Hi Peter!
On Wed, Sep 11, 2013 at 7:19 AM, Peter Luschny <[email protected]> wrote:
> Consider
>
> (F1) sqrt(1+x^3)/x
> (F2) sqrt(1+1/x^3)*sqrt(x)
>
> According to Mathematica's online integrator
>
> (I1) integral F1 dx = (2/3)*(sqrt(x^3+1)-arctanh(sqrt(x^3+1)))
> (I2) integral F2 dx =
> (2*sqrt(1/x^3+1)*x^(3/2)*(sqrt(x^3+1)-arctanh(sqrt(x^3+1))))/(3*sqrt(x^3+1))
>
> SymPy Live computes (I1) as
> (S) 2*x**(3/2)/(3*sqrt(1 + x**(-3))) - 2*asinh(x**(-3/2))/3 +
> 2/(3*x**(3/2)*sqrt(1 + x**(-3)))
>
> SymPy Live timed out with (I2). SymPy 0.7.3 computes (I2) as
> (S) 2*x**(3/2)/(3*sqrt(1 + x**(-3))) - 2*asinh(x**(-3/2))/3 +
> 2/(3*x**(3/2)*sqrt(1 + x**(-3)))
>
> The derivative of (S) is (F2) and not (F1). So I am inclined to
> say that SymPy computes (I1) not correctly.
Thanks for reporting this. Here is what I tried:
In [1]: x = Symbol("x", real=True)
In [2]: f = integrate(sqrt(1+x**3)/x, x)
In [3]: e = f.diff(x).simplify().expand().factor().cancel()
In [4]: print e
(x**3 + 1)/(x**(5/2)*sqrt(1 + x**(-3)))
It's kind of a pain to simplify "f", but at the end, the expression [4] is equal
to sqrt(1+x^3)/x, as you can check by hand, at least for x > 0.
How did you make it equal to (F2)?
Ondrej
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