Evaluate $\int^\infty_0 t^{a+b-1}(t+1)^{-b-1} U(a+2,a-b+2,ct)dt$ 
Evaluate
  $$
\int^\infty_0 t^{a+b-1}\left(t+1\right)^{-b-1} U\left(a+2,a-b+2,ct\right)dt
$$
  under the condition $a>0$, $b>0$ and $c>0$, where $U(\cdot,\cdot,\cdot)$ denotes the confluent hypergeometric function of the second kind. All of $a$, $b$ and $c$ are noninteger, but the stronger constraint on the range of $b$ is accepted.

Mathematica found the following form with the condition $b>\frac{1}{2}$, but I believe the simpler or shorter expression exists.
$$
\frac{2 c^{1-a} \, _2F_2(3,b+1;2-a,3-b;-c)}{a \left(a^2-1\right) (b-2) (b-1) b}-\frac{\sqrt{\pi } 4^{b-1} \csc (\pi  b) \Gamma
   \left(b-\frac{1}{2}\right) c^{-a+b-1} \Gamma (a-b+1) \, _2F_2(b+1,2 b-1;b-1,b-a;-c)}{b \Gamma (a+2) \Gamma (b-1)}+\frac{\pi  \csc (\pi  a)
   \Gamma (-a+b-1) \Gamma (a+b) \, _2F_2(a+2,a+b;a,a-b+2;-c)}{\Gamma (a) \Gamma (b+1)^2}.
$$

(Too big for comment)
It's probably a matter of taste whether this counts as simpler (and it certainly isn't shorter), but the second confluent hypergeometric function can be reduced to a linear combination of confluent hypergeometric functions of smaller rank by working through their respective integral representations: 
$$\begin{align}
\small{{_2F_2}{\left(b+1,2b-1;b-1,b-a;-c\right)}}
&=\small{\frac{\Gamma{\left(b-a\right)}}{\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b-2}(1-t)^{-a-b}{_1F_1}{\left(b+1;b-1;-ct\right)}\,\mathrm{d}t}\\
&=\frac{c^2\,\Gamma{\left(b-a\right)}}{(b-1)b\,\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b}(1-t)^{-a-b}e^{-ct}\,\mathrm{d}t\\
&~~~~~ -\frac{2c\,\Gamma{\left(b-a\right)}}{(b-1)\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}\frac{t^{2b-1}}{(1-t)^{a+b}}e^{-ct}\,\mathrm{d}t\\
&~~~~~ +\frac{\Gamma{\left(b-a\right)}}{\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b-2}(1-t)^{-a-b}e^{-ct}\,\mathrm{d}t\\
&=\frac{2(2b-1)c^2}{(a-b-1)(a-b)(b-1)}\,{_1F_1}{\left(2b+1;2-a+b;-c\right)}\\
&~~~~~ -\frac{2c(2b-1)}{(b-1)(b-a)}\,{_1F_1}{\left(2b;1-a+b;-c\right)}\\
&~~~~~ +{_1F_1}{\left(2b-1;b-a;-c\right)}.\\
\end{align}$$
The third hypergeometric term can be similarly "simplified" in the same manner. However, it seems the first hypergeometric term isn't as straightforward...
