3.667 \(\int x \sqrt {1-x^2} \cos ^{-1}(x)^2 \, dx\)

Optimal. Leaf size=66 \[ \frac {2}{9} x^3 \cos ^{-1}(x)+\frac {2}{27} \left (1-x^2\right )^{3/2}+\frac {4 \sqrt {1-x^2}}{9}-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {2}{3} x \cos ^{-1}(x) \]

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Rubi [A]  time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4678, 4646, 444, 43} \[ \frac {2}{27} \left (1-x^2\right )^{3/2}+\frac {4 \sqrt {1-x^2}}{9}+\frac {2}{9} x^3 \cos ^{-1}(x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {2}{3} x \cos ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[1 - x^2]*ArcCos[x]^2,x]

[Out]

(4*Sqrt[1 - x^2])/9 + (2*(1 - x^2)^(3/2))/27 - (2*x*ArcCos[x])/3 + (2*x^3*ArcCos[x])/9 - ((1 - x^2)^(3/2)*ArcC
os[x]^2)/3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 4646

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2)
^p, x]}, Dist[a + b*ArcCos[c*x], u, x] + Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; F
reeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 4678

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

Rubi steps

\begin {align*} \int x \sqrt {1-x^2} \cos ^{-1}(x)^2 \, dx &=-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {2}{3} \int \left (1-x^2\right ) \cos ^{-1}(x) \, dx\\ &=-\frac {2}{3} x \cos ^{-1}(x)+\frac {2}{9} x^3 \cos ^{-1}(x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {2}{3} \int \frac {x \left (1-\frac {x^2}{3}\right )}{\sqrt {1-x^2}} \, dx\\ &=-\frac {2}{3} x \cos ^{-1}(x)+\frac {2}{9} x^3 \cos ^{-1}(x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1-\frac {x}{3}}{\sqrt {1-x}} \, dx,x,x^2\right )\\ &=-\frac {2}{3} x \cos ^{-1}(x)+\frac {2}{9} x^3 \cos ^{-1}(x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2-\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {2}{3 \sqrt {1-x}}+\frac {\sqrt {1-x}}{3}\right ) \, dx,x,x^2\right )\\ &=\frac {4 \sqrt {1-x^2}}{9}+\frac {2}{27} \left (1-x^2\right )^{3/2}-\frac {2}{3} x \cos ^{-1}(x)+\frac {2}{9} x^3 \cos ^{-1}(x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 50, normalized size = 0.76 \[ \frac {1}{27} \left (-2 \sqrt {1-x^2} \left (x^2-7\right )-9 \left (1-x^2\right )^{3/2} \cos ^{-1}(x)^2+6 x \left (x^2-3\right ) \cos ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[1 - x^2]*ArcCos[x]^2,x]

[Out]

(-2*Sqrt[1 - x^2]*(-7 + x^2) + 6*x*(-3 + x^2)*ArcCos[x] - 9*(1 - x^2)^(3/2)*ArcCos[x]^2)/27

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {1-x^2} \cos ^{-1}(x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x*Sqrt[1 - x^2]*ArcCos[x]^2,x]

[Out]

Could not integrate

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fricas [A]  time = 1.15, size = 41, normalized size = 0.62 \[ \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arccos \relax (x) + \frac {1}{27} \, {\left (9 \, {\left (x^{2} - 1\right )} \arccos \relax (x)^{2} - 2 \, x^{2} + 14\right )} \sqrt {-x^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(x^3 - 3*x)*arccos(x) + 1/27*(9*(x^2 - 1)*arccos(x)^2 - 2*x^2 + 14)*sqrt(-x^2 + 1)

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giac [A]  time = 1.05, size = 53, normalized size = 0.80 \[ \frac {2}{9} \, x^{3} \arccos \relax (x) - \frac {1}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arccos \relax (x)^{2} - \frac {2}{27} \, \sqrt {-x^{2} + 1} x^{2} - \frac {2}{3} \, x \arccos \relax (x) + \frac {14}{27} \, \sqrt {-x^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*x^3*arccos(x) - 1/3*(-x^2 + 1)^(3/2)*arccos(x)^2 - 2/27*sqrt(-x^2 + 1)*x^2 - 2/3*x*arccos(x) + 14/27*sqrt(
-x^2 + 1)

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maple [C]  time = 0.32, size = 158, normalized size = 2.39

method	result	size
default	\(-\frac {\left (6 i \arccos \relax (x )+9 \arccos \relax (x )^{2}-2\right ) \left (4 i x^{3}-4 \sqrt {-x^{2}+1}\, x^{2}-3 i x +\sqrt {-x^{2}+1}\right )}{216}+\frac {\left (\arccos \relax (x )^{2}-2+2 i \arccos \relax (x )\right ) \left (i x -\sqrt {-x^{2}+1}\right )}{8}-\frac {\left (\arccos \relax (x )^{2}-2-2 i \arccos \relax (x )\right ) \left (i x +\sqrt {-x^{2}+1}\right )}{8}+\frac {\left (-6 i \arccos \relax (x )+9 \arccos \relax (x )^{2}-2\right ) \left (4 i x^{3}+4 \sqrt {-x^{2}+1}\, x^{2}-3 i x -\sqrt {-x^{2}+1}\right )}{216}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/216*(6*I*arccos(x)+9*arccos(x)^2-2)*(4*I*x^3-4*(-x^2+1)^(1/2)*x^2-3*I*x+(-x^2+1)^(1/2))+1/8*(arccos(x)^2-2+
2*I*arccos(x))*(I*x-(-x^2+1)^(1/2))-1/8*(arccos(x)^2-2-2*I*arccos(x))*(I*x+(-x^2+1)^(1/2))+1/216*(-6*I*arccos(
x)+9*arccos(x)^2-2)*(4*I*x^3+4*(-x^2+1)^(1/2)*x^2-3*I*x-(-x^2+1)^(1/2))

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maxima [A]  time = 0.96, size = 52, normalized size = 0.79 \[ -\frac {1}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arccos \relax (x)^{2} - \frac {2}{27} \, \sqrt {-x^{2} + 1} x^{2} + \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arccos \relax (x) + \frac {14}{27} \, \sqrt {-x^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-x^2 + 1)^(3/2)*arccos(x)^2 - 2/27*sqrt(-x^2 + 1)*x^2 + 2/9*(x^3 - 3*x)*arccos(x) + 14/27*sqrt(-x^2 + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,{\mathrm {acos}\relax (x)}^2\,\sqrt {1-x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acos(x)^2*(1 - x^2)^(1/2),x)

[Out]

int(x*acos(x)^2*(1 - x^2)^(1/2), x)

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sympy [A]  time = 3.04, size = 78, normalized size = 1.18 \[ \frac {2 x^{3} \operatorname {acos}{\relax (x )}}{9} + \frac {x^{2} \sqrt {1 - x^{2}} \operatorname {acos}^{2}{\relax (x )}}{3} - \frac {2 x^{2} \sqrt {1 - x^{2}}}{27} - \frac {2 x \operatorname {acos}{\relax (x )}}{3} - \frac {\sqrt {1 - x^{2}} \operatorname {acos}^{2}{\relax (x )}}{3} + \frac {14 \sqrt {1 - x^{2}}}{27} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**3*acos(x)/9 + x**2*sqrt(1 - x**2)*acos(x)**2/3 - 2*x**2*sqrt(1 - x**2)/27 - 2*x*acos(x)/3 - sqrt(1 - x**2
)*acos(x)**2/3 + 14*sqrt(1 - x**2)/27

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