Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts? I have a series of six integrals. The first, say, is 
\begin{equation}
\int_ 0^{\frac {1} {16}} - 
 48 q^{3/2}\sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
Mathematica returns this as unperformed, as well as the subproblem
\begin{equation}
\int_ 0^{\frac {1} {16}} \sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
However, high-precision numerical integration shows that the latter integral equals $\frac{25}{204}$.
So, is this last result indicative that the original integral can be solved, in one way of another, by integration-by-parts? (I don't immediately see how.)
The other five integrands in my series--all exhibiting similar numerical phenomenon (upon division by $\frac{25}{204}$, after integration over $[0,\frac{1}{16}]$)--are 
\begin{equation}
\left\{-\frac{2352}{5} q^{5/2} \sqrt{18 q-2 \sqrt{17 q+4} \sqrt{q}+4},-\frac{96}{5}
   \sqrt{2} \sqrt{q} \sqrt{9 q-\sqrt{17 q+4} \sqrt{q}+2},-144 \sqrt{2} q^{3/2} \sqrt{9
   q-\sqrt{17 q+4} \sqrt{q}+2},\frac{96}{5} \sqrt{2} q \sqrt{(-17 q-4) \left(-9 q+\sqrt{q
   (17 q+4)}-2\right)},\frac{432}{5} \sqrt{2} q^2 \sqrt{(-17 q-4) \left(-9 q+\sqrt{q (17
   q+4)}-2\right)}\right\}.
\end{equation}
Needless to say, I am interested in the evaluation of the six integrals through any appropriate methodology. (Actually, I am specifically interested in the sum of the six integrals, more so  than each one itself.)

Yes, employing the transformations indicated by Vasily Mitch, the results of the six integrations are 
\begin{equation}
\left\{\frac{1133}{9248}-\frac{384 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{289
   \sqrt{17}},\frac{147 \left(655360 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)-1431757\right)}{855255040},\frac{3
   \left(256 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)-901\right)}{2890},\frac{3399}{9248}-\frac{1152 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{289
   \sqrt{17}},\frac{88043-12288 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{393040},\frac{1728
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{4913
   \sqrt{17}}-\frac{593721}{50309120}\right\}.
\end{equation}
Summing these six results, together with another seven (earlier obtained without the need for the transformations), gives me my desired ("bound-entanglement") probability result of 
\begin{equation}
\frac{24 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{17 \sqrt{17}}-\frac{91}{544} \approx 0.002187222381.
\end{equation}
So, to get back to the title of my question, "Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts?"--perhaps the answer is in the negative, but moot, in any case. 
Serendipity!
