3.7.48 \(\int \frac {1}{\sqrt {1-x} \sqrt {x}} \, dx\)

Optimal. Leaf size=8 \[ -\sin ^{-1}(1-2 x) \]

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Rubi [A]  time = 0.00, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {53, 619, 216} \begin {gather*} -\sin ^{-1}(1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - x]*Sqrt[x]),x]

[Out]

-ArcSin[1 - 2*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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.50 \begin {gather*} -2 \sin ^{-1}\left (\sqrt {1-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - x]*Sqrt[x]),x]

[Out]

-2*ArcSin[Sqrt[1 - x]]

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IntegrateAlgebraic [A]  time = 0.04, size = 8, normalized size = 1.00 \begin {gather*} 2 \sin ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[1 - x]*Sqrt[x]),x]

[Out]

2*ArcSin[Sqrt[x]]

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fricas [B]  time = 1.26, size = 14, normalized size = 1.75 \begin {gather*} -2 \, \arctan \left (\frac {\sqrt {-x + 1}}{\sqrt {x}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan(sqrt(-x + 1)/sqrt(x))

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giac [A]  time = 1.09, size = 6, normalized size = 0.75 \begin {gather*} 2 \, \arcsin \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arcsin(sqrt(x))

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maple [B]  time = 0.00, size = 27, normalized size = 3.38 \begin {gather*} \frac {\sqrt {\left (-x +1\right ) x}\, \arcsin \left (2 x -1\right )}{\sqrt {-x +1}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x+1)^(1/2)/x^(1/2),x)

[Out]

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

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maxima [B]  time = 2.95, size = 14, normalized size = 1.75 \begin {gather*} -2 \, \arctan \left (\frac {\sqrt {-x + 1}}{\sqrt {x}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan(sqrt(-x + 1)/sqrt(x))

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mupad [B]  time = 0.05, size = 16, normalized size = 2.00 \begin {gather*} -4\,\mathrm {atan}\left (\frac {\sqrt {1-x}-1}{\sqrt {x}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(1 - x)^(1/2)),x)

[Out]

-4*atan(((1 - x)^(1/2) - 1)/x^(1/2))

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sympy [A]  time = 0.97, size = 20, normalized size = 2.50 \begin {gather*} \begin {cases} - 2 i \operatorname {acosh}{\left (\sqrt {x} \right )} & \text {for}\: \left |{x}\right | > 1 \\2 \operatorname {asin}{\left (\sqrt {x} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*I*acosh(sqrt(x)), Abs(x) > 1), (2*asin(sqrt(x)), True))

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