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

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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {53, 619, 216} \begin {gather*} -\frac {\sin ^{-1}\left (\frac {1}{3} (-2 b x-1)\right )}{b} \end {gather*}

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 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-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*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.38 \begin {gather*} -\frac {2 \sin ^{-1}\left (\frac {\sqrt {1-b x}}{\sqrt {3}}\right )}{b} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.05, size = 26, normalized size = 1.62 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-b x}}{\sqrt {b x+2}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.07, size = 43, normalized size = 2.69 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 1.05, size = 18, normalized size = 1.12 \begin {gather*} \frac {2 \, \arcsin \left (\frac {1}{3} \, \sqrt {3} \sqrt {b x + 2}\right )}{b} \end {gather*}

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

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maple [B]  time = 0.01, size = 66, normalized size = 4.12 \begin {gather*} \frac {\sqrt {\left (-b x +1\right ) \left (b x +2\right )}\, \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1}{2 b}\right )}{\sqrt {-b^{2} x^{2}-b x +2}}\right )}{\sqrt {-b x +1}\, \sqrt {b x +2}\, \sqrt {b^{2}}} \end {gather*}

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 [A]  time = 3.01, size = 19, normalized size = 1.19 \begin {gather*} -\frac {\arcsin \left (-\frac {2 \, b^{2} x + b}{3 \, b}\right )}{b} \end {gather*}

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]

-arcsin(-1/3*(2*b^2*x + b)/b)/b

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mupad [B]  time = 0.32, size = 40, normalized size = 2.50 \begin {gather*} -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {2}-\sqrt {b\,x+2}\right )}{\left (\sqrt {1-b\,x}-1\right )\,\sqrt {b^2}}\right )}{\sqrt {b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*atan((b*(2^(1/2) - (b*x + 2)^(1/2)))/(((1 - b*x)^(1/2) - 1)*(b^2)^(1/2))))/(b^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- b x + 1} \sqrt {b x + 2}}\, dx \end {gather*}

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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