\(\int \sqrt {1-x^2} \arccos (x) \, dx\) [652]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 34 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {x^2}{4}+\frac {1}{2} x \sqrt {1-x^2} \arccos (x)-\frac {\arccos (x)^2}{4} \]

[Out]

1/4*x^2-1/4*arccos(x)^2+1/2*x*arccos(x)*(-x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4742, 4738, 30} \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {1}{2} \sqrt {1-x^2} x \arccos (x)-\frac {\arccos (x)^2}{4}+\frac {x^2}{4} \]

[In]

Int[Sqrt[1 - x^2]*ArcCos[x],x]

[Out]

x^2/4 + (x*Sqrt[1 - x^2]*ArcCos[x])/2 - ArcCos[x]^2/4

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4738

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-(b*c*(n + 1))^(-1)
)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && E
qQ[c^2*d + e, 0] && NeQ[n, -1]

Rule 4742

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[x*Sqrt[d + e*x^2]*((
a + b*ArcCos[c*x])^n/2), x] + (Dist[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcCos[c*x])^n/S
qrt[1 - c^2*x^2], x], x] + Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[x*(a + b*ArcCos[c*x])^(
n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x \sqrt {1-x^2} \arccos (x)+\frac {\int x \, dx}{2}+\frac {1}{2} \int \frac {\arccos (x)}{\sqrt {1-x^2}} \, dx \\ & = \frac {x^2}{4}+\frac {1}{2} x \sqrt {1-x^2} \arccos (x)-\frac {\arccos (x)^2}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {1}{4} \left (x^2+2 x \sqrt {1-x^2} \arccos (x)-\arccos (x)^2\right ) \]

[In]

Integrate[Sqrt[1 - x^2]*ArcCos[x],x]

[Out]

(x^2 + 2*x*Sqrt[1 - x^2]*ArcCos[x] - ArcCos[x]^2)/4

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method result size
default \(-\frac {\arccos \left (x \right ) \left (-x \sqrt {-x^{2}+1}+\arccos \left (x \right )\right )}{2}+\frac {\arccos \left (x \right )^{2}}{4}+\frac {x^{2}}{4}-\frac {1}{4}\) \(33\)

[In]

int(arccos(x)*(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*arccos(x)*(-x*(-x^2+1)^(1/2)+arccos(x))+1/4*arccos(x)^2+1/4*x^2-1/4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {1}{2} \, \sqrt {-x^{2} + 1} x \arccos \left (x\right ) + \frac {1}{4} \, x^{2} - \frac {1}{4} \, \arccos \left (x\right )^{2} \]

[In]

integrate(arccos(x)*(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(-x^2 + 1)*x*arccos(x) + 1/4*x^2 - 1/4*arccos(x)^2

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {x^{2}}{4} + \left (\frac {x \sqrt {1 - x^{2}}}{2} + \frac {\operatorname {asin}{\left (x \right )}}{2}\right ) \operatorname {acos}{\left (x \right )} + \frac {\operatorname {asin}^{2}{\left (x \right )}}{4} \]

[In]

integrate(acos(x)*(-x**2+1)**(1/2),x)

[Out]

x**2/4 + (x*sqrt(1 - x**2)/2 + asin(x)/2)*acos(x) + asin(x)**2/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {1}{4} \, x^{2} + \frac {1}{2} \, {\left (\sqrt {-x^{2} + 1} x + \arcsin \left (x\right )\right )} \arccos \left (x\right ) + \frac {1}{4} \, \arcsin \left (x\right )^{2} \]

[In]

integrate(arccos(x)*(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

1/4*x^2 + 1/2*(sqrt(-x^2 + 1)*x + arcsin(x))*arccos(x) + 1/4*arcsin(x)^2

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\frac {1}{2} \, \sqrt {-x^{2} + 1} x \arccos \left (x\right ) + \frac {1}{4} \, x^{2} - \frac {1}{4} \, \arccos \left (x\right )^{2} - \frac {1}{8} \]

[In]

integrate(arccos(x)*(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-x^2 + 1)*x*arccos(x) + 1/4*x^2 - 1/4*arccos(x)^2 - 1/8

Mupad [F(-1)]

Timed out. \[ \int \sqrt {1-x^2} \arccos (x) \, dx=\int \mathrm {acos}\left (x\right )\,\sqrt {1-x^2} \,d x \]

[In]

int(acos(x)*(1 - x^2)^(1/2),x)

[Out]

int(acos(x)*(1 - x^2)^(1/2), x)