{a}{(a^2+b^2)^{3/2}}.$$
Then plugging in numbers $a=-in$, $b=\sqrt{1-n^2}$ (analytic continuation), we have
$$\int_0^\infty nx\,e^{inx}J_0(x\sqrt{1-n^2})dx=\pm\frac{n^2}{(2n^2-1)^{3/2}}.$$
The power $3/2$ yields two branches with the $\pm$ sign and needs some discussions.