Question

integrate root (secx+1)

Answer 1

there seems no closed expression for this integration.

Answer 2

What about this:
\[\int\sqrt{\sec x+1}\;dx\cdot\frac{\sqrt{\sec x-1}}{\sqrt{\sec x-1}}\\
\int\frac{\sqrt{\sec^2 x-1}}{\sqrt{\sec x-1}}\;dx\\
\int\frac{\tan x}{\sqrt{\sec x-1}}\;dx\\
\int\frac{\tan x}{\sqrt{\sec x-1}}\;dx\cdot\frac{\sec x}{\sec x}\\
\int\frac{\sec x\tan x}{\sec x\sqrt{\sec x-1}}\;dx\]
\[u=\sec x\\
du=\sec x\tan x\;dx\]
\[\int\frac{1}{u\sqrt{u-1}}\;du\]
Still no closed form? I'm not sure.

Answer 3

looks like wolfram was giving me wrong result, Here is result from Maple

Answer 4

looks like the result from the maple is correct ... differentiating and simplifying it yields
\[ \frac{(1+cos(x))}{(sqrt((1+cos(x))/cos(x))*cos(x))} \]
which is equal to the integrand.

Answer 5

looks like I shouldn't rely too much on software ... using the result from above by @SithsAndGiggles yields
\[ \int \sqrt{1 + \sec x} dx = 2 \tan^{-1} \left( \sec x - 1\right) + c\]
differentiating which you get the result, your integrand. But the problem is
\[ \tan x = \sqrt{ \sec^2x - 1 }\] is a conditional expression where tan is positive. I am not sure if it should be the result.