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For standard Brownian motion and $\alpha ,t\ge 0$, $\beta >0$, the title gives the joint Laplace transform of the terminal square and the integrated square of Brownian motion. At $\beta =0$ the answer is $(1+2\alpha t{)}^{-1/2}$. To prove the formula, use the stochastic exponentialOn every finite interval $[0,T]$, $0<Z_t\le e^{\beta T/2}$, so the bounded local martingale criterion makes it a martingale. The Girsanov theorem changes $B$ into the zero-started Ornstein-Uhlenbeck process $dB=dW-\beta Bdt$ under the terminal density $Z_t$. Its terminal variance is $v=(1-e^{-2\beta t})/(2\beta )$. Multiplying $Z_t$ by $\exp [(\beta /2-\alpha )B_t^2-\beta t/2]$ recovers the integrand on the left of the title. The Gaussian quadratic exponential moment then gives $e^{-\beta t/2}[1-(\beta -2\alpha )v{]}^{-1/2}$, which simplifies to the title formula. At $\alpha =0$ this reduces to the Laplace transform of the integrated square of Brownian motion after Brownian scaling.

The independent squared-normal series for the integrated square of Brownian motion gives the product ${\prod }_n(1+2\lambda /((n-1/2{)}^2{\pi }^2){)}^{-1/2}$. The half-integer cosine product reduces this to $(\cosh \sqrt{2\lambda }{)}^{-1/2}$. Expanding at zero gives mean $1/2$ and variance $1/3$.