# Revision history [back]

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.