How to integrate $\ln \big( b + \sqrt{b^2 + c^2 + x^2}\,\big)$?

Question

I am looking to demonstrate the following result. Any ideas are much appreciated. $$ \begin{align}\int \ln \left( b + \sqrt{b^2 + c^2 + x^2}\right) dx = &\;x \ln \left( b + \sqrt{b^2 +c^2 +x^2}\right) +b \ln \left(2x + 2\sqrt{ b^2 +c^2 +x^2} \right)\\&\; - c \arctan \left(\frac{ b x} { c \sqrt{b^2 +c^2 +x^2}}\right) + c \arctan \left(\frac{x}{c}\right) -x \end{align}$$

Answer 1

Since integral to function $f(x)$ is such function $F(x)$ so that $F'(x) = f(x) + C$ for $C \in \mathbb{R}$ you can find derivative of your result and compare it to the original task.

Answer 2

From the first term at the right let's try integrations by parts : \begin{align} I(b,c)&:=\int \ln \left( b + \sqrt{b^2 + c^2 + x^2}\right) dx-x \ln \left( b + \sqrt{b^2 +c^2 +x^2}\right) \\ &= -\int x\frac{1}{b + \sqrt{b^2 +c^2 +x^2}}\frac{2\,x}{2\sqrt{b^2 +c^2 +x^2}}\,dx\\ &= \int \frac{{b - \sqrt{b^2 +c^2 +x^2}}}{c^2 +x^2}\frac{x^2}{\sqrt{b^2 +c^2 +x^2}}\,dx\\ &= b\int \frac{c^2+x^2-c^2}{(c^2 +x^2)\sqrt{b^2 +c^2 +x^2}}\,dx-\int\frac{x^2}{c^2 +x^2}\,dx\\ &= b\int \frac{dx}{\sqrt{b^2 +c^2 +x^2}}-\int \frac{b\,c^2}{(c^2 +x^2)\sqrt{b^2 +c^2 +x^2}}\,dx+\int\frac{c^2-(c^2+x^2)}{c^2 +x^2}\,dx\\ &= b\,\ln \left(2x + 2\sqrt{ b^2 +c^2 +x^2} \right) - c\,\arctan \frac{ b x} { c \sqrt{b^2 +c^2 +x^2}} + c\, \arctan \frac{x}{c} -x\\ \end{align} The three last ones should be more direct !