3.1.61 \(\int x \text {ArcCos}(\sqrt {x}) \, dx\) [61]

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

[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 = 0.01, 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 = {4927, 12, 52, 55, 633, 222} \begin {gather*} \frac {1}{2} x^2 \text {ArcCos}\left (\sqrt {x}\right )-\frac {3}{32} \text {ArcSin}(1-2 x)-\frac {1}{8} \sqrt {1-x} x^{3/2}-\frac {3}{16} \sqrt {1-x} \sqrt {x} \end {gather*}

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 52

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 55

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 222

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 633

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 4927

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} \text {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 = 0.02, size = 41, normalized size = 0.68 \begin {gather*} \frac {1}{16} \left (-\sqrt {-((-1+x) x)} (3+2 x)+8 x^2 \text {ArcCos}\left (\sqrt {x}\right )+3 \text {ArcSin}\left (\sqrt {x}\right )\right ) \end {gather*}

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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Maple [A]
time = 0.01, size = 41, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {x^{\frac {3}{2}} \sqrt {1-x}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \arcsin \left (\sqrt {x}\right )}{16}\)	\(41\)
default	\(\frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {x^{\frac {3}{2}} \sqrt {1-x}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \arcsin \left (\sqrt {x}\right )}{16}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.47, size = 40, normalized size = 0.67 \begin {gather*} \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 ) \end {gather*}

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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Fricas [A]
time = 1.64, size = 31, normalized size = 0.52 \begin {gather*} -\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 ) \end {gather*}

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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Sympy [A]
time = 2.14, size = 66, normalized size = 1.10 \begin {gather*} \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}\: \sqrt {x} > -1 \wedge \sqrt {x} < 1 \end {cases}}{2} \end {gather*}

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,
(sqrt(x) > -1) & (sqrt(x) < 1)))/2

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Giac [A]
time = 0.44, size = 40, normalized size = 0.67 \begin {gather*} \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 ) \end {gather*}

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

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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