3.1537 \(\int \frac{1}{\sqrt{2-b x} \sqrt{2+b x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0033367, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {41, 216} \[ \frac{\sin ^{-1}\left (\frac{b x}{2}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[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{2-b x} \sqrt{2+b x}} \, dx &=\int \frac{1}{\sqrt{4-b^2 x^2}} \, dx\\ &=\frac{\sin ^{-1}\left (\frac{b x}{2}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0076968, size = 11, normalized size = 1. \[ \frac{\sin ^{-1}\left (\frac{b x}{2}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 56, normalized size = 5.1 \begin{align*}{\sqrt{ \left ( -bx+2 \right ) \left ( bx+2 \right ) }\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}+4}}}} \right ){\frac{1}{\sqrt{-bx+2}}}{\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+2)^(1/2)/(b*x+2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.429, size = 24, normalized size = 2.18 \begin{align*} \frac{\arcsin \left (\frac{b^{2} x}{2 \, \sqrt{b^{2}}}\right )}{\sqrt{b^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(1/2*b^2*x/sqrt(b^2))/sqrt(b^2)

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Fricas [B]  time = 2.02964, size = 74, normalized size = 6.73 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{b x + 2} \sqrt{-b x + 2} - 2}{b x}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 3.18199, size = 76, normalized size = 6.91 \begin{align*} - \frac{i{G_{6, 6}^{6, 2}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} & \frac{1}{2}, \frac{1}{2}, 1, 1 \\0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 0 & \end{matrix} \middle |{\frac{4}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} b} + \frac{{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, 1 & \\- \frac{1}{4}, \frac{1}{4} & - \frac{1}{2}, 0, 0, 0 \end{matrix} \middle |{\frac{4 e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), 4/(b**2*x**2))/(4*pi**(3/2)*b) + me
ijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), 4*exp_polar(-2*I*pi)/(b**2*x**2))/(4
*pi**(3/2)*b)

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Giac [A]  time = 1.07294, size = 20, normalized size = 1.82 \begin{align*} \frac{2 \, \arcsin \left (\frac{1}{2} \, \sqrt{b x + 2}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arcsin(1/2*sqrt(b*x + 2))/b