Could randomisation make for a better book cipher? Suppose that the file linux.words is used as the "book".  This is a publicly available list of just shy of half a million words.  And suppose that this book is used, not in its native alphabetic order, but in some randomised order.  The number of possible random arrangements is a decimal which is over two million decimal digits long. The cipher text will be a stream of numbers which index into the chosen random arrangement of the "book".  Does this approach make for an improved book cipher?
From the Wikipedia page on Book ciphers:


  Essentially, the code version of a "book cipher" is just like any
  other code, but one in which the trouble of preparing and distributing
  the codebook has been eliminated by using an existing text. However
  this means, as well as being attacked by all the usual means employed
  against other codes or ciphers, partial solutions may help the
  cryptanalyst to guess other codewords, or even to break the code
  completely by identifying the key text. This is, however, not the only
  way a book cipher may be broken. It is still susceptible to other
  methods of cryptanalysis, and as such is quite easily broken, even
  without sophisticated means, without the cryptanalyst having any idea
  what book the cipher is keyed to.


So you would effectively have a very weird book that would be unknown to the attacker, but that isn't enough to make it secure.  In particular, (link)


  If you know 1,000 words, you will be between a functional beginner and
  conversational level in English.  In most of the world’s languages,
  500 words will be more than enough to get you through any tourist
  situations and everyday introductions.


Randomizing the word order wouldn't change the frequency clues.  Not being able to guess the book would add a small amount of additional security, but not enough to give it any meaningful improvement over a traditional book cipher.