Is there analytical solution to the following integral? I have an integral
$$I=\int_0^{a}\frac{a-x}{c^m+(R^2+x^2)^{m/2}}\,\mathrm{d}x~,$$
where $a&gt;0, R&gt;0$ and $m&gt;0$ are reals. Is there any analytical solution possible for this integral. Solution in terms of Hypergeometric function would do.
Well, you can split the integral into two pieces, as a first thing.

$$a\int_0^a \frac{\text{d}x}{b + (R^2 + x^2)^{\beta}} - \int_0^a \frac{x\ \text{d}x}{b + (R^2 + x^2)^{\beta}}$$

Where I called $b = c^m$ and $\beta = m/2$.

The second integral is analytically solvable in terms of Hypergeometric function as you sighed for. By setting

$$R^2 + x^2 = z ~~~~~~~ \text{d}z = 2x\ \text{d}x$$

We can easily rewrite the second integral as

$$-\frac{1}{2}\int_{R^2}^{R^2+a^2} \frac{\text{d}z}{b + z^{\beta}}$$

whose solution is

$$-\frac{1}{2(\beta-1)} \left\{\left| R\right| ^{2-2 \beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left| R\right| ^{-2 \beta}\right)-\left(\frac{1}{b}\right)^{-1/\beta} b^{-1/\beta} \left(a^2+R^2\right)^{1-\beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left(a^2+R^2\right)^{-\beta}\right)\right\}$$

For what concerns the first part, well it's tougher than what it could seem. 
I am still thinking about it but, for example, we can play a bit.

For example, if we assumed that $b &lt;&lt; (R^2 + x^2)^{\beta}$, then the first integral would approximate to

$$a\int_0^a \frac{\text{d}x}{(R^2 + x^2)^{\beta}}$$

whose solution is well known and again in terms of Hypergeometric function:

$$a R^{-2 \beta} \, _2F_1\left(\frac{1}{2},\beta;\frac{3}{2};-\frac{a^2}{R^2}\right)$$

So you have a part with an exact solution, and another one with an approximate solution. 
Of course we could play the same by considering 

$$b &gt;&gt; (R^2+x^2)^{\beta}$$

But that integral would be trivial then.
Well, you can split the integral into two pieces, as a first thing.

$$a\int_0^a \frac{\text{d}x}{b + (R^2 + x^2)^{\beta}} - \int_0^a \frac{x\ \text{d}x}{b + (R^2 + x^2)^{\beta}}$$

Where I called $b = c^m$ and $\beta = m/2$.

The second integral is analytically solvable in terms of Hypergeometric function as you sighed for. By setting

$$R^2 + x^2 = z ~~~~~~~ \text{d}z = 2x\ \text{d}x$$

We can easily rewrite the second integral as

$$-\frac{1}{2}\int_{R^2}^{R^2+a^2} \frac{\text{d}z}{b + z^{\beta}}$$

whose solution is

$$-\frac{1}{2(\beta-1)} \left\{\left| R\right| ^{2-2 \beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left| R\right| ^{-2 \beta}\right)-\left(\frac{1}{b}\right)^{-1/\beta} b^{-1/\beta} \left(a^2+R^2\right)^{1-\beta} \, _2F_1\left(1,\frac{\beta-1}{\beta};2-\frac{1}{\beta};-b \left(a^2+R^2\right)^{-\beta}\right)\right\}$$

For what concerns the first part, well it's tougher than what it could seem. 
I am still thinking about it but, for example, we can play a bit.

For example, if we assumed that $b &lt;&lt; (R^2 + x^2)^{\beta}$, then the first integral would approximate to

$$a\int_0^a \frac{\text{d}x}{(R^2 + x^2)^{\beta}}$$

whose solution is well known and again in terms of Hypergeometric function:

$$a R^{-2 \beta} \, _2F_1\left(\frac{1}{2},\beta;\frac{3}{2};-\frac{a^2}{R^2}\right)$$

So you have a part with an exact solution, and another one with an approximate solution. 
Of course we could play the same by considering 

$$b &gt;&gt; (R^2+x^2)^{\beta}$$

But that integral would be trivial then.