Tough integral to recover Ornstein-Uhlenbeck solution.

Question

I am working on a tough integral that should recover the Ornstein-Uhlenbeck process solution with $x_0 = 0$. It is quite nasty and I need some guidance. The integral I wish to compute, ignoring normalisation constants, is given by \begin{equation} I = \int_{x_1^2}^{\infty} \cfrac{\exp(-\mu t/4D) \cos\left(\tfrac{1}{2D} \int_{0}^{x_1} \sqrt{\mu -x^2}dx \right)}{\sqrt{\sqrt{\mu}\sqrt{\mu - x_1^2}}} \mathop{d \mu}. \end{equation} The integral inside the cosine function can be done exactly and one gets an unwelcome $arcsin$ term. Moreover, generally for integrals like this, I would write it as the real part of an imaginary exponent and manipulate the exponent into a quadratic form and use the result of the beautiful gaussian integral. My first method of attack is to define a new variable, $\xi^2 = \mu - x_1^2$ so that the integral spans the positive real line.
Using a relationship between $\arcsin$ and $\arctan$, I believe that one has \begin{equation} I = \int_{0}^{\infty} \exp(-\tfrac{\xi^2 t}{4D}) \cos\left(\tfrac{1}{4D}(\xi^2 + x_1^2) \arctan(\tfrac{x_1}{\xi})+ \tfrac{x_1 \xi}{4D} \right) \sqrt{\xi} (\xi^2 + x_1^2)^{-\tfrac{1}{4}} d \xi, \end{equation} where I have omitted writing the term $2 \exp(-\tfrac{x_1^2 t}{4D})$ as it has been absorbed into the normalisation counter-part.
Using the result $\arctan(\tfrac{y}{x}) = \cfrac{i}{2} \log \left(\tfrac{x - iy}{x+iy} \right)$, then \begin{equation} I = \Re{\int_{0}^{\infty}} \exp(-\tfrac{\xi^2 t}{4D})\exp \left(\tfrac{i}{4D}(\xi^2 + x_1^2) \arctan(\tfrac{x_1}{\xi})+ \tfrac{i x_1 \xi}{4D} \right) \sqrt{\xi} (\xi^2 + x_1^2)^{-\tfrac{1}{4}} d \xi, \end{equation} so \begin{equation} I = \Re \int_{0}^{\infty} \exp(-\tfrac{t}{4D}((\xi-\tfrac{x_1 i}{2t})^2 + \tfrac{x_1^2}{4t^2}) \left(\cfrac{\xi - ix_1}{\xi + ix_1} \right)^{-\tfrac{\xi^2 + x_1^2}{8D}} \sqrt{\xi} (\xi^2 + x_1^2)^{-\tfrac{1}{4}} d \xi. \end{equation} Now the vestiges of a Gaussian integral are present.