3.1538 \(\int \frac{1}{\sqrt{1-b x} \sqrt{2+b x}} \, dx\)

Optimal. Leaf size=16 \[ -\frac{\sin ^{-1}\left (\frac{1}{3} (-2 b x-1)\right )}{b} \]

[Out]

-(ArcSin[(-1 - 2*b*x)/3]/b)

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Rubi [A]  time = 0.0133358, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {53, 619, 216} \[ -\frac{\sin ^{-1}\left (\frac{1}{3} (-2 b x-1)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcSin[(-1 - 2*b*x)/3]/b)

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 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 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{1-b x} \sqrt{2+b x}} \, dx &=\int \frac{1}{\sqrt{2-b x-b^2 x^2}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{9 b^2}}} \, dx,x,-b-2 b^2 x\right )}{3 b^2}\\ &=-\frac{\sin ^{-1}\left (\frac{1}{3} (-1-2 b x)\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0124177, size = 22, normalized size = 1.38 \[ -\frac{2 \sin ^{-1}\left (\frac{\sqrt{1-b x}}{\sqrt{3}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcSin[Sqrt[1 - b*x]/Sqrt[3]])/b

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Maple [B]  time = 0.007, size = 66, normalized size = 4.1 \begin{align*}{\sqrt{ \left ( -bx+1 \right ) \left ( bx+2 \right ) }\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{1}{2\,b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-bx+2}}}} \right ){\frac{1}{\sqrt{-bx+1}}}{\frac{1}{\sqrt{bx+2}}}{\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01138, size = 104, normalized size = 6.5 \begin{align*} -\frac{\arctan \left (\frac{{\left (2 \, b x + 1\right )} \sqrt{b x + 2} \sqrt{-b x + 1}}{2 \,{\left (b^{2} x^{2} + b x - 2\right )}}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- b x + 1} \sqrt{b x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-b*x + 1)*sqrt(b*x + 2)), x)

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Giac [A]  time = 1.07573, size = 24, normalized size = 1.5 \begin{align*} \frac{2 \, \arcsin \left (\frac{1}{3} \, \sqrt{3} \sqrt{b x + 2}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arcsin(1/3*sqrt(3)*sqrt(b*x + 2))/b