3.1536 \(\int \frac{1}{\sqrt{3-b x} \sqrt{2+b x}} \, dx\)

Optimal. Leaf size=16 \[ -\frac{\sin ^{-1}\left (\frac{1}{5} (1-2 b x)\right )}{b} \]

[Out]

-(ArcSin[(1 - 2*b*x)/5]/b)

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Rubi [A]  time = 0.0137068, 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} \[ -\frac{\sin ^{-1}\left (\frac{1}{5} (1-2 b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[3 - b*x]*Sqrt[2 + b*x]),x]

[Out]

-(ArcSin[(1 - 2*b*x)/5]/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{3-b x} \sqrt{2+b x}} \, dx &=\int \frac{1}{\sqrt{6+b x-b^2 x^2}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{25 b^2}}} \, dx,x,b-2 b^2 x\right )}{5 b^2}\\ &=-\frac{\sin ^{-1}\left (\frac{1}{5} (1-2 b x)\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.01148, size = 22, normalized size = 1.38 \[ -\frac{2 \sin ^{-1}\left (\frac{\sqrt{3-b x}}{\sqrt{5}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcSin[Sqrt[3 - b*x]/Sqrt[5]])/b

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Maple [B]  time = 0.008, size = 65, normalized size = 4.1 \begin{align*}{\sqrt{ \left ( -bx+3 \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+6}}}} \right ){\frac{1}{\sqrt{-bx+3}}}{\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+3)^(1/2)/(b*x+2)^(1/2),x)

[Out]

((-b*x+3)*(b*x+2))^(1/2)/(-b*x+3)^(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+6
)^(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+3)^(1/2)/(b*x+2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01378, 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 + 3}}{2 \,{\left (b^{2} x^{2} - b x - 6\right )}}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x+3)^(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 + 3)/(b^2*x^2 - b*x - 6))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- b x + 3} \sqrt{b x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-b*x + 3)*sqrt(b*x + 2)), x)

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Giac [A]  time = 1.07897, size = 24, normalized size = 1.5 \begin{align*} \frac{2 \, \arcsin \left (\frac{1}{5} \, \sqrt{5} \sqrt{b x + 2}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arcsin(1/5*sqrt(5)*sqrt(b*x + 2))/b