calculus - Evaluate $int_{0}^{infty} frac{1}{sqrt{x(1+e^x)}}mathrm dx$

I would like to evaluate:

$$\int_{0}^{\infty} \frac{1}{\sqrt{x(1+e^x)}}\mathrm dx$$
Given that I can't find $\int \frac{1}{\sqrt{x(1+e^x)}}\mathrm dx$, a substitution is needed: I tried $$u=\sqrt{x(1+e^x)}$$ and $$u=\frac{1}{\sqrt{x(1+e^x)}}$$ but I could not get rid of $x$ in the new integral....
Do you have ideas of substitution?

Answer

$$\begin{align} \int_0^\infty\frac{1}{\sqrt{x(1+e^x)}}\mathrm{d}x &=2\int_0^\infty\frac{1}{\sqrt{1+e^{x^2}}}\mathrm{d}x\\ &=2\int_0^\infty(1+e^{-x^2})^{-1/2}e^{-x^2/2}\;\mathrm{d}x\\ &=2\int_0^\infty\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}e^{(2k+1)x^2/2}\;\mathrm{d}x\\ &=\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}\sqrt{\frac{2\pi}{2k+1}} \end{align}$$