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

Optimal. Leaf size=60 \[ -\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {3}{32} \sin ^{-1}(1-2 x) \]

[Out]

1/2*x^2*arccos(x^(1/2))+3/32*arcsin(-1+2*x)-1/8*x^(3/2)*(1-x)^(1/2)-3/16*(1-x)^(1/2)*x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4843, 12, 50, 53, 619, 216} \[ -\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {3}{32} \sin ^{-1}(1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - x]*Sqrt[x])/16 - (Sqrt[1 - x]*x^(3/2))/8 + (x^2*ArcCos[Sqrt[x]])/2 - (3*ArcSin[1 - 2*x])/32

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 \cos ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )+\frac {1}{2} \int \frac {x^{3/2}}{2 \sqrt {1-x}} \, dx\\ &=\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )+\frac {1}{4} \int \frac {x^{3/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )+\frac {3}{16} \int \frac {\sqrt {x}}{\sqrt {1-x}} \, dx\\ &=-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )+\frac {3}{32} \int \frac {1}{\sqrt {1-x} \sqrt {x}} \, dx\\ &=-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )+\frac {3}{32} \int \frac {1}{\sqrt {x-x^2}} \, dx\\ &=-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )-\frac {3}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1-2 x\right )\\ &=-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \cos ^{-1}\left (\sqrt {x}\right )-\frac {3}{32} \sin ^{-1}(1-2 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.68 \[ \frac {1}{16} \left (8 x^2 \cos ^{-1}\left (\sqrt {x}\right )-\sqrt {-((x-1) x)} (2 x+3)+3 \sin ^{-1}\left (\sqrt {x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[-((-1 + x)*x)]*(3 + 2*x)) + 8*x^2*ArcCos[Sqrt[x]] + 3*ArcSin[Sqrt[x]])/16

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fricas [A]  time = 0.43, size = 31, normalized size = 0.52 \[ -\frac {1}{16} \, {\left (2 \, x + 3\right )} \sqrt {x} \sqrt {-x + 1} + \frac {1}{16} \, {\left (8 \, x^{2} - 3\right )} \arccos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*x + 3)*sqrt(x)*sqrt(-x + 1) + 1/16*(8*x^2 - 3)*arccos(sqrt(x))

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giac [A]  time = 0.19, size = 40, normalized size = 0.67 \[ \frac {1}{2} \, x^{2} \arccos \left (\sqrt {x}\right ) - \frac {1}{8} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {3}{16} \, \sqrt {x} \sqrt {-x + 1} - \frac {3}{16} \, \arccos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccos(sqrt(x)) - 1/8*x^(3/2)*sqrt(-x + 1) - 3/16*sqrt(x)*sqrt(-x + 1) - 3/16*arccos(sqrt(x))

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maple [A]  time = 0.00, size = 41, normalized size = 0.68 \[ \frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {x^{\frac {3}{2}} \sqrt {1-x}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \arcsin \left (\sqrt {x}\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccos(x^(1/2))-1/8*x^(3/2)*(1-x)^(1/2)-3/16*(1-x)^(1/2)*x^(1/2)+3/16*arcsin(x^(1/2))

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maxima [A]  time = 0.40, size = 40, normalized size = 0.67 \[ \frac {1}{2} \, x^{2} \arccos \left (\sqrt {x}\right ) - \frac {1}{8} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {3}{16} \, \sqrt {x} \sqrt {-x + 1} + \frac {3}{16} \, \arcsin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccos(sqrt(x)) - 1/8*x^(3/2)*sqrt(-x + 1) - 3/16*sqrt(x)*sqrt(-x + 1) + 3/16*arcsin(sqrt(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*acos(x^(1/2)), x)

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sympy [A]  time = 3.86, size = 58, normalized size = 0.97 \[ \frac {x^{2} \operatorname {acos}{\left (\sqrt {x} \right )}}{2} + \frac {\begin {cases} \frac {\sqrt {x} \left (1 - 2 x\right ) \sqrt {1 - x}}{8} - \frac {\sqrt {x} \sqrt {1 - x}}{2} + \frac {3 \operatorname {asin}{\left (\sqrt {x} \right )}}{8} & \text {for}\: x \geq 0 \wedge x < 1 \end {cases}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*acos(sqrt(x))/2 + Piecewise((sqrt(x)*(1 - 2*x)*sqrt(1 - x)/8 - sqrt(x)*sqrt(1 - x)/2 + 3*asin(sqrt(x))/8,
(x >= 0) & (x < 1)))/2

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