Simplifying the following mathematical expression using a computer?

Question

I have this following beastly expression typed up very nicely in LaTeX formatting, as you can see. What is the easiest way that I can get a computer to simplify this expression for me? I have zero programming experience. I installed sagemath but it seems pretty complicated.

$W_{(1,1)}(t,v)=\frac{-t^{-2k}v^k}{3}(\frac{v^{\frac{3}{2}}-v^{\frac{-3}{2}}}{t^{\frac{3}{2}}-t^{\frac{-3}{2}}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})+\frac{t^{-2k}v^k}{4}(\frac{v-v^{-1}}{t-t^{-1}})^2+\frac{t^{-2k}v^k}{12}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})-\frac{t^{-k}v^k}{4}(\frac{v^2-v^{-2}}{t^2-t^{-2}}) + \frac{t^{-k}v^k}{8}(\frac{v-v^{-1}}{t-t^{-1}})^2+\frac{t^{-k}v^k}{4}(\frac{v-v^{-1}}{t-t^{-1}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^2-\frac{t^{-k}v^k}{8}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^4+\frac{-v^kt^{k}}{4}(\frac{v^2-v^{-2}}{t^2-t^{-2}})+\frac{v^kt^{k}}{3}(\frac{v^{\frac{3}{2}}-v^{\frac{-3}{2}}}{t^{\frac{3}{2}}-t^{\frac{-3}{2}}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})+\frac{v^kt^{k}}{8}(\frac{v-v^{-1}}{t-t^{-1}})^2-\frac{v^kt^{k}}{4}(\frac{v-v^{-1}}{t-t^{-1}})(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^2+\frac{v^kt^{k}}{24}(\frac{v^{\frac{1}{2}}-v^{\frac{-1}{2}}}{t^{\frac{1}{2}}-t^{\frac{-1}{2}}})^4$

Answer 1

Here's an all-form-and-no-substance simplification that merely introduces the notation $$X_p:= \frac{v^p-v^{-p}}{t^p-t^{-p}}$$ and factors-out powers of $t$ and $v$ from certain groups:

$$\begin{align} W_{(1,1)}(t,v)&=\frac{t^{-2k}v^k}{12}\left( -4X_{3/2}X_{1/2}+3X_1^2+X_{1/2}^4\right) \\[4pt] &+\frac{t^{-k} v^k}{8}\left(-2X_2 + X_1^2+2X_1X_{1/2}^2-X_{1/2}^4\right) \\[4pt] &+\frac{t^k v^k}{24}\left(-6X_2 +8X_{3/2}X_{1/2}+3X_1^2-6X_1X_{1/2}^2+X_{1/2}^4\right) \end{align} \tag{1}$$

That the $X$-subscripts in each term (interpreting exponents as multipliers) sum to $2$ serves as a nice sanity-check. (OP's original rendering of the expression omitted the exponent on the $X_{1/2}$ term in the first line. The subscript check helped identify this error. Check the edit history for a version of this answer that included consideration of the erroneous expression.)

Dropping $(1)$ into Mathematica makes a bit of a mess. To de-messify it slightly, define $$s := \sqrt{t} \qquad u := \sqrt{v}$$ so that half-powers of $t$ and $v$ become integer powers of $s$ and $u$. Then, simplifying $(1)$ grouping-by-grouping gives $$\begin{align} W_{(1,1)}(s^2,u^2) = &-s^{-4k+6}u^{2k-4} \frac{(u^2-1)^2 (s^2-u^2) (s^2 u^2-1)}{(s^2-1) (s^4-1)^2 (s^6-1)} \\[4pt] &-s^{-2k+4}u^{2k-4}\frac{\left(u^2-1\right) \left(s^2-u^2\right) \left(s^2 u^2-1\right)(s^4-u^2)}{(s^2-1)^2(s^4-1) (s^8-1)} \\[4pt] &-s^{2k+4} u^{2k-4}\frac{(u^2-1)(s^2-u^2)(s^4-u^2)(s^6-u^2)}{(s^2-1) (s^4-1) (s^6-1) (s^8-1)} \\[4pt] =&\phantom{-}\frac{u^{2k}}{s^{4k}}\;\frac{u_1^2 m_1 m_{-1}}{s_1 s_2^2 s_3} +\frac{u^{2k}}{s^{2k}}\;\frac{u_1 m_1 m_2m_{-1} }{s_1^2 s_2 s_4} -s^{2k}u^{2k}\;\frac{u_1 m_1 m_2 m_3}{s_1 s_2 s_3 s_4} \\[4pt] =&\phantom{-}\frac{u^{2k}}{s^{4k}}\;\frac{ u_1m_1(s_1 s_4 u_1 m_{-1} +s^{2k}s_2 s_3 m_2 m_{-1} -s^{6k}s_1 s_2 m_2 m_3) }{s_1^2s_2^2s_3 s_4} \end{align}$$ where $$s_p:=s^p-s^{-p} \qquad u_p:=u^p-u^{-p} \qquad m_p := \frac{s^p}{u}-\frac{u}{s^p}$$

Further defining $r_p := s^p+s^{-p}$, we have $$\begin{align} s_3 &= s^3-s^{-3}= (s-s^{-1})(s^2+1+s^{-2}) = s_1 (r_2+1) \\ s_4 &= s^4-s^{-4}= (s^2-s^{-2})(s^2+s^{-2}) = s_2 r_2 \end{align}$$ which allows us to write

$$W_{(1,1)}(s^2,u^2)=\frac{u^{2k}}{s^{4k}}\;\frac{ u_1m_1(r_2 u_1 m_{-1} +s^{2k} (r_2+1) m_2 m_{-1} -s^{6k} m_2 m_3) }{s_1 s_2 s_3 s_4} $$

There may be even cleaner ways to write the expression, but this is as far as I'll go.

Answer 2

Wolfram|Alpha supports Latex. Unfortunately, the engine didn't seem to understand your query, perhaps because of its length. If possible, I would suggest splitting the expression up into chunks, and entering them one by one into Wolfram|Alpha.