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

Optimal. Leaf size=78 \[ -\frac {1}{18} \sqrt {1-x} x^{5/2}-\frac {5}{72} \sqrt {1-x} x^{3/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{96} \sin ^{-1}(1-2 x) \]

[Out]

1/3*x^3*arccos(x^(1/2))+5/96*arcsin(-1+2*x)-5/72*x^(3/2)*(1-x)^(1/2)-1/18*x^(5/2)*(1-x)^(1/2)-5/48*(1-x)^(1/2)
*x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4843, 12, 50, 53, 619, 216} \[ -\frac {1}{18} \sqrt {1-x} x^{5/2}-\frac {5}{72} \sqrt {1-x} x^{3/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{96} \sin ^{-1}(1-2 x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCos[Sqrt[x]],x]

[Out]

(-5*Sqrt[1 - x]*Sqrt[x])/48 - (5*Sqrt[1 - x]*x^(3/2))/72 - (Sqrt[1 - x]*x^(5/2))/18 + (x^3*ArcCos[Sqrt[x]])/3
- (5*ArcSin[1 - 2*x])/96

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 53

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 4843

Int[((a_.) + ArcCos[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcCos[
u]))/(d*(m + 1)), x] + Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int x^2 \cos ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {1}{3} \int \frac {x^{5/2}}{2 \sqrt {1-x}} \, dx\\ &=\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {1}{6} \int \frac {x^{5/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {5}{36} \int \frac {x^{3/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {5}{72} \sqrt {1-x} x^{3/2}-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {5}{48} \int \frac {\sqrt {x}}{\sqrt {1-x}} \, dx\\ &=-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{72} \sqrt {1-x} x^{3/2}-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {5}{96} \int \frac {1}{\sqrt {1-x} \sqrt {x}} \, dx\\ &=-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{72} \sqrt {1-x} x^{3/2}-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )+\frac {5}{96} \int \frac {1}{\sqrt {x-x^2}} \, dx\\ &=-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{72} \sqrt {1-x} x^{3/2}-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )-\frac {5}{96} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1-2 x\right )\\ &=-\frac {5}{48} \sqrt {1-x} \sqrt {x}-\frac {5}{72} \sqrt {1-x} x^{3/2}-\frac {1}{18} \sqrt {1-x} x^{5/2}+\frac {1}{3} x^3 \cos ^{-1}\left (\sqrt {x}\right )-\frac {5}{96} \sin ^{-1}(1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 46, normalized size = 0.59 \[ \frac {1}{144} \left (48 x^3 \cos ^{-1}\left (\sqrt {x}\right )-\sqrt {-((x-1) x)} \left (8 x^2+10 x+15\right )+15 \sin ^{-1}\left (\sqrt {x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[-((-1 + x)*x)]*(15 + 10*x + 8*x^2)) + 48*x^3*ArcCos[Sqrt[x]] + 15*ArcSin[Sqrt[x]])/144

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fricas [A]  time = 0.43, size = 36, normalized size = 0.46 \[ -\frac {1}{144} \, {\left (8 \, x^{2} + 10 \, x + 15\right )} \sqrt {x} \sqrt {-x + 1} + \frac {1}{48} \, {\left (16 \, x^{3} - 5\right )} \arccos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/144*(8*x^2 + 10*x + 15)*sqrt(x)*sqrt(-x + 1) + 1/48*(16*x^3 - 5)*arccos(sqrt(x))

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giac [A]  time = 3.96, size = 52, normalized size = 0.67 \[ \frac {1}{3} \, x^{3} \arccos \left (\sqrt {x}\right ) - \frac {1}{18} \, x^{\frac {5}{2}} \sqrt {-x + 1} - \frac {5}{72} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {5}{48} \, \sqrt {x} \sqrt {-x + 1} - \frac {5}{48} \, \arccos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccos(sqrt(x)) - 1/18*x^(5/2)*sqrt(-x + 1) - 5/72*x^(3/2)*sqrt(-x + 1) - 5/48*sqrt(x)*sqrt(-x + 1) -
5/48*arccos(sqrt(x))

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maple [A]  time = 0.00, size = 53, normalized size = 0.68 \[ \frac {x^{3} \arccos \left (\sqrt {x}\right )}{3}-\frac {x^{\frac {5}{2}} \sqrt {1-x}}{18}-\frac {5 x^{\frac {3}{2}} \sqrt {1-x}}{72}-\frac {5 \sqrt {1-x}\, \sqrt {x}}{48}+\frac {5 \arcsin \left (\sqrt {x}\right )}{48} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccos(x^(1/2)),x)

[Out]

1/3*x^3*arccos(x^(1/2))-1/18*x^(5/2)*(1-x)^(1/2)-5/72*x^(3/2)*(1-x)^(1/2)-5/48*(1-x)^(1/2)*x^(1/2)+5/48*arcsin
(x^(1/2))

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maxima [A]  time = 0.41, size = 52, normalized size = 0.67 \[ \frac {1}{3} \, x^{3} \arccos \left (\sqrt {x}\right ) - \frac {1}{18} \, x^{\frac {5}{2}} \sqrt {-x + 1} - \frac {5}{72} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {5}{48} \, \sqrt {x} \sqrt {-x + 1} + \frac {5}{48} \, \arcsin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccos(sqrt(x)) - 1/18*x^(5/2)*sqrt(-x + 1) - 5/72*x^(3/2)*sqrt(-x + 1) - 5/48*sqrt(x)*sqrt(-x + 1) +
5/48*arcsin(sqrt(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {acos}\left (\sqrt {x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(x^(1/2)),x)

[Out]

int(x^2*acos(x^(1/2)), x)

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sympy [A]  time = 8.09, size = 73, normalized size = 0.94 \[ \frac {x^{3} \operatorname {acos}{\left (\sqrt {x} \right )}}{3} + \frac {\begin {cases} \frac {x^{\frac {3}{2}} \left (1 - x\right )^{\frac {3}{2}}}{6} + \frac {3 \sqrt {x} \left (1 - 2 x\right ) \sqrt {1 - x}}{16} - \frac {\sqrt {x} \sqrt {1 - x}}{2} + \frac {5 \operatorname {asin}{\left (\sqrt {x} \right )}}{16} & \text {for}\: x \geq 0 \wedge x < 1 \end {cases}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*acos(sqrt(x))/3 + Piecewise((x**(3/2)*(1 - x)**(3/2)/6 + 3*sqrt(x)*(1 - 2*x)*sqrt(1 - x)/16 - sqrt(x)*sqr
t(1 - x)/2 + 5*asin(sqrt(x))/16, (x >= 0) & (x < 1)))/3

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