How to get variables out of square roots and collect in order to simplify

Question

f = 2 Sqrt[b^3 (1 + b) (1 - b + 2 a^2 b) c] + 8 Sqrt[b^5 (1 + b) (1 - b + 2 a^2 b) c] + 
10 Sqrt[b^7 (1 + b) (1 - b + 2 a^2 b) c] - 10 Sqrt[b^11 (1 + b) (1 - b + 2 a^2 b) c] - 
8 Sqrt[b^13 (1 + b) (1 - b + 2 a^2 b) c] - 2 Sqrt[b^15 (1 + b) (1 - b + 2 a^2 b) c]

If we write (this is the best after I read so many questions in this page):

FullSimplify[f, Assumptions -> {a > 1, b > 1, c > 0,
  Element[a, Reals], Element[b, Reals], Element[c, Reals]}, 
    ComplexityFunction -> LeafCount]

We obtain (just each coefficient is reduced by a factor 2):

(Sqrt[b^3 (1 + b) (1 - b + 2 a^2 b) c] + 4 Sqrt[b^5 (1 + b) (1 - b + 2 a^2 b) c] + 
5 Sqrt[b^7 (1 + b) (1 - b + 2 a^2 b) c] - 5 Sqrt[b^11 (1 + b) (1 - b + 2 a^2 b) c] - 
4 Sqrt[b^13 (1 + b) (1 - b + 2 a^2 b) c] - Sqrt[b^15 (1 + b) (1 - b + 2 a^2 b) c])

We can do this (we see the common part Sqrt[b (1 + b) (1 + (-1 + 2 a^2) b) c]):

FullSimplify[(2 b + 8 b^2 + 10 b^3 - 10 b^5 - 8 b^6 - 2 b^7) Sqrt[b (1 + b)  
(1 - b + 2 a^2 b) c], {b > 1}]

and Mathematica out is:

-2 (-1 + b) b (1 + b)^5 Sqrt[b (1 + b) (1 + (-1 + 2 a^2) b) c]

But the problem is that this is a little part of a huge expression and doing this manualy may take weeks or more.

The optimal expresion would be:

-2 b (b - 1) (b + 1)^5 Sqrt[b c (b + 1) (1 + b (2 a^2 - 1))]

But I will be hilarious with the previous.

Suplemental question:

As prescribed in the second response by bbgodfrey, we obtain our objective! Thanks you very much!

But permit me to go far away, if we put bbgodfrey's solution in a more complex expression like:

ff = (a (2 a (-2 + a^2) b^(3/2) - 2 a^3 b^(5/2) - 2 a (-2 + a^2) b^(7/2) + 2 a^3 b^(9/2) + 
Sqrt[2] (-1 + a^2) b Sqrt[(-1 + b) (1 + b) (1 - b + 2 a^2 b) c] - 6 Sqrt[2](-1 + a^2) b^2 
Sqrt[(-1 + b) (1 + b) (1 - b + 2 a^2 b) c] + Sqrt[2] (-1 + a^2) b^3 Sqrt[(-1 + b) (1 + b) 
(1 - b + 2 a^2 b) c] + (4 a (Sqrt[b^3 (1 + b) (1 - b + 2 a^2 b) c] + 3 Sqrt[b^5 
(1 + b) (1 - b + 2 a^2 b) c] + 2 Sqrt[b^7 (1 + b) (1 - b + 2 a^2 b) c] - 2 Sqrt[b^9 
(1 + b) (1 - b + 2 a^2 b) c] - 3 Sqrt[b^11 (1 + b) (1 - b + 2 a^2 b) c] - Sqrt[b^13 
(1 + b) (1 - b + 2 a^2 b) c]))/(1 + b)^3 + b (-2 a Sqrt[b] + 2 a (-4 + a^2) b^(3/2) - 
2 a^3 b^(5/2) - 2 a (-4 + a^2) b^(7/2) + 2 (a + a^3) b^(9/2) + Sqrt[2] (-1 + a^2) b 
Sqrt[(-1 + b) (1 + b) (1 - b + 2 a^2 b) c] - 6 Sqrt[2] (-1 + a^2) b^2 Sqrt[(-1 + b) (1 + b)
(1 - b + 2 a^2 b) c] + Sqrt[2] (-1 + a^2) b^3 Sqrt[(-1 + b) (1 + b) (1 - b + 2 a^2 b) c] + 
4 a (Sqrt[b^3 (1 + b) (1 - b + 2 a^2 b) c] - Sqrt[b^7 (1 + b) (1 - b + 2 a^2 b) c]))))
/(4 (-1 + b) b^2 (1 + b) (2 a^3 Sqrt[b] + 2 a^3 b^(3/2) + Sqrt[2] (1 + a^2) Sqrt[(-1 + b)
(1 + b) (1 + (-1 + 2 a^2) b) c]))

FullSimplify[ff /. Sqrt[z1___ z3_^n_ z2___] -> z3^(n/2) Sqrt[z1 z2]]

Mathematica out is:

(a (2 a^3 (-1 + b)^2 Sqrt[b] (1 + b) - Sqrt[2] (1 + (-6 + b) b) Sqrt[(1 + (-1 + 2 a^2) b) 
(-1 + b^2) c] + Sqrt[2] a^2 (1 + (-6 + b) b) Sqrt[(1 + (-1 + 2 a^2) b) (-1 + b^2) c] + 
2 a Sqrt[b] (-1 + b^2) (3 + b - 2 Sqrt[(1 + b) (1 - b + 2 a^2 b) c])))/(4 (-1 + b) b 
(2 a^3 Sqrt[b] (1 + b) + Sqrt[2] Sqrt[(1 + (-1 + 2 a^2) b) (-1 + b^2) c] + Sqrt[2] 
a^2 Sqrt[(1 + (-1 + 2 a^2) b) (-1 + b^2) c]))

When we expect:

(a (2 a Sqrt[b] (3 + a^2 (-1 + b) + b) (-1 + b^2) + Sqrt[2] (-1 + a^2) (1 + (-6 + b) b) 
Sqrt[(1 + (-1 + 2 a^2) b) (-1 + b^2) c] - 2 a (-1 + b^2) 2 Sqrt[b (1 + b) (1 + (-1 + 2 a^2)
b) c]))/(4 (-1 + b) b (2 a^3 Sqrt[b] (1 + b) + Sqrt[2] (1 + a^2) Sqrt[(1 + (-1 + 2 a^2) b)
(-1 + b^2) c]))

Why can we automatize Mathematica to collect and reduce the expression length?

We are very interested in adding rules to accomplish a full simplification. What is the estructure of this sentences? We would be very grateful if anybody could explain or give information (other questions, tutorials...) about this.

Thanks to all the people who answers, comments, edits and reads!

Answer 1

The following works for expressions of the form in the question but, of course, not for all expressions.

Simplify[f /. Sqrt[Times[z1___, b^n_, z2___]] -> b^((n - 1)/2) Sqrt[Times[z1, b, z2]]]
(* -2 (-1 + b) b (1 + b)^5 Sqrt[b (1 + b) (1 + (-1 + 2 a^2) b) c] *)

If the variety of expressions to the simplified is modest, more rules can be developed along these lines to accomplish a full simplification.

Addendum

A simpler and more general solution is

Simplify[f /. Sqrt[z1___ z3_^n_ z2___] -> z3^(n/2) Sqrt[z1 z2]]
{* -2 (-1 + b) b^(3/2) (1 + b)^5 Sqrt[(1 + b) (1 + (-1 + 2 a^2) b) c] *}

Second Addendum

The following largely but not entirely satisfies the goal of the addition to the question.

FullSimplify[ff /. b -> d^2 /. Sqrt[z1___ z3_^n_ z2___] -> z3^(n/2) Sqrt[z1 z2], d > 0];
FullSimplify[
  Collect[Numerator[%136], Sqrt[c (1 + (-1 + 2 a^2) d^2) (-1 + d^4)], Simplify]/
  Collect[Denominator[%136], Sqrt[c (1 + (-1 + 2 a^2) d^2) (-1 + d^4)], Simplify] /. 
  d -> Sqrt[b], b > 0]
(* (a (Sqrt[2] (-1 + a^2) (1 + (-6 + b) b) Sqrt[(1 + (-1 + 2 a^2) b) (-1 +  b^2) c] + 
    2 a Sqrt[b] (-1 + b^2) (3 + a^2 (-1 + b) + b - 
    2 Sqrt[(1 + b) (1 - b + 2 a^2 b) c])))/
   (8 a^3 b^(3/2) (-1 + b^2) + 4 Sqrt[2] (1 + a^2) (-1 + b) b 
    Sqrt[(1 + (-1 + 2 a^2) b) (-1 + b^2) c]) *)

The denominator could be further simplified by using a ComplexityFunction that penalizes terms that the OP wishes to eliminate. However, at some point the approach becomes so specific to the problem that it is equivalent to performing the simplification by hand.

Answer 2

Let us take

z = Sqrt[b (1 + b) (1 + (-1 + 2 a^2) b) c]; k = {a > 0, b > 0, c > 0};

So

z * FullSimplify[f/z, Assumptions -> k]

-2 (-1 + b) b (1 + b)^5 Sqrt[b (1 + b) (1 + (-1 + 2 a^2) b) c]