3.1556 \(\int \frac {1}{\sqrt {3-2 x} \sqrt {x}} \, dx\)

Optimal. Leaf size=20 \[ \sqrt {2} \sin ^{-1}\left (\sqrt {\frac {2}{3}} \sqrt {x}\right ) \]

[Out]

arcsin(1/3*6^(1/2)*x^(1/2))*2^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {54, 216} \[ \sqrt {2} \sin ^{-1}\left (\sqrt {\frac {2}{3}} \sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[3 - 2*x]*Sqrt[x]),x]

[Out]

Sqrt[2]*ArcSin[Sqrt[2/3]*Sqrt[x]]

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {3-2 x} \sqrt {x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-2 x^2}} \, dx,x,\sqrt {x}\right )\\ &=\sqrt {2} \sin ^{-1}\left (\sqrt {\frac {2}{3}} \sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ \sqrt {2} \sin ^{-1}\left (\sqrt {\frac {2}{3}} \sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[3 - 2*x]*Sqrt[x]),x]

[Out]

Sqrt[2]*ArcSin[Sqrt[2/3]*Sqrt[x]]

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fricas [A]  time = 0.43, size = 21, normalized size = 1.05 \[ -\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x + 3}}{2 \, \sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(1/2)/x^(1/2),x, algorithm="fricas")

[Out]

-sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-2*x + 3)/sqrt(x))

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giac [A]  time = 1.09, size = 13, normalized size = 0.65 \[ \sqrt {2} \arcsin \left (\frac {1}{3} \, \sqrt {6} \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(1/2)/x^(1/2),x, algorithm="giac")

[Out]

sqrt(2)*arcsin(1/3*sqrt(6)*sqrt(x))

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maple [B]  time = 0.01, size = 31, normalized size = 1.55 \[ \frac {\sqrt {\left (-2 x +3\right ) x}\, \sqrt {2}\, \arcsin \left (\frac {4 x}{3}-1\right )}{2 \sqrt {-2 x +3}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-2*x+3)^(1/2)/x^(1/2),x)

[Out]

1/2*((-2*x+3)*x)^(1/2)/(-2*x+3)^(1/2)/x^(1/2)*2^(1/2)*arcsin(4/3*x-1)

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maxima [A]  time = 3.12, size = 21, normalized size = 1.05 \[ -\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x + 3}}{2 \, \sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)^(1/2)/x^(1/2),x, algorithm="maxima")

[Out]

-sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-2*x + 3)/sqrt(x))

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mupad [B]  time = 0.30, size = 27, normalized size = 1.35 \[ 2\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (\sqrt {3}-\sqrt {3-2\,x}\right )}{2\,\sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(3 - 2*x)^(1/2)),x)

[Out]

2*2^(1/2)*atan((2^(1/2)*(3^(1/2) - (3 - 2*x)^(1/2)))/(2*x^(1/2)))

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sympy [A]  time = 1.00, size = 44, normalized size = 2.20 \[ \begin {cases} - \sqrt {2} i \operatorname {acosh}{\left (\frac {\sqrt {6} \sqrt {x}}{3} \right )} & \text {for}\: \frac {2 \left |{x}\right |}{3} > 1 \\\sqrt {2} \operatorname {asin}{\left (\frac {\sqrt {6} \sqrt {x}}{3} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-2*x)**(1/2)/x**(1/2),x)

[Out]

Piecewise((-sqrt(2)*I*acosh(sqrt(6)*sqrt(x)/3), 2*Abs(x)/3 > 1), (sqrt(2)*asin(sqrt(6)*sqrt(x)/3), True))

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