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answered 2019-07-24 15:17:59 +0100

Emmanuel Charpentier gravatar image

Well, in this case, let's look at your expression's antiderivative (Sage is known to have some infelicities in definite integration). Sage's default integrator (i. e. maxima) needs to know if $c^2-1>0$ ; account for that as well as for other Sage's integrators :

(c,t)=var("c, t")
E=sqrt(1-t^2)*(t+c)
Sols={}
with assuming(c>-1,c<1):Sols.update({"Mn":E.integrate(t)})
with assuming(c<-1):Sols.update({"Mp":E.integrate(t)})
Sols.update({"Mf":E.integrate(t, algorithm="fricas")})
Sols.update({"Mg":E.integrate(t, algorithm="giac")})
Sols.update({"Mm":mathematica.Integrate(E,t).sage()}
Sols.update({"Ms":E.integrate(t, algorithm="sympy")}))

We have now six expressions (four pairwise distinct expressions) for a primitive of your expression $E$:

Sols
{'Mn': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mp': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mf': 1/6*(3*c*t^5 + 2*t^6 - 15*c*t^3 - 12*t^4 + 12*c*t + 12*t^2 - 6*(3*c*t^2 - (c*t^2 - 4*c)*sqrt(-t^2 + 1) - 4*c)*arctan((sqrt(-t^2 + 1) - 1)/t) + 3*(3*c*t^3 + 2*t^4 - 4*c*t - 4*t^2)*sqrt(-t^2 + 1))/(3*t^2 - (t^2 - 4)*sqrt(-t^2 + 1) - 4),
 'Mg': 1/2*c*arcsin(t) + 1/6*((3*c + 2*t)*t - 2)*sqrt(-t^2 + 1),
 'Mm': 1/2*c*arcsin(t) + 1/6*(3*c*t + 2*t^2 - 2)*sqrt(-t^2 + 1),
 'Ms': 1/2*sqrt(-t^2 + 1)*c*t + 1/3*sqrt(-t^2 + 1)*t^2 + 1/2*c*arcsin(t) - 1/3*sqrt(-t^2 + 1)}

More easily seen and understood via \LaTeX:

$$ \begin{align} Mn &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mp &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mf &: \frac{3 \, c t^{5} + 2 \, t^{6} - 15 \, c t^{3} - 12 \, t^{4} + 12 \, c t + 12 \, t^{2} - 6 \, {\left(3 \, c t^{2} - {\left(c t^{2} - 4 \, c\right)} \sqrt{-t^{2} + 1} - 4 \, c\right)} \arctan\left(\frac{\sqrt{-t^{2} + 1} - 1}{t}\right) + 3 \, {\left(3 \, c t^{3} + 2 \, t^{4} - 4 \, c t - 4 \, t^{2}\right)} \sqrt{-t^{2} + 1}}{6 \, {\left(3 \, t^{2} - {\left(t^{2} - 4\right)} \sqrt{-t^{2} + 1} - 4\right)}}\\Mg &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left({\left(3 \, c + 2 \, t\right)} t - 2\right)} \sqrt{-t^{2} + 1}\\Mm &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left(3 \, c t + 2 \, t^{2} - 2\right)} \sqrt{-t^{2} + 1}\\Ms &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{3} \, \sqrt{-t^{2} + 1} t^{2} + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, \sqrt{-t^{2} + 1} \end{align}$$

(one can ensure that maxima's solution for $c>1$ is identical to the one found for $c<-1$:

with assuming(c>1): bool(E.integrate(t)==Sols.get("Mp"))
True).

But all these expressions turn out to derivate to $E$:

[(Sols.get(u).diff(t)/E).canonicalize_radical() for u in Sols.keys()]
[1, 1, 1, 1, 1, 1]

Different integrators have different strategies for uintegration, leading to different primitives. Barring discontinuities, these for expression should differ by a constant.

Well, in this case, let's look at your expression's antiderivative (Sage is known to have some infelicities in definite integration). Sage's default integrator (i. e. maxima) needs to know if $c^2-1>0$ ; account for that as well as for other Sage's integrators :

(c,t)=var("c, t")
E=sqrt(1-t^2)*(t+c)
Sols={}
with assuming(c>-1,c<1):Sols.update({"Mn":E.integrate(t)})
with assuming(c<-1):Sols.update({"Mp":E.integrate(t)})
Sols.update({"Mf":E.integrate(t, algorithm="fricas")})
Sols.update({"Mg":E.integrate(t, algorithm="giac")})
Sols.update({"Mm":mathematica.Integrate(E,t).sage()}
Sols.update({"Ms":E.integrate(t, algorithm="sympy")}))

We have now six expressions (four pairwise distinct expressions) for a primitive of your expression $E$:

Sols
{'Mn': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mp': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mf': 1/6*(3*c*t^5 + 2*t^6 - 15*c*t^3 - 12*t^4 + 12*c*t + 12*t^2 - 6*(3*c*t^2 - (c*t^2 - 4*c)*sqrt(-t^2 + 1) - 4*c)*arctan((sqrt(-t^2 + 1) - 1)/t) + 3*(3*c*t^3 + 2*t^4 - 4*c*t - 4*t^2)*sqrt(-t^2 + 1))/(3*t^2 - (t^2 - 4)*sqrt(-t^2 + 1) - 4),
 'Mg': 1/2*c*arcsin(t) + 1/6*((3*c + 2*t)*t - 2)*sqrt(-t^2 + 1),
 'Mm': 1/2*c*arcsin(t) + 1/6*(3*c*t + 2*t^2 - 2)*sqrt(-t^2 + 1),
 'Ms': 1/2*sqrt(-t^2 + 1)*c*t + 1/3*sqrt(-t^2 + 1)*t^2 + 1/2*c*arcsin(t) - 1/3*sqrt(-t^2 + 1)}

More easily seen and understood via \LaTeX:

$$ \begin{align} Mn &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mp &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mf &: \frac{3 \, c t^{5} + 2 \, t^{6} - 15 \, c t^{3} - 12 \, t^{4} + 12 \, c t + 12 \, t^{2} - 6 \, {\left(3 \, c t^{2} - {\left(c t^{2} - 4 \, c\right)} \sqrt{-t^{2} + 1} - 4 \, c\right)} \arctan\left(\frac{\sqrt{-t^{2} + 1} - 1}{t}\right) + 3 \, {\left(3 \, c t^{3} + 2 \, t^{4} - 4 \, c t - 4 \, t^{2}\right)} \sqrt{-t^{2} + 1}}{6 \, {\left(3 \, t^{2} - {\left(t^{2} - 4\right)} \sqrt{-t^{2} + 1} - 4\right)}}\\Mg &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left({\left(3 \, c + 2 \, t\right)} t - 2\right)} \sqrt{-t^{2} + 1}\\Mm &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left(3 \, c t + 2 \, t^{2} - 2\right)} \sqrt{-t^{2} + 1}\\Ms &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{3} \, \sqrt{-t^{2} + 1} t^{2} + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, \sqrt{-t^{2} + 1} \end{align}$$

(one can ensure that maxima's solution for $c>1$ is identical to the one found for $c<-1$:

with assuming(c>1): bool(E.integrate(t)==Sols.get("Mp"))
True).

But all these expressions turn out to derivate to $E$:

[(Sols.get(u).diff(t)/E).canonicalize_radical() for u in Sols.keys()]
[1, 1, 1, 1, 1, 1]

Different integrators have different strategies for uintegration, integration, leading to different primitives. Barring discontinuities, these for four expression should differ by a constant.

Well, in this case, let's look at your expression's antiderivative (Sage is known to have some infelicities in definite integration). Sage's default integrator (i. e. maxima) needs to know if $c^2-1>0$ ; account for that as well as for other Sage's integrators :

(c,t)=var("c, t")
E=sqrt(1-t^2)*(t+c)
Sols={}
with assuming(c>-1,c<1):Sols.update({"Mn":E.integrate(t)})
with assuming(c<-1):Sols.update({"Mp":E.integrate(t)})
Sols.update({"Mf":E.integrate(t, algorithm="fricas")})
Sols.update({"Mg":E.integrate(t, algorithm="giac")})
Sols.update({"Mm":mathematica.Integrate(E,t).sage()}
Sols.update({"Ms":E.integrate(t, algorithm="sympy")}))

We have now six expressions (four pairwise distinct expressions) for a primitive of your expression $E$:

Sols
{'Mn': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mp': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mf': 1/6*(3*c*t^5 + 2*t^6 - 15*c*t^3 - 12*t^4 + 12*c*t + 12*t^2 - 6*(3*c*t^2 - (c*t^2 - 4*c)*sqrt(-t^2 + 1) - 4*c)*arctan((sqrt(-t^2 + 1) - 1)/t) + 3*(3*c*t^3 + 2*t^4 - 4*c*t - 4*t^2)*sqrt(-t^2 + 1))/(3*t^2 - (t^2 - 4)*sqrt(-t^2 + 1) - 4),
 'Mg': 1/2*c*arcsin(t) + 1/6*((3*c + 2*t)*t - 2)*sqrt(-t^2 + 1),
 'Mm': 1/2*c*arcsin(t) + 1/6*(3*c*t + 2*t^2 - 2)*sqrt(-t^2 + 1),
 'Ms': 1/2*sqrt(-t^2 + 1)*c*t + 1/3*sqrt(-t^2 + 1)*t^2 + 1/2*c*arcsin(t) - 1/3*sqrt(-t^2 + 1)}

More easily seen and understood via \LaTeX:

$$ \begin{align} Mn &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mp &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mf &: \frac{3 \, c t^{5} + 2 \, t^{6} - 15 \, c t^{3} - 12 \, t^{4} + 12 \, c t + 12 \, t^{2} - 6 \, {\left(3 \, c t^{2} - {\left(c t^{2} - 4 \, c\right)} \sqrt{-t^{2} + 1} - 4 \, c\right)} \arctan\left(\frac{\sqrt{-t^{2} + 1} - 1}{t}\right) + 3 \, {\left(3 \, c t^{3} + 2 \, t^{4} - 4 \, c t - 4 \, t^{2}\right)} \sqrt{-t^{2} + 1}}{6 \, {\left(3 \, t^{2} - {\left(t^{2} - 4\right)} \sqrt{-t^{2} + 1} - 4\right)}}\\Mg &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left({\left(3 \, c + 2 \, t\right)} t - 2\right)} \sqrt{-t^{2} + 1}\\Mm &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left(3 \, c t + 2 \, t^{2} - 2\right)} \sqrt{-t^{2} + 1}\\Ms &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{3} \, \sqrt{-t^{2} + 1} t^{2} + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, \sqrt{-t^{2} + 1} \end{align}$$

(one can ensure that maxima's solution for $c>1$ is identical to the one found for $c<-1$:

with assuming(c>1): bool(E.integrate(t)==Sols.get("Mp"))
True).

But all these expressions turn out to derivate to $E$:

[(Sols.get(u).diff(t)/E).canonicalize_radical() for u in Sols.keys()]
[1, 1, 1, 1, 1, 1]

Different integrators have different strategies for integration, leading to different primitives. Barring discontinuities, these four expression expressions should differ by a constant.

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