\(\int x \sqrt {1-x^2} \arccos (x)^2 \, dx\) [667]

Optimal result

Integrand size = 17, antiderivative size = 66 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {4 \sqrt {1-x^2}}{9}+\frac {2}{27} \left (1-x^2\right )^{3/2}-\frac {2}{3} x \arccos (x)+\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2 \]

[Out]

2/27*(-x^2+1)^(3/2)-2/3*x*arccos(x)+2/9*x^3*arccos(x)-1/3*(-x^2+1)^(3/2)*arccos(x)^2+4/9*(-x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4768, 4740, 455, 45} \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2-\frac {2}{3} x \arccos (x)+\frac {2}{27} \left (1-x^2\right )^{3/2}+\frac {4 \sqrt {1-x^2}}{9} \]

[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 45

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 455

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 4740

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 4768

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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*} \text {integral}& = -\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2-\frac {2}{3} \int \left (1-x^2\right ) \arccos (x) \, dx \\ & = -\frac {2}{3} x \arccos (x)+\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2-\frac {2}{3} \int \frac {x \left (1-\frac {x^2}{3}\right )}{\sqrt {1-x^2}} \, dx \\ & = -\frac {2}{3} x \arccos (x)+\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2-\frac {1}{3} \text {Subst}\left (\int \frac {1-\frac {x}{3}}{\sqrt {1-x}} \, dx,x,x^2\right ) \\ & = -\frac {2}{3} x \arccos (x)+\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2-\frac {1}{3} \text {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 \arccos (x)+\frac {2}{9} x^3 \arccos (x)-\frac {1}{3} \left (1-x^2\right )^{3/2} \arccos (x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {1}{27} \left (-2 \sqrt {1-x^2} \left (-7+x^2\right )+6 x \left (-3+x^2\right ) \arccos (x)-9 \left (1-x^2\right )^{3/2} \arccos (x)^2\right ) \]

[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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\left (x^{2}-1\right ) \sqrt {-x^{2}+1}\, \arccos \left (x \right )^{2}}{3}+\frac {2 \arccos \left (x \right ) \left (x^{2}-3\right ) x}{9}-\frac {2 \left (x^{2}-1\right ) \sqrt {-x^{2}+1}}{27}+\frac {4 \sqrt {-x^{2}+1}}{9}\)	\(59\)

[In]

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

[Out]

1/3*(x^2-1)*(-x^2+1)^(1/2)*arccos(x)^2+2/9*arccos(x)*(x^2-3)*x-2/27*(x^2-1)*(-x^2+1)^(1/2)+4/9*(-x^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.62 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arccos \left (x\right ) + \frac {1}{27} \, {\left (9 \, {\left (x^{2} - 1\right )} \arccos \left (x\right )^{2} - 2 \, x^{2} + 14\right )} \sqrt {-x^{2} + 1} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {2 x^{3} \operatorname {acos}{\left (x \right )}}{9} + \frac {x^{2} \sqrt {1 - x^{2}} \operatorname {acos}^{2}{\left (x \right )}}{3} - \frac {2 x^{2} \sqrt {1 - x^{2}}}{27} - \frac {2 x \operatorname {acos}{\left (x \right )}}{3} - \frac {\sqrt {1 - x^{2}} \operatorname {acos}^{2}{\left (x \right )}}{3} + \frac {14 \sqrt {1 - x^{2}}}{27} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.79 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=-\frac {1}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arccos \left (x\right )^{2} - \frac {2}{27} \, \sqrt {-x^{2} + 1} x^{2} + \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arccos \left (x\right ) + \frac {14}{27} \, \sqrt {-x^{2} + 1} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int x \sqrt {1-x^2} \arccos (x)^2 \, dx=\frac {2}{9} \, x^{3} \arccos \left (x\right ) - \frac {1}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arccos \left (x\right )^{2} - \frac {2}{27} \, \sqrt {-x^{2} + 1} x^{2} - \frac {2}{3} \, x \arccos \left (x\right ) + \frac {14}{27} \, \sqrt {-x^{2} + 1} \]

[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)

Mupad [F(-1)]

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

[In]

int(x*acos(x)^2*(1 - x^2)^(1/2),x)

[Out]

int(x*acos(x)^2*(1 - x^2)^(1/2), x)