• To: mathgroup at smc.vnet.net
• From: BobHanlon at aol.com
• Date: Sun, 30 Apr 2000 21:13:39 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```f1[x_] := Evaluate[Simplify[Integrate[(x^2*(x - 1))^(1/3), x]]];
f2[x_] := (f1[x] // PowerExpand // FullSimplify);
f3[x_] := (f1[x] // FunctionExpand);
f4[x_] := (f3[x] // PowerExpand // FullSimplify);

(#[x] &  /@ {f1, f2, f3, f4}) // ColumnForm

{f1[x] == f3[x], f2[x] == f4[x]} // FullSimplify

{True, True}

To compute the functions you need to give them numeric arguments

Table[f1[x], {x, -5., 5., .3}]

Note that the output values are complex.

Use of PowerExpand appears to be valid:

And @@ Table[f1[x] == f2[x] == f3[x] == f4[x], {x, -5., 5., .3}]

True

And @@ Flatten[
Table[f1[x + I*y] == f2[x + I*y] == f3[x + I*y] == f4[x + I*y], {x, -5.,
5., .3}, {y, -5., 5., .3}]]

True

Plot[Abs[#[x]], {x, -3/2, 3/2}] & /@ {f1, f2};

In a message dated 4/25/2000 2:01:46 AM, zeno at magicnet.net writes:

>I symbolically integrated the function..(x^2*(x-1))^(1/3) with respect
>to
>x.
>
>There is in the answer... Hypergeometric2F1[2/3,2/3,5/3,x]
>
>I can do nothing more with that..it just returns it. A Hypergeometric2F1
>with different parameters like Hypergeometric2F1[2,2,5,x] gives an answer.
>I
>am using version 3. Is Mathematica unable to compute it?
>
>I can get the Integral with out the Hypergeometric function on the TI-92+,
>(it gives the answer in a different for using Tan, etc.) but I still would
>like to work with the Mathematica answer.
>

Bob

BobHanlon at aol.com

```

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