Evaluate $\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$ Question:

Solve the integral
  $$
\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx
$$

My solution:
Multiply both the numerator and denominator by $\sqrt{x+1}-\sqrt{x}$. This changes the integral to
$$
\begin{align}
\int&\left(1+x+\sqrt{1+x^2}\right)\left(\sqrt{x+1}-\sqrt{x}\right)\,dx\\
&= \int{(1+x)^{3/2}}\,dx-\int{\sqrt{(1+x)(1+x^2)}}\,dx-\int{\sqrt{x}(1+x)}\,dx-\int{\sqrt{x}\sqrt{1+x^2}}\,dx\\
& = \frac{2}{5}(1+x)^{5/2}-I-\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}-J
\end{align}
$$
where $I = \int{(1+x)^{1/2}(1+x^2)^{1/2}}\,dx$ and $J = \int{x^{1/2}(1+x^2)^{1/2}}\,dx$.
How can I solve the integrals $I$ and $J$?

One can be brought to the form $\int u^a (1-u)^b \mathrm{d}u$ which is discussed in this answer of mine:
$$\begin{eqnarray}
  J &=& \int \sqrt{x} \sqrt{1+x^2} \mathrm{d}x \stackrel{u=x^2}{=} \frac{1}{2} \int u^{-1/4} (1+u)^{1/2} \mathrm{d} u \\ &=& \frac{2}{3} u^{3/4} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -u\right) +\text{const.} = \frac{2}{3} x^{3/2} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -x^2\right) +\text{const.}
\end{eqnarray}
$$
The other integral is an elliptic integral:
$$\begin{eqnarray}
  I &=& \frac{4}{15} \sqrt{2\alpha} \left(\alpha \operatorname{F}\left( \arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right) -\frac{1}{\alpha} \operatorname{E}\left(\arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right)\right) \\ && + \frac{2}{15} \sqrt{1+x}\sqrt{1+x^2} \left(3x+1 + \frac{4}{x+\alpha}\right) + \text{const.}
\end{eqnarray}
$$
where $\alpha = \sqrt{2}+1$ and
$$
   \operatorname{E}\left(\phi, m\right) = \int_0^\phi \sqrt{1-m \sin^2\varphi}\, \mathrm{d}\varphi, \quad \operatorname{F}\left(\phi, m\right) = \int_0^\phi \frac{\mathrm{d}\varphi}{\sqrt{1-m \sin^2\varphi}}
$$
It can be evaluated using the Jacobi elliptic functions substitution:
$$
    \operatorname{sn}\left(t, -\alpha^2\right) = \frac{x-\frac{1}{\alpha}}{x+\alpha}
$$
as described in Byrd and Friedman.
