The inverse Laplace transform of $ s^{3/2}-a-bs \over s^{3/2}+a+bs$ How can I solve the inverse Laplace transform as below:
$$\mathscr{L}^{-1}\left( s^{3/2}-a-bs \over s^{3/2}+a+bs \right) $$
where a and b are constants.
Hint: we can consider
$${ s^{3/2}-a-bs \over s^{3/2}+a+bs} = {1 \over {1+as^{-3/2}+bs^{-1/2}}}-{a \over {s^{3/2}+bs+a}}-{b \over {s^{1/2}+as^{-1}+b}}$$
I applied residue theorem to each, and at the end I got 3 integrals which I could not solve, they are given by:
$${1\over \pi} \int_{0}^{\infty}\left({{ax^{-3/2}-bx^{-1/2}}\over {1+(ax^{-3/2}-bx^{-1/2})^2}}\right) \exp(-xt)\,\mathrm{d} x, $$
$${a\over \pi} \int_{0}^{\infty}\left({{x^{3/2}}\over {x^3+(a-bx)^2}}\right) \exp(-xt)\,\mathrm{d} x, $$
and
$${-b\over \pi} \int_{0}^{\infty}\left({{x^{1/2}}\over {x+(b-ax^{-1})^2}}\right) \exp(-xt)\,\mathrm{d} x. $$
Thanks!
 A: By rationalizing the denominator, we have
$\mathcal{L}^{-1}\left\{\dfrac{s^\frac{3}{2}-a-bs}{s^\frac{3}{2}+a+bs}\right\}$
$=\mathcal{L}^{-1}\left\{\dfrac{(s^\frac{3}{2}-a-bs)^2}{(s^\frac{3}{2}+a+bs)(s^\frac{3}{2}-a-bs)}\right\}$
$=\mathcal{L}^{-1}\left\{\dfrac{s^3-2s^\frac{3}{2}(bs+a)+b^2s^2+2abs+a^2}{s^3-b^2s^2-2abs-a^2}\right\}$
$=\mathcal{L}^{-1}\{1\}-\mathcal{L}^{-1}\left\{\dfrac{2s^\frac{3}{2}(bs+a)}{s^3-b^2s^2-2abs-a^2}\right\}+\mathcal{L}^{-1}\left\{\dfrac{2(b^2s^2+2abs+a^2)}{s^3-b^2s^2-2abs-a^2}\right\}$
The first term is the inverse Laplace transform of $1$ and it is equal to $\delta(t)$ .
The third term is the inverse Laplace transform of rational function.
The second term can be considered by this way:
$\mathcal{L}^{-1}\left\{\dfrac{2s^\frac{3}{2}(bs+a)}{s^3-b^2s^2-2abs-a^2}\right\}$
$=\mathcal{L}^{-1}\left\{\dfrac{1}{\sqrt s}\times\dfrac{2s^2(bs+a)}{s^3-b^2s^2-2abs-a^2}\right\}$
$=\dfrac{1}{\sqrt{\pi t}}*\mathcal{L}^{-1}\left\{\dfrac{2bs^3+2as^2}{s^3-b^2s^2-2abs-a^2}\right\}$
