= Solution
Put $c=\beta/2-\alpha$. The change of measure and part (a)(iii) yield
$$
L(t,\alpha,\beta)=e^{-\beta t/2}\mathbb E_{\mathbb Q_t}e^{cB_t^2}.
$$
For a centered <normal random variable> $G$ of <variance> $v>0$, a <Gaussian integral> gives the <Gaussian quadratic exponential moment>
$$
\mathbb Ee^{cG^2}=(1-2cv)^{-1/2},\qquad 1-2cv>0.
$$
Here $c\leq\beta/2$ and $\beta v_t<1/2$ for finite positive $t$, so the required positivity holds, including when $c>0$. Thus
$$
L(t,\alpha,\beta)
=\left[e^{\beta t}\left(1-(\beta-2\alpha)
\frac{1-e^{-2\beta t}}{2\beta}\right)\right]^{-1/2}.
$$
Expanding the two exponentials in terms of <hyperbolic functions> gives
$$
\boxed{L(t,\alpha,\beta)
=\left(\cosh(\beta t)+\frac{2\alpha}{\beta}\sinh(\beta t)\right)^{-1/2}\quad(\beta>0).}
$$
At $t=0$ this is one. As $\beta\downarrow0$, the denominator tends to $1+2\alpha t$, recovering part (a)(i). At $\alpha=0$ it gives the <Laplace transform of the integrated square of Brownian motion>. The joint terminal-and-integral transform is the <Joint Laplace transform of Brownian square and its time integral>.
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