3.4.63 \(\int \text {ArcSin}(\sqrt {x}) \, dx\) [363]

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

[Out]

-1/4*arcsin(-1+2*x)+x*arcsin(x^(1/2))+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 = {4924, 12, 52, 55, 633, 222} \begin {gather*} \frac {1}{4} \text {ArcSin}(1-2 x)+x \text {ArcSin}\left (\sqrt {x}\right )+\frac {1}{2} \sqrt {1-x} \sqrt {x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[Sqrt[x]],x]

[Out]

(Sqrt[1 - x]*Sqrt[x])/2 + ArcSin[1 - 2*x]/4 + x*ArcSin[Sqrt[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 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 4924

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[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 \sin ^{-1}\left (\sqrt {x}\right ) \, dx &=x \sin ^{-1}\left (\sqrt {x}\right )-\int \frac {\sqrt {x}}{2 \sqrt {1-x}} \, dx\\ &=x \sin ^{-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 \sin ^{-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 \sin ^{-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 \sin ^{-1}\left (\sqrt {x}\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1-2 x\right )\\ &=\frac {1}{2} \sqrt {1-x} \sqrt {x}+\frac {1}{4} \sin ^{-1}(1-2 x)+x \sin ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 46, normalized size = 1.24 \begin {gather*} x \text {ArcSin}\left (\sqrt {x}\right )+\frac {1}{2} \left (\sqrt {-((-1+x) x)}-2 \text {ArcTan}\left (\frac {\sqrt {x}}{-1+\sqrt {1-x}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[Sqrt[x]],x]

[Out]

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

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Maple [A]
time = 0.01, size = 26, normalized size = 0.70

method	result	size
derivativedivides	\(x \arcsin \left (\sqrt {x}\right )+\frac {\sqrt {1-x}\, \sqrt {x}}{2}-\frac {\arcsin \left (\sqrt {x}\right )}{2}\)	\(26\)
default	\(x \arcsin \left (\sqrt {x}\right )+\frac {\sqrt {1-x}\, \sqrt {x}}{2}-\frac {\arcsin \left (\sqrt {x}\right )}{2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arcsin(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 = 0.48, size = 25, normalized size = 0.68 \begin {gather*} x \arcsin \left (\sqrt {x}\right ) + \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} - \frac {1}{2} \, \arcsin \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.90, size = 24, normalized size = 0.65 \begin {gather*} \frac {1}{2} \, {\left (2 \, x - 1\right )} \arcsin \left (\sqrt {x}\right ) + \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 29, normalized size = 0.78 \begin {gather*} \frac {\sqrt {x} \sqrt {1 - x}}{2} + x \operatorname {asin}{\left (\sqrt {x} \right )} - \frac {\operatorname {asin}{\left (\sqrt {x} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 27, normalized size = 0.73 \begin {gather*} {\left (x - 1\right )} \arcsin \left (\sqrt {x}\right ) + \frac {1}{2} \, \sqrt {x} \sqrt {-x + 1} + \frac {1}{2} \, \arcsin \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.74, size = 37, normalized size = 1.00 \begin {gather*} x\,\mathrm {asin}\left (\sqrt {x}\right )-\mathrm {atan}\left (\frac {\sqrt {x}}{\sqrt {1-x}-1}\right )+\frac {\sqrt {x}\,\sqrt {1-x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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