3.546 \(\int \frac{x^2}{\sqrt{9-4 x^2}} \, dx\)

Optimal. Leaf size=27 \[ \frac{9}{16} \sin ^{-1}\left (\frac{2 x}{3}\right )-\frac{1}{8} x \sqrt{9-4 x^2} \]

[Out]

-(x*Sqrt[9 - 4*x^2])/8 + (9*ArcSin[(2*x)/3])/16

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Rubi [A]  time = 0.0050559, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {321, 216} \[ \frac{9}{16} \sin ^{-1}\left (\frac{2 x}{3}\right )-\frac{1}{8} x \sqrt{9-4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[9 - 4*x^2],x]

[Out]

-(x*Sqrt[9 - 4*x^2])/8 + (9*ArcSin[(2*x)/3])/16

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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{x^2}{\sqrt{9-4 x^2}} \, dx &=-\frac{1}{8} x \sqrt{9-4 x^2}+\frac{9}{8} \int \frac{1}{\sqrt{9-4 x^2}} \, dx\\ &=-\frac{1}{8} x \sqrt{9-4 x^2}+\frac{9}{16} \sin ^{-1}\left (\frac{2 x}{3}\right )\\ \end{align*}

Mathematica [A]  time = 0.0052729, size = 27, normalized size = 1. \[ \frac{9}{16} \sin ^{-1}\left (\frac{2 x}{3}\right )-\frac{1}{8} x \sqrt{9-4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[9 - 4*x^2],x]

[Out]

-(x*Sqrt[9 - 4*x^2])/8 + (9*ArcSin[(2*x)/3])/16

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Maple [A]  time = 0.003, size = 20, normalized size = 0.7 \begin{align*}{\frac{9}{16}\arcsin \left ({\frac{2\,x}{3}} \right ) }-{\frac{x}{8}\sqrt{-4\,{x}^{2}+9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-4*x^2+9)^(1/2),x)

[Out]

9/16*arcsin(2/3*x)-1/8*x*(-4*x^2+9)^(1/2)

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Maxima [A]  time = 2.57791, size = 26, normalized size = 0.96 \begin{align*} -\frac{1}{8} \, \sqrt{-4 \, x^{2} + 9} x + \frac{9}{16} \, \arcsin \left (\frac{2}{3} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+9)^(1/2),x, algorithm="maxima")

[Out]

-1/8*sqrt(-4*x^2 + 9)*x + 9/16*arcsin(2/3*x)

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Fricas [A]  time = 1.2887, size = 92, normalized size = 3.41 \begin{align*} -\frac{1}{8} \, \sqrt{-4 \, x^{2} + 9} x - \frac{9}{8} \, \arctan \left (\frac{\sqrt{-4 \, x^{2} + 9} - 3}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+9)^(1/2),x, algorithm="fricas")

[Out]

-1/8*sqrt(-4*x^2 + 9)*x - 9/8*arctan(1/2*(sqrt(-4*x^2 + 9) - 3)/x)

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Sympy [A]  time = 0.202437, size = 22, normalized size = 0.81 \begin{align*} - \frac{x \sqrt{9 - 4 x^{2}}}{8} + \frac{9 \operatorname{asin}{\left (\frac{2 x}{3} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-4*x**2+9)**(1/2),x)

[Out]

-x*sqrt(9 - 4*x**2)/8 + 9*asin(2*x/3)/16

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Giac [A]  time = 2.96208, size = 26, normalized size = 0.96 \begin{align*} -\frac{1}{8} \, \sqrt{-4 \, x^{2} + 9} x + \frac{9}{16} \, \arcsin \left (\frac{2}{3} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+9)^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(-4*x^2 + 9)*x + 9/16*arcsin(2/3*x)