Output the juggler sequence

Julia, 64 50 48 42 32 30 bytes

g(x)=[x;x<3||g(x^(x%2+.5)÷1)]

This is a recursive function that accepts an integer and returns a float array.

We build an array by concatenating the input with the next term of the sequence, computed as x to the power of its parity plus 1/2. This gives us either x^1/2 or x^1+1/2 = x^3/2. Integer division by 1 gets the floor. When the condition x < 3 is true, the final element will be a Boolean rather than a numeric value, but since the array is not of type Any, this is cast to have the same type as the rest of the array.

Saved 14 bytes thanks to Dennis!

Jelly, 12 11 10 bytes

*BṪ×½Ḟµ’п

Thanks to @Sp3000 for golfing off 1 byte!

Try it online!

How it works

*BṪ×½Ḟµ’п    Main link. Input: n

*B            Elevate n to all of its digits in base 2.
  Ṫ           Extract the last power.
              This yields n ** (n % 2).
   ×½         Multiply with sqrt(n). This yields n ** (n % 2 + 0.5).
     Ḟ        Floor.

      µ       Push the previous chain on the stack and begin a new, monadic chain.
        п    Repeat the previous chain while...
       ’        n - 1 is non-zero.
              Collect all intermediate results in an array.

JavaScript (ES7), 45 33 bytes

f=n=>n<2?n:n+","+f(n**(.5+n%2)|0)

Explanation

Recursive approach. Returns a comma-separated string of numbers.

f=n=>
  n<2?n:          // stop when n == 1
  n               // return n at the start of the list
  +","+f(         // add the rest of the sequence to the list
    n**(.5+n%2)|0 // juggler algorithm
  )

Test

** not used in test for browser compatibility.

f=n=>n<2?n:n+","+f(Math.pow(n,.5+n%2)|0)

<input type="number" oninput="result.textContent=f(+this.value)" />
<pre id="result"></pre>