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]

(-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

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Rubi [A]  time = 0.028212, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0375544, size = 46, normalized size = 0.59 \[ \frac{1}{144} \left (-\sqrt{-(x-1) x} \left (8 x^2+10 x+15\right )+48 x^3 \cos ^{-1}\left (\sqrt{x}\right )+15 \sin ^{-1}\left (\sqrt{x}\right )\right ) \]

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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Maple [A]  time = 0.004, size = 53, normalized size = 0.7 \begin{align*}{\frac{{x}^{3}}{3}\arccos \left ( \sqrt{x} \right ) }-{\frac{1}{18}{x}^{{\frac{5}{2}}}\sqrt{1-x}}-{\frac{5}{72}{x}^{{\frac{3}{2}}}\sqrt{1-x}}-{\frac{5}{48}\sqrt{1-x}\sqrt{x}}+{\frac{5}{48}\arcsin \left ( \sqrt{x} \right ) } \end{align*}

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 = 1.43558, size = 70, normalized size = 0.9 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 2.63229, size = 115, normalized size = 1.47 \begin{align*} -\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 ) \end{align*}

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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Sympy [A]  time = 10.1623, size = 73, normalized size = 0.94 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.12811, size = 70, normalized size = 0.9 \begin{align*} \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 ) \end{align*}

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))