Simple integral raises AttributeError

Eric Gourgoulhon

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Dec 4, 2019, 10:38:34 AM12/4/19

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Hi,

In Sage 9.0.beta8 we have

sage: a = var('a')
sage: integrate(1/(x^4 + x^2 + a), x)
...

AttributeError: 'RootSum' object has no attribute '_sage_'

The same error occurs in Sage 8.9, but not in Sage 8.8 (and below). In Sage 8.8, we have instead:

sage: a = var('a')
sage: integrate(1/(x^4 + x^2 + a), x)
integrate(1/(x^4 + x^2 + a), x)

Note that RootSum is a SymPy object. Actually, in Sage 9.0.beta8, forcing the algorithm to 'maxima' yields the same result as in Sage 8.8:

sage: integrate(1/(x^4 + x^2 + a), x, algorithm='maxima')
integrate(1/(x^4 + x^2 + a), x)

So it seems that since Sage 8.9, when integrate() is not capable to find an answer via Maxima, it tries SymPy but is not capable to translate the result back to Sage. I could not find a ticket about this. Shall I open one?

Eric.

PS: for the record, a primitive of 1/(x^4 + x^2 + a) is

sage: b = sqrt(1 - 4*a)

sage: f = sqrt(2)/b*(arctan(sqrt(2)*x/sqrt(1 - b))/sqrt(1 - b) - arctan(sqrt(2)*x/sqrt(1 + b))/sqrt(1 + b))

as we can check:

sage: diff(f, x).simplify_full()
1/(x^4 + x^2 + a)

Dima Pasechnik

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Dec 4, 2019, 12:37:15 PM12/4/19

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sure, please open a ticket.

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Emmanuel Charpentier

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Dec 4, 2019, 2:14:48 PM12/4/19

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The old (> 5 years...) Trac#16816 ticket is germane...

BTW, I'm not sure that this would be very helpful:

sage: foo=sympy.integrate(*[sympy.sympify(u) for u in (1/(x^4+x^2+a), x)]); foo

RootSum(_t**4*(256*a**3 - 128*a**2 + 16*a) + _t**2*(4 - 16*a) + 1, Lambda(_t, _t*log(32*_t**3*a**2 - 8*_t**3*a + 4*_t*a - 2*_t + x)))

sage: foo.doit()._sage_().factor().simplify_full()

1/4*sqrt(2)*(sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(1/4*sqrt(2)*((4*a^2 - a)*((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) + 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) - sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(-1/4*sqrt(2)*((4*a^2 - a)*((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) - 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) + sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(1/4*sqrt(2)*((4*a^2 - a)*((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) + 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) - sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(-1/4*sqrt(2)*((4*a^2 - a)*((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) - 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))))

Cheers ! ...

However, Sage can come to the rescue. We can compute the root sum *in Sage*, thus getting a "more interesting" result:

sage: S1=sum([u[0]^u[1] for u in foo.args[0]._sage_().roots()]).expand().simplify(); S1

1/2

And (manually) use this result in the second expression. Here, I'm stuck, because I do not (yet) understand how the lambda expression, second argument of RooSum, is supposed to be used.

BTW, fricas can integrate this expression:

sage: IF=f(x).integrate(x,algorithm="fricas").expand().factor().simplify_full();

....: IF

1/4*sqrt(2)*(sqrt(((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 1)/(4*a^2 - a))*log(1/2*sqrt(2)*(2*sqrt(2)*x + ((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 4*a - 1)*sqrt(((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 1)/(4*a^2 - a)))) - sqrt(((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 1)/(4*a^2 - a))*log(1/2*sqrt(2)*(2*sqrt(2)*x - ((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 4*a - 1)*sqrt(((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) + 1)/(4*a^2 - a)))) - sqrt(-((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 1)/(4*a^2 - a))*log(1/2*sqrt(2)*(2*sqrt(2)*x + ((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 4*a + 1)*sqrt(-((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 1)/(4*a^2 - a)))) + sqrt(-((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 1)/(4*a^2 - a))*log(1/2*sqrt(2)*(2*sqrt(2)*x - ((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 4*a + 1)*sqrt(-((4*a^2 - a)*sqrt(-1/(4*a^3 - a^2)) - 1)/(4*a^2 - a)))))

And fiddling with sympy.cse suggests that the reference given might be correct:

sage: sympy.cse(sympy.sympify(IF), optimizations="basic")

([(x0, sqrt(2)),

(x1, 4*a),

(x2, x1 - 1),

(x3, 1/x2),

(x4, a*x2*sqrt(-x3/a**2)),

(x5, x4 + 1),

(x6, x3/a),

(x7, sqrt(x5*x6)),

(x8, 2*x*x0),

(x9, x7*(x2 + x4)),

(x10, x0/2),

(x11, sqrt(-x6*(x4 - 1))),

(x12, x11*(-x1 + x5))],

[-x0*(-x11*log(x10*(-x12 + x8)) + x11*log(x10*(x12 + x8)) + x7*log(x10*(x8 - x9)) - x7*log(x10*(x8 + x9)))/4])

HTH,

Emmanuel Charpentier

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Dec 4, 2019, 2:17:51 PM12/4/19

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And, BTW, another data point:

sage: f(x).integrate(x, algorithm="mathematica_free")

-1/2*sqrt(2)*(sqrt(-sqrt(-4*a + 1) + 1)*arctan(sqrt(2)*x/sqrt(sqrt(-4*a + 1) + 1)) - sqrt(sqrt(-4*a + 1) + 1)*arctan(sqrt(2)*x/sqrt(-sqrt(-4*a + 1) + 1)))/sqrt(-(4*a - 1)*a)

HTH,

Eric Gourgoulhon

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Dec 4, 2019, 5:11:56 PM12/4/19

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Le mercredi 4 décembre 2019 18:37:15 UTC+1, Dima Pasechnik a écrit :

sure, please open a ticket.

This is now

https://trac.sagemath.org/ticket/28842

Please review.

Eric.

Eric Gourgoulhon

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Dec 4, 2019, 5:17:36 PM12/4/19

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Le mercredi 4 décembre 2019 20:14:48 UTC+1, Emmanuel Charpentier a écrit :

The old (> 5 years...) Trac#16816 ticket is germane...

Thanks for pointing out this ticket!

In #28842, I propose a quick fix until #16816 is solved, i.e. until sum of roots exist in Sage.

This fix restores the behavior prior to Sage 8.9, namely it returns the unevaluated integral.

Eric.

Dima Pasechnik

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Dec 4, 2019, 5:21:39 PM12/4/19

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On Wed, Dec 4, 2019 at 7:14 PM Emmanuel Charpentier
<emanuel.c...@gmail.com> wrote:
>
> The old (> 5 years...) Trac#16816 ticket is germane...
>
> BTW, I'm not sure that this would be very helpful:
>
> sage: foo=sympy.integrate(*[sympy.sympify(u) for u in (1/(x^4+x^2+a), x)]); foo
> RootSum(_t**4*(256*a**3 - 128*a**2 + 16*a) + _t**2*(4 - 16*a) + 1, Lambda(_t, _t*log(32*_t**3*a**2 - 8*_t**3*a + 4*_t*a - 2*_t + x)))
> sage: foo.doit()._sage_().factor().simplify_full()
> 1/4*sqrt(2)*(sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(1/4*sqrt(2)*((4*a^2 - a)*((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) + 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) - sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(-1/4*sqrt(2)*((4*a^2 - a)*((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) - 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a + sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) + sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(1/4*sqrt(2)*((4*a^2 - a)*((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) + 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))) - sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))*log(-1/4*sqrt(2)*((4*a^2 - a)*((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a))^(3/2) - 2*sqrt(2)*x + 2*(2*a - 1)*sqrt((4*a - sqrt(-64*a^3 + 48*a^2 - 12*a + 1) - 1)/(16*a^3 - 8*a^2 + a)))))
>
> Cheers ! ...
>
> However, Sage can come to the rescue. We can compute the root sum *in Sage*, thus getting a "more interesting" result:
>
> sage: S1=sum([u[0]^u[1] for u in foo.args[0]._sage_().roots()]).expand().simplify(); S1
> 1/2
>
> And (manually) use this result in the second expression. Here, I'm stuck, because I do not (yet) understand how the lambda expression, second argument of RooSum, is supposed to be used.

RootSum(f(t), Lambda(y,g(y)) means

sum_{y: f(y)=0} g(y)

so we sum the function g() in Lambda over all the roots of the 1st
argument, f().

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Emmanuel Charpentier

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Dec 6, 2019, 10:28:40 PM12/6/19

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Okay. That means that we need

a wrapper for RootSum, AND
a wrapper for Lambda.

ISTR that Mathematica has a similar setup.A rare occasion to kill two birds wit the same (two) stones. However, Mathematica's "pure functions" are, as far as I can tell, totally unknown to Sage...

Thank you, Dima !

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rjf

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Dec 8, 2019, 9:16:10 PM12/8/19

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FYI:

In Maxima, integrate_use_rootsof: true

changes the result from an unevaluated integral to

lsum(log(x-%r1)/(4*%r1^3+2*%r1),%r1,rootsof(a+%r1^4+%r1^2,%r1))

which is fairly compact, and to the point, but

which may or may not be what you want. The inside of rootsof

can be picked out and the 4 roots solved in terms of radicals

by Maxima's solve().

If the expression rootsof() is replaced by a list of the roots

returns by solve, { something like substpart( map(rhs, solve( Z,3,1,x)), Z 3), lsum }

you get a large explicit mess of nested radicals and logs.

That Mathematical result with arctan is nicer that the explicit

radical stuff.

RJF

Emmanuel Charpentier

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Dec 9, 2019, 6:32:42 AM12/9/19

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Le lundi 9 décembre 2019 03:16:10 UTC+1, rjf a écrit :

FYI:

In Maxima, integrate_use_rootsof: true
changes the result from an unevaluated integral to
lsum(log(x-%r1)/(4*%r1^3+2*%r1),%r1,rootsof(a+%r1^4+%r1^2,%r1))

which is fairly compact, and to the point, but
which may or may not be what you want. The inside of rootsof
can be picked out and the 4 roots solved in terms of radicals

by Maxima's solve(). *

Aciddentally ! For example, $x^5+x^3+1$ doesn’t have a solution in terms of radicals. Trying to solve it in Maxima fails ; and %solve gives a list of numerical roots.

If the expression rootsof() is replaced by a list of the roots
returns by solve, { something like substpart( map(rhs, solve( Z,3,1,x)), Z 3), lsum }
you get a large explicit mess of nested radicals and logs.

Indeed. And one may add that this solution has some drawbacks problems: computing only one value of the resultt primitive takes several seconds, trying to plot it is pointless…

That Mathematical result with arctan is nicer that the explicit
radical stuff.

“Nicer” depends of what you want to do with the result. In some cases, the approximate answer obtained via %solve is preferable (e. g. plotting). In other cases, the explicit log/roots mess is preferable (e. g. trying to prove that the value of the primitive is real for all real values of the aurument).

The “exact” answers given by Sage via (x^5+x^3+1).roots(ring=QQbar) may be preferable: a radical expression, if it exists, may be obtained via the radical_expression() method, and the precision is not limited to any specific floating-point approximation ; some equalities may be provable using them.

OTOH, such values won’t pass to Maxima, and the resulting answers cannot be further simplified…

The more general way to do this is :

remark that the polynoimial $x^5+x^3+1$ has five (possibly confounded) roots, and name them $r_1,\dots,r_5$ ;
rewrite it as $\displaystyle\prod_{r\in r_1,\dots,\r_5}(x-r)$ ,
integrate using this rewriting,
substitute $r_1\dots,r_5$ as needed by your various uses of the result.

HTH,

Thierry

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Dec 9, 2019, 8:55:25 AM12/9/19

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Hi,

On Wed, Dec 04, 2019 at 07:38:34AM -0800, Eric Gourgoulhon wrote:
> Hi,
>
> In Sage 9.0.beta8 we have
>
> sage: a = var('a')
> sage: integrate(1/(x^4 + x^2 + a), x)
> ...
> AttributeError: 'RootSum' object has no attribute '_sage_'
>
> The same error occurs in Sage 8.9, but not in Sage 8.8 (and below). In Sage
> 8.8, we have instead:
>
> sage: a = var('a')
> sage: integrate(1/(x^4 + x^2 + a), x)
> integrate(1/(x^4 + x^2 + a), x)
>
> Note that RootSum is a SymPy object. Actually, in Sage 9.0.beta8, forcing
> the algorithm to 'maxima' yields the same result as in Sage 8.8:
>
> sage: integrate(1/(x^4 + x^2 + a), x, algorithm='maxima')
> integrate(1/(x^4 + x^2 + a), x)
>
> So it seems that since Sage 8.9, when integrate() is not capable to find an
> answer via Maxima, it tries SymPy but is not capable to translate the
> result back to Sage. I could not find a ticket about this. Shall I open one?

For what it worth, the change was done at
https://trac.sagemath.org/ticket/27958

Ciao,
Thierry

> Eric.
>
> PS: for the record, a primitive of 1/(x^4 + x^2 + a) is
>
> sage: b = sqrt(1 - 4*a)
> sage: f = sqrt(2)/b*(arctan(sqrt(2)*x/sqrt(1 - b))/sqrt(1 - b) -
> arctan(sqrt(2)*x/sqrt(1 + b))/sqrt(1 + b))
>
> as we can check:
>
> sage: diff(f, x).simplify_full()
> 1/(x^4 + x^2 + a)
>
>

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Dima Pasechnik

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Dec 9, 2019, 12:14:53 PM12/9/19

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On Mon, Dec 9, 2019 at 1:55 PM Thierry <sage-goo...@lma.metelu.net> wrote:
>
> Hi,
>
> On Wed, Dec 04, 2019 at 07:38:34AM -0800, Eric Gourgoulhon wrote:
> > Hi,
> >
> > In Sage 9.0.beta8 we have
> >
> > sage: a = var('a')
> > sage: integrate(1/(x^4 + x^2 + a), x)
> > ...
> > AttributeError: 'RootSum' object has no attribute '_sage_'
> >
> > The same error occurs in Sage 8.9, but not in Sage 8.8 (and below). In Sage
> > 8.8, we have instead:
> >
> > sage: a = var('a')
> > sage: integrate(1/(x^4 + x^2 + a), x)
> > integrate(1/(x^4 + x^2 + a), x)

Fricas just returns the answer in radicals, if you call
sage: integrate(1/(x^4 + x^2 + a), x, algorithm='fricas')

with
sage: integrate(1/(x^5 + x^3 + a), x, algorithm='fricas')
one gets an error, for a reason similar to the RootSum thing for sympy
(i.e. it can express the integral as a sum over roots of a polynomial, it's
just Sage doesn't know how to parse it)

> To view this discussion on the web visit https://groups.google.com/d/msgid/sage-devel/20191209135522.n7ymn5yfpcurdppq%40metelu.net.

Emmanuel Charpentier

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Dec 9, 2019, 5:35:05 PM12/9/19

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Well... My solution has problems,at list with 8.9.beta9 and also on Sagecell, which is currently at 9.0) . Consider :

L=flatten([[u[0]]*u[1] for u in (x^5+x^3+1).roots(ring=QQbar)]); print("L = ",L)

P=prod([x-u for u in L]);print("P = ",P)

Rac=[var("r_{}".format(u)) for u in range(len(L))];print("R&c = ",Rac)

D=dict(zip(Rac,L));print("D = ",D)

Pr=prod([x-u for u in Rac]).expand();print("Pr = ",Pr)

%time print(Pr.coefficient(x,2).subs(D)==0)

But %time print(bool(Pr.coefficient(x,2).subs(D)==0)) never returns. Other trials show that trying to use SR methods on expressions including QQbar coeficients fails.