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

Optimal. Leaf size=11 \[ \frac {\sin ^{-1}(2 b x+3)}{b} \]

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Rubi [A]  time = 0.01, antiderivative size = 11, 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}(2 b x+3)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 0.01, size = 49, normalized size = 4.45 \begin {gather*} \frac {2 \sqrt {b x+1} \sqrt {b x+2} \sinh ^{-1}\left (\sqrt {b x+1}\right )}{b \sqrt {-((b x+1) (b x+2))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.05, size = 26, normalized size = 2.36 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {-b x-1}}{\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 = 1.09, size = 44, normalized size = 4.00 \begin {gather*} -\frac {\arctan \left (\frac {{\left (2 \, b x + 3\right )} \sqrt {b x + 2} \sqrt {-b x - 1}}{2 \, {\left (b^{2} x^{2} + 3 \, 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 + 3)*sqrt(b*x + 2)*sqrt(-b*x - 1)/(b^2*x^2 + 3*b*x + 2))/b

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giac [A]  time = 1.17, size = 13, normalized size = 1.18 \begin {gather*} \frac {2 \, \arcsin \left (\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(sqrt(b*x + 2))/b

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maple [B]  time = 0.01, size = 66, normalized size = 6.00 \begin {gather*} \frac {\sqrt {\left (-b x -1\right ) \left (b x +2\right )}\, \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {3}{2 b}\right )}{\sqrt {-b^{2} x^{2}-3 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+3/2/b)/(-b^2*x^2-3*b*x
-2)^(1/2))

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maxima [A]  time = 3.01, size = 21, normalized size = 1.91 \begin {gather*} -\frac {\arcsin \left (-\frac {2 \, b^{2} x + 3 \, b}{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(-(2*b^2*x + 3*b)/b)/b

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mupad [B]  time = 0.30, size = 41, normalized size = 3.73 \begin {gather*} \frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {-b\,x-1}-\mathrm {i}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {b^2}}\right )}{\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]

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