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

Optimal. Leaf size=37 \[ -\frac {1}{2} \sqrt {1-x} \sqrt {x}-\frac {1}{4} \sin ^{-1}(1-2 x)+x \cos ^{-1}\left (\sqrt {x}\right ) \]

[Out]

x*arccos(x^(1/2))+1/4*arcsin(-1+2*x)-1/2*(1-x)^(1/2)*x^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4841, 12, 50, 53, 619, 216} \[ -\frac {1}{2} \sqrt {1-x} \sqrt {x}-\frac {1}{4} \sin ^{-1}(1-2 x)+x \cos ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - x]*Sqrt[x])/2 + x*ArcCos[Sqrt[x]] - ArcSin[1 - 2*x]/4

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 4841

Int[ArcCos[u_], x_Symbol] :> Simp[x*ArcCos[u], x] + Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.03 \[ \frac {1}{2} \left (-\sqrt {-((x-1) x)}-\sin ^{-1}\left (\sqrt {1-x}\right )\right )+x \cos ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcCos[Sqrt[x]] + (-Sqrt[-((-1 + x)*x)] - ArcSin[Sqrt[1 - x]])/2

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fricas [A]  time = 0.44, size = 24, normalized size = 0.65 \[ \frac {1}{2} \, {\left (2 \, x - 1\right )} \arccos \left (\sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x - 1)*arccos(sqrt(x)) - 1/2*sqrt(x)*sqrt(-x + 1)

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giac [A]  time = 1.73, size = 25, normalized size = 0.68 \[ x \arccos \left (\sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} - \frac {1}{2} \, \arccos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccos(sqrt(x)) - 1/2*sqrt(x)*sqrt(-x + 1) - 1/2*arccos(sqrt(x))

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maple [A]  time = 0.00, size = 26, normalized size = 0.70 \[ x \arccos \left (\sqrt {x}\right )-\frac {\sqrt {1-x}\, \sqrt {x}}{2}+\frac {\arcsin \left (\sqrt {x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 25, normalized size = 0.68 \[ x \arccos \left (\sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} + \frac {1}{2} \, \arcsin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccos(sqrt(x)) - 1/2*sqrt(x)*sqrt(-x + 1) + 1/2*arcsin(sqrt(x))

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mupad [B]  time = 0.64, size = 35, normalized size = 0.95 \[ \mathrm {atan}\left (\frac {\sqrt {x}}{\sqrt {1-x}-1}\right )+x\,\mathrm {acos}\left (\sqrt {x}\right )-\frac {\sqrt {x}\,\sqrt {1-x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(x^(1/2)/((1 - x)^(1/2) - 1)) + x*acos(x^(1/2)) - (x^(1/2)*(1 - x)^(1/2))/2

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sympy [A]  time = 0.26, size = 29, normalized size = 0.78 \[ - \frac {\sqrt {x} \sqrt {1 - x}}{2} + x \operatorname {acos}{\left (\sqrt {x} \right )} - \frac {\operatorname {acos}{\left (\sqrt {x} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(x)*sqrt(1 - x)/2 + x*acos(sqrt(x)) - acos(sqrt(x))/2

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