3.632 \(\int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx\)

Optimal. Leaf size=24 \[ \frac{2 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{\sqrt{b}} \]

[Out]

(2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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Rubi [A]  time = 0.0067891, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {54, 216} \[ \frac{2 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*Sqrt[2 - b*x]),x]

[Out]

(2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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

Mathematica [A]  time = 0.0044428, size = 24, normalized size = 1. \[ \frac{2 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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Maple [B]  time = 0.003, size = 50, normalized size = 2.1 \begin{align*}{\sqrt{ \left ( -bx+2 \right ) x}\arctan \left ({\sqrt{b} \left ( x-{b}^{-1} \right ){\frac{1}{\sqrt{-b{x}^{2}+2\,x}}}} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+2}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)/b^(1/2)*arctan(b^(1/2)*(x-1/b)/(-b*x^2+2*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48754, size = 159, normalized size = 6.62 \begin{align*} \left [-\frac{\sqrt{-b} \log \left (-b x + \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right )}{b}, -\frac{2 \, \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right )}{\sqrt{b}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1)/b, -2*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/sqrt
(b)]

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Sympy [A]  time = 1.22131, size = 58, normalized size = 2.42 \begin{align*} \begin{cases} - \frac{2 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{\sqrt{b}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\\frac{2 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{\sqrt{b}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError