How to express an expression with only ArcTan and ArcTanh?

Question

I have an expression which is simply

(j/k) x^(j/k) LerchPhi[x,1,j/k)] 

where 0 < j < k.

Manually I have been able (tedious work) to obtain only ArcTan's and ArcTanh's. How could I ask Mathematica to do this automatically in the most compact form ?

Answer 1

This should get you started. For some reason Mathematica returns LerchPhi[x, 1, j/k] unevaluated if the third argument is a symbol. So let's assume $j = 1$ and $k = 4$ (it does meet your requirement $0 < j < k$ ).

Let's first define some complexity function:

cf[k_][e_] :=  k Count[e, _Log | _ArcCot | _ArcCoth, Infinity] + LeafCount[e]

Now we simplify:

FullSimplify[ FunctionExpand[(1/4) x^(1/4) LerchPhi[x, 1, 1/4]], 
              ComplexityFunction -> cf[#]] & /@ Range[4]

{1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]), 1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]),
 1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)]),   1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)])}

Looks like it converged fast to a nicely simplified expression in terms of only ArcTan and ArcTanh

1/2 (ArcTan[x^(1/4)] + ArcTanh[x^(1/4)])

Admittedly, for this lone case, FunctionExpand will do just fine, but if you're trying to generalize this, you'll need to use the ComplexityFunction. To see this try:

FunctionExpand[(1/8) x^(1/8) LerchPhi[x, 1, 1/8]]

You'll get your answer in terms of Log only

x^(1/8) ( -(Log[1 - x^(1/8)]/(8 x^(1/8))) 
          + (I Log[1 - I x^(1/8)])/(8 x^(1/8)) 
          - (I Log[1 + I x^(1/8)])/(8 x^(1/8)) 
          + Log[1 + x^(1/8)]/(8 x^(1/8)) 
          - ((-1)^(1/4) Log[1 - E^(-((I π)/4)) x^(1/8)])/(8 x^(1/8)) 
          + ((-1)^(3/4) Log[1 - E^((I π)/4) x^(1/8)])/(8 x^(1/8)) 
          - ((-1)^(3/4) Log[1 - E^(-((3 I π)/4)) x^(1/8)])/(8 x^(1/8)) 
          + ((-1)^(1/4) Log[1 - E^((3 I π)/4) x^(1/8)])/(8 x^(1/8)))

But if we apply our cf:

FullSimplify[FunctionExpand[(1/8) x^(1/8) LerchPhi[x, 1, 1/8]], ComplexityFunction -> cf[1]]

We obtain:

1/4 (ArcTan[x^(1/8)] + ArcTanh[x^(1/8)] - (-1)^( 3/4) (ArcTan[(-1)^(1/4) x^(1/8)] 
+ ArcTanh[(-1)^(1/4) x^(1/8)]))

Again, in terms of only ArcTan and ArcTanh.