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

Optimal. Leaf size=63 \[ \frac{1}{6} \sqrt{9-4 x^2} x^5-\frac{3}{32} \sqrt{9-4 x^2} x^3-\frac{81}{256} \sqrt{9-4 x^2} x+\frac{729}{512} \sin ^{-1}\left (\frac{2 x}{3}\right ) \]

[Out]

(-81*x*Sqrt[9 - 4*x^2])/256 - (3*x^3*Sqrt[9 - 4*x^2])/32 + (x^5*Sqrt[9 - 4*x^2])/6 + (729*ArcSin[(2*x)/3])/512

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Rubi [A]  time = 0.0161792, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {279, 321, 216} \[ \frac{1}{6} \sqrt{9-4 x^2} x^5-\frac{3}{32} \sqrt{9-4 x^2} x^3-\frac{81}{256} \sqrt{9-4 x^2} x+\frac{729}{512} \sin ^{-1}\left (\frac{2 x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[9 - 4*x^2],x]

[Out]

(-81*x*Sqrt[9 - 4*x^2])/256 - (3*x^3*Sqrt[9 - 4*x^2])/32 + (x^5*Sqrt[9 - 4*x^2])/6 + (729*ArcSin[(2*x)/3])/512

Rule 279

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

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 x^4 \sqrt{9-4 x^2} \, dx &=\frac{1}{6} x^5 \sqrt{9-4 x^2}+\frac{3}{2} \int \frac{x^4}{\sqrt{9-4 x^2}} \, dx\\ &=-\frac{3}{32} x^3 \sqrt{9-4 x^2}+\frac{1}{6} x^5 \sqrt{9-4 x^2}+\frac{81}{32} \int \frac{x^2}{\sqrt{9-4 x^2}} \, dx\\ &=-\frac{81}{256} x \sqrt{9-4 x^2}-\frac{3}{32} x^3 \sqrt{9-4 x^2}+\frac{1}{6} x^5 \sqrt{9-4 x^2}+\frac{729}{256} \int \frac{1}{\sqrt{9-4 x^2}} \, dx\\ &=-\frac{81}{256} x \sqrt{9-4 x^2}-\frac{3}{32} x^3 \sqrt{9-4 x^2}+\frac{1}{6} x^5 \sqrt{9-4 x^2}+\frac{729}{512} \sin ^{-1}\left (\frac{2 x}{3}\right )\\ \end{align*}

Mathematica [A]  time = 0.0117474, size = 39, normalized size = 0.62 \[ \frac{1}{768} x \sqrt{9-4 x^2} \left (128 x^4-72 x^2-243\right )+\frac{729}{512} \sin ^{-1}\left (\frac{2 x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[9 - 4*x^2],x]

[Out]

(x*Sqrt[9 - 4*x^2]*(-243 - 72*x^2 + 128*x^4))/768 + (729*ArcSin[(2*x)/3])/512

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Maple [A]  time = 0.006, size = 46, normalized size = 0.7 \begin{align*} -{\frac{{x}^{3}}{24} \left ( -4\,{x}^{2}+9 \right ) ^{{\frac{3}{2}}}}-{\frac{9\,x}{128} \left ( -4\,{x}^{2}+9 \right ) ^{{\frac{3}{2}}}}+{\frac{81\,x}{256}\sqrt{-4\,{x}^{2}+9}}+{\frac{729}{512}\arcsin \left ({\frac{2\,x}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*x^3*(-4*x^2+9)^(3/2)-9/128*x*(-4*x^2+9)^(3/2)+81/256*x*(-4*x^2+9)^(1/2)+729/512*arcsin(2/3*x)

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Maxima [A]  time = 4.07798, size = 61, normalized size = 0.97 \begin{align*} -\frac{1}{24} \,{\left (-4 \, x^{2} + 9\right )}^{\frac{3}{2}} x^{3} - \frac{9}{128} \,{\left (-4 \, x^{2} + 9\right )}^{\frac{3}{2}} x + \frac{81}{256} \, \sqrt{-4 \, x^{2} + 9} x + \frac{729}{512} \, \arcsin \left (\frac{2}{3} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(-4*x^2 + 9)^(3/2)*x^3 - 9/128*(-4*x^2 + 9)^(3/2)*x + 81/256*sqrt(-4*x^2 + 9)*x + 729/512*arcsin(2/3*x)

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Fricas [A]  time = 1.54162, size = 132, normalized size = 2.1 \begin{align*} \frac{1}{768} \,{\left (128 \, x^{5} - 72 \, x^{3} - 243 \, x\right )} \sqrt{-4 \, x^{2} + 9} - \frac{729}{256} \, \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^4*(-4*x^2+9)^(1/2),x, algorithm="fricas")

[Out]

1/768*(128*x^5 - 72*x^3 - 243*x)*sqrt(-4*x^2 + 9) - 729/256*arctan(1/2*(sqrt(-4*x^2 + 9) - 3)/x)

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Sympy [A]  time = 4.51964, size = 167, normalized size = 2.65 \begin{align*} \begin{cases} \frac{2 i x^{7}}{3 \sqrt{4 x^{2} - 9}} - \frac{15 i x^{5}}{8 \sqrt{4 x^{2} - 9}} - \frac{27 i x^{3}}{64 \sqrt{4 x^{2} - 9}} + \frac{729 i x}{256 \sqrt{4 x^{2} - 9}} - \frac{729 i \operatorname{acosh}{\left (\frac{2 x}{3} \right )}}{512} & \text{for}\: \frac{4 \left |{x^{2}}\right |}{9} > 1 \\- \frac{2 x^{7}}{3 \sqrt{9 - 4 x^{2}}} + \frac{15 x^{5}}{8 \sqrt{9 - 4 x^{2}}} + \frac{27 x^{3}}{64 \sqrt{9 - 4 x^{2}}} - \frac{729 x}{256 \sqrt{9 - 4 x^{2}}} + \frac{729 \operatorname{asin}{\left (\frac{2 x}{3} \right )}}{512} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*I*x**7/(3*sqrt(4*x**2 - 9)) - 15*I*x**5/(8*sqrt(4*x**2 - 9)) - 27*I*x**3/(64*sqrt(4*x**2 - 9)) +
729*I*x/(256*sqrt(4*x**2 - 9)) - 729*I*acosh(2*x/3)/512, 4*Abs(x**2)/9 > 1), (-2*x**7/(3*sqrt(9 - 4*x**2)) + 1
5*x**5/(8*sqrt(9 - 4*x**2)) + 27*x**3/(64*sqrt(9 - 4*x**2)) - 729*x/(256*sqrt(9 - 4*x**2)) + 729*asin(2*x/3)/5
12, True))

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Giac [A]  time = 2.5577, size = 45, normalized size = 0.71 \begin{align*} \frac{1}{768} \,{\left (8 \,{\left (16 \, x^{2} - 9\right )} x^{2} - 243\right )} \sqrt{-4 \, x^{2} + 9} x + \frac{729}{512} \, \arcsin \left (\frac{2}{3} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(8*(16*x^2 - 9)*x^2 - 243)*sqrt(-4*x^2 + 9)*x + 729/512*arcsin(2/3*x)