Integration of $\sqrt{ x^2 - x\cdot \sin(x) + 4}$ How to integrate 

$$\sqrt{ x^2 - x\cdot \sin(x) + 4}$$

with respect to $dx$?

I tried solving this using substitution,  but it doesn't seem to work.  Can somebody give me a little direction on how to  proceed.
Even using special functions, the antiderivative does not seem to be possible.

For the integral, series expansion around $x=0$ could be a possible solution since 
$$\sqrt{ x^2 - x\, \sin(x) + 4}=2+\frac{x^4}{24}-\frac{x^6}{480}-\frac{31 x^8}{80640}+\frac{31
   x^{10}}{725760}+\frac{2213 x^{12}}{319334400}+O\left(x^{14}\right)$$ Trying for
$$I_a=\int_0^a\sqrt{ x^2 - x\, \sin(x) + 4}\,dx$$ we would get
$$\left(
\begin{array}{ccc}
 a &amp; \text{approximation} &amp; \text{exact} \\
 0.1 &amp; 0.20000 &amp; 0.20000 \\
 0.2 &amp; 0.40000 &amp; 0.40000 \\
 0.3 &amp; 0.60002 &amp; 0.60002 \\
 0.4 &amp; 0.80009 &amp; 0.80009 \\
 0.5 &amp; 1.00026 &amp; 1.00026 \\
 0.6 &amp; 1.20064 &amp; 1.20064 \\
 0.7 &amp; 1.40137 &amp; 1.40137 \\
 0.8 &amp; 1.60266 &amp; 1.60266 \\
 0.9 &amp; 1.80476 &amp; 1.80476 \\
 1.0 &amp; 2.00800 &amp; 2.00800 \\
 1.1 &amp; 2.21275 &amp; 2.21275 \\
 1.2 &amp; 2.41948 &amp; 2.41948 \\
 1.3 &amp; 2.62871 &amp; 2.62870 \\
 1.4 &amp; 2.84100 &amp; 2.84098 \\
 1.5 &amp; 3.05699 &amp; 3.05695 \\
 1.6 &amp; 3.27738 &amp; 3.27727 \\
 1.7 &amp; 3.50290 &amp; 3.50264 \\
 1.8 &amp; 3.73438 &amp; 3.73377 \\
 1.9 &amp; 3.97272 &amp; 3.97137 \\
 2.0 &amp; 4.21902 &amp; 4.21616
\end{array}
\right)$$

Edit

The above approximation is not too bad for "small" values of $a$. We can make if better if we consider
$$J_{k,a}=\int_{2k \pi}^{2k\pi+a}\sqrt{ x^2 - x\, \sin(x) + 4}\,dx$$ Chanking variable $x=2k \pi+y$
$$J_{k,a}=\int_{0}^{a}\sqrt{4 \pi ^2 k^2+4 \pi  k y-(2 \pi  k+y) \sin (y)+y^2+4}\,dy$$ and using the Taylor expansion of $\sin(y)$, the integrand becomes
$$2 \sqrt{\pi ^2 k^2+1}\left(1+\frac{\pi  k y}{4 (\pi ^2 k^2+1)}-\frac{\left(\pi ^2 k^2\right) y^2}{32 \left(\pi
   ^2 k^2+1\right)^2}+\frac{\pi  k \left(16 \pi ^4 k^4+35 \pi ^2 k^2+16\right)
   y^3}{384 \left(\pi ^2 k^2+1\right)^3}+\frac{\left(64 \pi ^6 k^6+241 \pi ^4
   k^4+320 \pi ^2 k^2+128\right) y^4}{6144 \left(\pi ^2
   k^2+1\right)^4}\right)+O\left(y^5\right)$$ for acceptable results as show below
$$\left(
\begin{array}{cccc}
 k &amp; a &amp;\text{approximation} &amp; \text{exact} \\
 1 &amp; 0.25 &amp; 1.66333 &amp; 1.66333 \\
 1 &amp; 0.50 &amp; 3.35705 &amp; 3.35704 \\
 1 &amp; 0.75 &amp; 5.08364 &amp; 5.08350 \\
 1 &amp; 1.00 &amp; 6.84782 &amp; 6.84701 \\
 1 &amp; 1.25 &amp; 8.65666 &amp; 8.65350 \\
 1 &amp; 1.50 &amp; 10.5197 &amp; 10.5102 \\
 1 &amp; 1.75 &amp; 12.4494 &amp; 12.4250 \\
 1 &amp; 2.00 &amp; 14.4607 &amp; 14.4060 \\
&amp; \\
 2 &amp; 0.25 &amp; 3.19659 &amp; 3.19659 \\
 2 &amp; 0.50 &amp; 6.42490  &amp; 6.42489 \\
 2 &amp; 0.75 &amp; 9.68763 &amp; 9.68749 \\
 2 &amp; 1.00 &amp; 12.9896 &amp; 12.9888 \\
 2 &amp; 1.25 &amp; 16.3377 &amp; 16.3348 \\
 2 &amp; 1.50 &amp; 19.7412 &amp; 19.7326 \\
 2 &amp; 1.75 &amp; 23.2117 &amp; 23.1901 \\
 2 &amp; 2.00 &amp; 26.7631 &amp; 26.7151 \\
&amp; \\
 3 &amp; 0.25 &amp; 4.75443 &amp; 4.75443 \\
 3 &amp; 0.50 &amp; 9.54087 &amp; 9.54086 \\
 3 &amp; 0.75 &amp; 14.3621 &amp; 14.3620 \\
 3 &amp; 1.00 &amp; 19.2230 &amp; 19.2223 \\
 3 &amp; 1.25 &amp; 24.1304 &amp; 24.1277 \\
 3 &amp; 1.50 &amp; 29.0935 &amp; 29.0853 \\
 3 &amp; 1.75 &amp; 34.1236 &amp; 34.1029 \\
 3 &amp; 2.00 &amp; 39.2341 &amp; 39.1883
\end{array}
\right)$$
You can try this substitution below. From the point I stopped on, it should be a matter of organizing things and doing basic stuff.

$$\eqalign{
  &amp; \int {\sqrt {{x^2} - x\sin x + 4} } dx = \int {\sqrt {4\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} } dx = 2\int {\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} dx  \cr 
  &amp; x = 2\tan \theta   \cr 
  &amp; {{dx} \over {d\theta }} = 2{\sec ^2}\theta   \cr 
  &amp; 2\int {\left( {{{4{{\tan }^2}\theta } \over 4} - {{4ta{n^2}\theta } \over {4\sqrt {4 + 4{{\tan }^2}\theta } }} + 1} \right)} 2{\sec ^2}\theta d\theta   \cr 
  &amp; 2\int {\left( {{{\tan }^2}\theta  - {{ta{n^2}\theta } \over {2\sec \theta }} + 1} \right)} 2{\sec ^2}\theta d\theta  \cr} $$

(sorry if I left anything wrong in the middle of all that!)