How to prove such a hyperbolic sine cosine related equality? [closed]

$$\ln \left(\frac{\left(1+\sqrt{5}\right)^2 \left(2+\sqrt{5}\right)}{4}\right)=\text{arcsinh }(2)+2 \text{ arccsch }(2)$$

closed as off-topic by Claude Leibovici, Watson, user91500, Chill2Macht, Daniel W. FarlowJul 23 '16 at 16:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Watson, user91500, Chill2Macht, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

$$\ln \left(\frac{\left(1+\sqrt{5}\right)^2 \left(2+\sqrt{5}\right)}{4}\right)=\ln\left(\left(1+\sqrt{5}\right)^2\left(2+\sqrt{5}\right)\right)-\ln (4)$$
$$=\ln\left(1+\sqrt{5}\right)^2+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\,\text{arcsinh}\, 2}-\ln 4$$
$$=\ln\frac{\left(1+\sqrt{5}\right)^2}{4}+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\, \text{arcsinh} \,2}$$
$$=\underbrace{2\ln\frac{\left(1+\sqrt{5}\right)}{2}}_{=\,2\,\text{arccsch}\, 2}+\underbrace{\ln\left(2+\sqrt{5}\right)}_{=\,\text{arcsinh} \,2}$$