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Michael Hardy
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Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray}\begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & & {}-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & {}-(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray}\begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & {}-3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & {}+9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & {}+(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & {}- 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray}\begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & {}- (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & {}+4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray}\begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & {} -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate (I'd be surprised if it hasn't been, but I don't know where to look).

Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate (I'd be surprised if it hasn't been, but I don't know where to look).

Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & & {}-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & {}-(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & {}-3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & {}+9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & {}+(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & {}- 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & {}- (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & {}+4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & {} -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate (I'd be surprised if it hasn't been, but I don't know where to look).

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Michael Engelhardt
Michael Engelhardt
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Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate (I'd be surprised if it hasn't been, but I don't know where to look).

Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate.

Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate (I'd be surprised if it hasn't been, but I don't know where to look).

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Michael Engelhardt
Michael Engelhardt
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Given a collection of free random variables $X_i $, to evaluate a trace of the form $$ \tau [ f_1 (X_{i_1 } ) f_2 (X_{i_2 } ) \ldots f_n (X_{i_n } ) ] $$ where $i_{j+1} \neq i_{j} $, consider the vanishing (by freeness) trace $$ \tau [ (f_1 (X_{i_1 } ) - \tau [f_1 (X_{i_1 }])\ (f_2 (X_{i_2 } ) - \tau [f_2 (X_{i_2 }]) \ldots (f_n (X_{i_n } ) - \tau [f_n (X_{i_n }]) ] = 0 $$ Multiplying this out, one can express the trace of a product of $n$ terms by traces of products of $(n-1)$, $(n-2)$, etc., terms, until the expression is reduced to one containing only individual moments.

Thus (using also cyclicity of the trace), \begin{eqnarray} \tau [XYXY] &=& 2\tau [X] \tau [X Y^2 ] +2\tau [Y] \tau[X^2 Y] -4\tau [X] \tau [Y] \tau [XY] \\ & &-(\tau[X])^2 \tau [Y^2 ] -(\tau[Y])^2 \tau [X^2 ]+4(\tau[X])^2 (\tau[Y])^2 \\ & & -(\tau[X])^2 (\tau[Y])^2 \\ &=& (\tau [X])^2 \tau [Y^2 ] + (\tau [Y])^2 \tau [X^2 ] -(\tau [X])^2 (\tau [Y])^2 \end{eqnarray} as noted in the OP, and similarly, \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [XYXY^2] + 3\tau [Y] \tau [X^2 YXY] -6\tau [X] \tau [Y] \tau [XYXY] \\ & & -3(\tau [X])^2 \tau [XY^3]-3\tau [X] \tau [Y] \tau [X^2 Y^2] -3(\tau [Y])^2 \tau [X^3 Y] \\ & & +9(\tau [X])^2 \tau [Y] \tau [XY^2] + 9\tau [X] (\tau [Y])^2 \tau [X^2 Y] +(\tau [X])^3 \tau [Y^3] \\ & & +(\tau [Y])^3 \tau [X^3] -9(\tau [X])^2 (\tau [Y])^2 \tau [XY] -3(\tau [X])^3 \tau [Y] \tau [Y^2] \\ & & - 3\tau [X] (\tau [Y])^3 \tau [X^2] +6(\tau [X])^3 (\tau [Y])^3 -(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} To evaluate this, we need \begin{eqnarray} \tau [XYXY^2 ] &=& 2\tau [X] \tau [XY^3] + \tau [Y] \tau [X^2 Y^2] +\tau [Y^2 ] \tau [X^2 Y] -2\tau [X] \tau [Y] \tau [XY^2] \\ & & - (\tau [X])^2 \tau [Y^3] -2\tau [X] \tau [Y^2 ] \tau [XY] - \tau [Y] \tau [Y^2 ] \tau [X^2 ] \\ & & +4(\tau [X])^2 \tau [Y] \tau [Y^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \\ &=& (\tau [X])^2 \tau [Y^3] + \tau [Y] \tau [Y^2 ] \tau [X^2 ] -(\tau [X])^2 \tau [Y] \tau [Y^2 ] \end{eqnarray} Inserting this, as well as the expression with $X$ and $Y$ exchanged, and the previous result for $\tau [XYXY]$, we end up with \begin{eqnarray} \tau [XYXYXY] &=& 3\tau [X] \tau [Y] \tau [X^2 ] \tau [Y^2 ] +(\tau [X])^3 \tau [Y^3] +(\tau [Y])^3 \tau [X^3] \\ & & -3(\tau [X])^3 \tau [Y] \tau [Y^2] -3(\tau [Y])^3 \tau [X] \tau [X^2] +2(\tau [X])^3 (\tau [Y])^3 \end{eqnarray} This seems feasible to automate.