Q:

A closed form expression for $\int_0^{\infty} (t^2+t^4)^n e^{-t^2-t^4}\,dt$

I was doing some computations for research purposes, which led me to this integral:
$$I(n) = \int_0^{\infty} (t^2+t^4)^n e^{-t^2-t^4}\,dt.$$
This is very suggestively written so as to employ a parametric differentiation technique as so:
$$\left(\frac{\partial^n}{\partial \alpha^n}\right)\int_0^{\infty}e^{-\alpha(t^2+t^4)}\,dt.$$
This integral has a nice, closed form expression:
$$\int_0^{\infty} e^{-\alpha(t^2+t^4)}\,dt = \frac{1}{4} e^{\frac{\alpha}{8}}K_{\frac{1}{4}}\left(\frac{\alpha}{8}\right),$$
where $K_{\nu}$ is the modified Bessel function of the second kind.  From here, I would have to employ $n$ differentiations which would be pretty messy to work out due to product rule.  $n$ applications of product rule does have a nice combinatorial expression but it is far from explicit.  Moreover, Bessel functions can get pretty complicated after differentiating so this seems like a bad approach. 
Instead I decided to run some examples on Mathematica and computed the first 22 of these and noticed a very surprising pattern.  In what follows $I_{\nu}$ is the modified Bessel function of the first kind. 
$$I(0) = \frac{1}{4} e^{\frac{1}{8}}K_{\frac{1}{4}}\left(\frac{1}{8}\right)$$
$$I(1) = \frac{1}{32} e^{\frac{1}{8}}\left(K_{\frac{1}{4}}\left(\frac{1}{8}\right) + K_{\frac{3}{4}}\left(\frac{1}{8}\right)\right)$$
$$I(2) = \frac{3}{128\sqrt{2}} e^{\frac{1}{8}}\pi \left(3 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + I_{\frac{1}{4}}\left(\frac{1}{8}\right) - I_{\frac{3}{4}}\left(\frac{1}{8}\right) + I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$
$$I(3) = \frac{1}{256\sqrt{2}} e^{\frac{1}{8}}\pi \left(39 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 17 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 14 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 14 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$
$$I(4) = \frac{1}{2048\sqrt{2}} e^{\frac{1}{8}} \pi \left(1029 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 367 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 349 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 349 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$
$$I(5) = \frac{9}{8192\sqrt{2}} e^{\frac{1}{8}} \pi \left(1953 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 619 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 643 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 643 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$
$$I(6) = \frac{1}{16384\sqrt{2}} e^{\frac{1}{8}} \pi \left(185157 I_{-\frac{1}{4}} \left(\frac{1}{8}\right) + 53131 I_{\frac{1}{4}}\left(\frac{1}{8}\right) - 59572 I_{\frac{3}{4}}\left(\frac{1}{8}\right) + 59572 I_{\frac{5}{4}}\left(\frac{1}{8}\right)\right)$$
Repeat ad nauseum.  Each of the terms in the denominator seems to be a power of $2$, the third and fourth terms seem to have the same coefficient (modulo a sign) and the signs are $+$, $+$, $-$, $+$.  The "nice" output seems to suggest to me that there is a closed-form expression for $I(n)$ in general but I haven't the slightest clue as to how to come up with it.  Can anyone shed some light on the matter? 
A PDF with more expressions can be found here. ( Mathematica output.) 

A:

Not a full answer, but a partial one:
Notice that the powers of two in the denominator are increasing in periodic repetitions of 3, 2, 1.  That is, in this case the exponents on the 2 are 2, 5, 7, 8, 11, 13, 14, 17, 19, 20.... This is sequence A047268 on the OEIS website, defined as the sequence of numbers congruent to {1, 3, 5} mod(6). 
I fiddled around with the other coefficients, however nothing useful turned out (no known sequences with the numbers themselves or their differences, products, sums...).  If you absolutely need a closed form expression for it, then you could just define three new sequences $S_1$, $S_2$ and $S_3$ as the coefficients of the first three modified Bessel functions of the first kind in your expression.  Then the closed form expression would be:
$I(n) = 2^{-\delta_{n + 2}}e^{\frac18}(K_{\frac14}(\frac18)+nK_{\frac34}(\frac18))$ $ n = 0,1$
$I(n) = 2^{-\frac{\delta_{n + 2}}{2}}e^{\frac18}\pi(S_1I_{-\frac{1}{4}}(\frac18) + S_2I_{\frac{1}{4}}(\frac18) - S_3I_{\frac{3}{4}}(\frac18) + S_3I_{\frac{5}{4}}(\frac18))$ $ n \geq 2$.
Where,
$\delta_n$ is sequence A047268,
$S_1, S_2, S_3$ are explicitly given by you.  Note that the coefficients are not simplified by being factored out, as is for example in I(2) where 3 is factored out.  You need to multiply them back in to get the sequences so there will always be a coefficient of 1 outside the numerator. 
Finally, if you really want to get compact, you can write the second expression as such:
$I(n) = 2^{-\frac{\delta_{n + 2}}{2}}e^{\frac18}\pi\sum\limits_{k = 1}^4 S_kI_{\frac{2k - 3}{2}}(\frac18)$.
Where $\delta_n$ and $S_1, S_2, S_3$ are defined as above, and $S_4 = -S_3$ (it appears implicitly in the summation). 
Hope this helps! 

A:

Let us make the change of variables $t=\sinh \frac{x}{4}$. Since 
$$t^2+t^4=\frac{\sinh^2\frac{x}{2}}{4}=\frac{\cosh x-1}{8}$$ 
and $dt=\frac14 \cosh\frac{x}{4}dx$, the initial integral can be rewritten as
\begin{align}
I(n)&=\frac{e^{\frac18}}{4^{n+1}}\int_0^{\infty}\cosh\frac{x}{4}\sinh^{2n}\frac{x}{2}e^{-\frac18\cosh x}dx=\\
&=\frac{e^{\frac18}}{2^{4n+3}}\int_{-\infty}^{\infty}e^{\frac{x}{4}}\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)^{2n}e^{-\frac18\cosh x}dx=\\
&=\frac{e^{\frac18}}{2^{4n+3}}\sum_{k=0}^{2n} \left(-1\right)^k {2n \choose k}\int_{-\infty}^{\infty}e^{(n-k+\frac{1}{4})x-\frac18\cosh x}dx=\\
&=\frac{e^{\frac18}}{2^{4n+2}}\sum_{k=0}^{2n} \left(-1\right)^k {2n \choose k}
K_{n-k+\frac14}\left(\frac18\right). 
\end{align}
At the last step, I use the integral representation $\displaystyle K_{\nu}(r)=\frac12\int_{-\infty}^{\infty}e^{-r\cosh x+\nu x}dx$.

Added: Using the recurrence relation $K_{\nu+1}(r)-K_{\nu-1}(r)=\frac{2\nu}{r}K_{\nu}(r)$, the last sum (1) can be expressed in terms of rational multiples of only two Macdonald functions, for example, $K_{\frac14}\left(\frac18\right)$ and $K_{\frac54}\left(\frac18\right)$. However it seems unlikely that the corresponding coefficients can be written in a nice way.