Is there any closed form for this integral? 
The picture above is from uninstallation tool of fake antivirus in Korea. The "official" uninstallation tool will not proceed anymore unless user input the correct answer. (Nobody succeeded this) Due to its preposterousness, the image has been used as "meme" for malwares.
$$\int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-x^2}} dx$$
Anyhow, is there any closed form for the result of this definite integral? How can one compute this without calculators like Wolframalpha?
 A: I have generated a few series for the integral based on the following:
$$
I = \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-x^2}} dx
$$
$$
I = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_{0}^{1/3} \frac{x^{2n}}{\sqrt{1-x^2}} dx
$$
letting $x^2=u$
$$
I = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_{0}^{1/9} \frac{u^n}{\sqrt{1-u}} \frac{du}{2\sqrt{u}}
$$
$$
I = \sum_{n=0}^\infty \frac{(-1)^n}{2n!}\int_{0}^{1/9} u^{n-\frac{1}{2}}(1-u)^{\frac{-1}{2}}\;du
$$
The incomplete beta function, a generalisation of the beta function, is defined as
$$
B(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,dt.
$$
$$
I = \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{n!}B\left(\frac{1}{9};n+\frac{1}{2},\frac{1}{2}\right)
$$
I can't get any further this route but it seems to hold numerically. It seems we can also write which seems to hold numerically
$$
I= \frac{1}{2\sqrt{\pi}}\sum_{n=0}^\infty \frac{\Gamma(n+\frac{1}{2})(\Gamma(n+\frac{1}{2})-\Gamma(n+\frac{1}{2},\frac{1}{9}))}{n!}
$$
which contains the incomplete gamma function. This comes from the Mellin transform
$$
\int_0^\infty \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-a x^2}} dx \; da =  \int_{0}^{1/3} \frac{\Gamma(s)\Gamma(\frac{1}{2}-s)}{\sqrt{\pi}}\frac{\exp(-x^2)}{(-x^2)^s} dx = \frac{(-1)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s) \left(\Gamma \left(\frac{1}{2}-s\right)-\Gamma \left(\frac{1}{2}-s,\frac{1}{9}\right)\right)}{2 \sqrt{\pi }}
$$
by doing a double Mellin transform I managed to generate the following sum which also seems to hold numerically
$$
I = \sum_{s=0}^\infty \frac{(-1)^s}{3^{1+2s}(1+2s)s!}\;_2F_1\left(\frac{1}{2},\frac{1}{2}+s;\frac{3}{2}+s;\frac{1}{9}\right)
$$
this regenerates the sum containing the incomplete gamma function if the sum from inside the hypergeometric function is swapped with the outside sum. 
We can also introduce a parameter and take the inverse Laplace transform
$$
I(a) = \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{a-x^2}} dx
$$
$$
\mathcal{L}^{-1}_{a \to s}[I(a)] = \int_0^\frac{1}{3} \frac{e^{-x^2+sx^2}}{\sqrt{\pi s}} \; dx = \frac{\mathrm{erfi}\left(\frac{\sqrt{s-1}}{3}\right)}{2\sqrt{s(s-1)}}
$$
so we can rewrite the original integral as
$$
I(a=1)=\frac{1}{2}\int_0^\infty \frac{\mathrm{erfi}\left(\frac{\sqrt{s-1}}{3}\right)}{\sqrt{s(s-1)}}e^{-s} \; ds
$$
which again seems to hold out numerically
A: $\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{1-x}}~d(\sqrt{x})$
$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{2\sqrt{x}\sqrt{1-x}}~dx$
$=\int_0^1\dfrac{e^{-\frac{x}{9}}}{2\sqrt{\dfrac{x}{9}}\sqrt{1-\dfrac{x}{9}}}~d\left(\dfrac{x}{9}\right)$
$=\int_0^1\dfrac{e^{-\frac{x}{9}}}{6\sqrt{x}\sqrt{1-\dfrac{x}{9}}}~dx$
$=\dfrac{1}{3}\Phi_{1}\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{3}{2};\dfrac{1}{9},-\dfrac{1}{9}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)
A: Only a half answer:
$$
\int_0^{\frac{1}{3}} \frac{\exp \left(-x^2\right)}{\sqrt{1-x^2}} \, dx=\int_0^{\frac{1}{9}} \frac{\exp (-x)}{2 \sqrt{(1-x) x}} \, dx=\int_0^{\infty } \frac{\exp
   \left(-\frac{1}{x+9}\right)}{2 (x+9) \sqrt{x+8}} \, dx=
$$
Using the Laplace Transform we can write:
$$
\int_0^{\infty } \left(\mathcal{L}_x^{-1}\left[\exp \left(-\frac{1}{x+9}\right)\right](s)\right) \left(\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)\right) \, ds=
$$
$$
\int_0^{\infty } \left(e^{-9 s} \left(-\frac{J_1\left(2 \sqrt{s}\right)}{\sqrt{s}}+\delta (s)\right)\right) \left(2 e^{9 s} \pi  \left(\frac{\text{erfc}\left(\sqrt{s}\right)}{2}-2
   T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)\right)\right) \, ds=
$$
$$
\int_0^{\infty } \left(-\frac{\pi  J_1\left(2 \sqrt{s}\right) \text{erfc}\left(\sqrt{s}\right)}{\sqrt{s}}+\pi  \delta (s) \text{erfc}\left(\sqrt{s}\right)+\frac{4 \pi  J_1\left(2
   \sqrt{s}\right) T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)}{\sqrt{s}}-4 \pi  \delta (s) T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)\right) \, ds=
$$
$$
-\tan ^{-1}\left(2 \sqrt{2}\right)+\frac{\pi  I_0\left(\frac{1}{2}\right)}{2 \sqrt{e}}+\int_0^{\infty } \frac{2 \pi  J_1\left(2 \sqrt{s}\right) T\left(\sqrt{2} \sqrt{s},2
   \sqrt{2}\right)}{\sqrt{s}} \, ds=
$$
$$
-\tan ^{-1}\left(2 \sqrt{2}\right)+\frac{\pi  I_0\left(\frac{1}{2}\right)}{2 \sqrt{e}}+4 \pi  \int_0^{\infty } J_1(2 x) T\left(\sqrt{2} x,2 \sqrt{2}\right) \, dx
$$
where T(x,a) is the Owen's T-function and
J(1,x) is Bessel function of the first kind
EDITED:
Maybe not exist closed form solution and answer is on window box:

noclose

