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

Optimal. Leaf size=10 \[ \frac {1}{3} \sin ^{-1}\left (\frac {3 x}{2}\right ) \]

[Out]

1/3*arcsin(3/2*x)

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {41, 216} \[ \frac {1}{3} \sin ^{-1}\left (\frac {3 x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[(3*x)/2]/3

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 {2-3 x} \sqrt {2+3 x}} \, dx &=\int \frac {1}{\sqrt {4-9 x^2}} \, dx\\ &=\frac {1}{3} \sin ^{-1}\left (\frac {3 x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ \frac {1}{3} \sin ^{-1}\left (\frac {3 x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[(3*x)/2]/3

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fricas [B]  time = 0.45, size = 25, normalized size = 2.50 \[ -\frac {2}{3} \, \arctan \left (\frac {\sqrt {3 \, x + 2} \sqrt {-3 \, x + 2} - 2}{3 \, x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*arctan(1/3*(sqrt(3*x + 2)*sqrt(-3*x + 2) - 2)/x)

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giac [A]  time = 0.37, size = 12, normalized size = 1.20 \[ \frac {2}{3} \, \arcsin \left (\frac {1}{2} \, \sqrt {3 \, x + 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*arcsin(1/2*sqrt(3*x + 2))

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maple [B]  time = 0.01, size = 34, normalized size = 3.40 \[ \frac {\sqrt {\left (-3 x +2\right ) \left (3 x +2\right )}\, \arcsin \left (\frac {3 x}{2}\right )}{3 \sqrt {-3 x +2}\, \sqrt {3 x +2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2-3*x)^(1/2)/(3*x+2)^(1/2),x)

[Out]

1/3*((2-3*x)*(3*x+2))^(1/2)/(2-3*x)^(1/2)/(3*x+2)^(1/2)*arcsin(3/2*x)

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maxima [A]  time = 1.93, size = 6, normalized size = 0.60 \[ \frac {1}{3} \, \arcsin \left (\frac {3}{2} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arcsin(3/2*x)

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mupad [B]  time = 0.15, size = 32, normalized size = 3.20 \[ -\frac {4\,\mathrm {atan}\left (\frac {\sqrt {2}-\sqrt {2-3\,x}}{\sqrt {2}-\sqrt {3\,x+2}}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2 - 3*x)^(1/2)*(3*x + 2)^(1/2)),x)

[Out]

-(4*atan((2^(1/2) - (2 - 3*x)^(1/2))/(2^(1/2) - (3*x + 2)^(1/2))))/3

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sympy [B]  time = 1.04, size = 51, normalized size = 5.10 \[ \begin {cases} - \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {3} \sqrt {x + \frac {2}{3}}}{2} \right )}}{3} & \text {for}\: \frac {3 \left |{x + \frac {2}{3}}\right |}{4} > 1 \\\frac {2 \operatorname {asin}{\left (\frac {\sqrt {3} \sqrt {x + \frac {2}{3}}}{2} \right )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*I*acosh(sqrt(3)*sqrt(x + 2/3)/2)/3, 3*Abs(x + 2/3)/4 > 1), (2*asin(sqrt(3)*sqrt(x + 2/3)/2)/3, T
rue))

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