3.545 \(\int \frac {x^3}{\sqrt {9-4 x^2}} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{48} \left (9-4 x^2\right )^{3/2}-\frac {9}{16} \sqrt {9-4 x^2} \]

[Out]

1/48*(-4*x^2+9)^(3/2)-9/16*(-4*x^2+9)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {1}{48} \left (9-4 x^2\right )^{3/2}-\frac {9}{16} \sqrt {9-4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[9 - 4*x^2],x]

[Out]

(-9*Sqrt[9 - 4*x^2])/16 + (9 - 4*x^2)^(3/2)/48

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {9-4 x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\sqrt {9-4 x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {9}{4 \sqrt {9-4 x}}-\frac {1}{4} \sqrt {9-4 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {9}{16} \sqrt {9-4 x^2}+\frac {1}{48} \left (9-4 x^2\right )^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.71 \[ -\frac {1}{24} \sqrt {9-4 x^2} \left (2 x^2+9\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[9 - 4*x^2],x]

[Out]

-1/24*(Sqrt[9 - 4*x^2]*(9 + 2*x^2))

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fricas [A]  time = 0.94, size = 18, normalized size = 0.58 \[ -\frac {1}{24} \, {\left (2 \, x^{2} + 9\right )} \sqrt {-4 \, x^{2} + 9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-4*x^2+9)^(1/2),x, algorithm="fricas")

[Out]

-1/24*(2*x^2 + 9)*sqrt(-4*x^2 + 9)

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giac [A]  time = 0.98, size = 23, normalized size = 0.74 \[ \frac {1}{48} \, {\left (-4 \, x^{2} + 9\right )}^{\frac {3}{2}} - \frac {9}{16} \, \sqrt {-4 \, x^{2} + 9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-4*x^2+9)^(1/2),x, algorithm="giac")

[Out]

1/48*(-4*x^2 + 9)^(3/2) - 9/16*sqrt(-4*x^2 + 9)

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maple [A]  time = 0.01, size = 29, normalized size = 0.94 \[ \frac {\left (2 x -3\right ) \left (2 x +3\right ) \left (2 x^{2}+9\right )}{24 \sqrt {-4 x^{2}+9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(-4*x^2+9)^(1/2),x)

[Out]

1/24*(2*x-3)*(2*x+3)*(2*x^2+9)/(-4*x^2+9)^(1/2)

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maxima [A]  time = 2.90, size = 26, normalized size = 0.84 \[ -\frac {1}{12} \, \sqrt {-4 \, x^{2} + 9} x^{2} - \frac {3}{8} \, \sqrt {-4 \, x^{2} + 9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-4*x^2+9)^(1/2),x, algorithm="maxima")

[Out]

-1/12*sqrt(-4*x^2 + 9)*x^2 - 3/8*sqrt(-4*x^2 + 9)

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mupad [B]  time = 0.02, size = 18, normalized size = 0.58 \[ -\frac {\sqrt {\frac {9}{4}-x^2}\,\left (\frac {x^2}{3}+\frac {3}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(9 - 4*x^2)^(1/2),x)

[Out]

-((9/4 - x^2)^(1/2)*(x^2/3 + 3/2))/2

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sympy [A]  time = 0.38, size = 29, normalized size = 0.94 \[ - \frac {x^{2} \sqrt {9 - 4 x^{2}}}{12} - \frac {3 \sqrt {9 - 4 x^{2}}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(-4*x**2+9)**(1/2),x)

[Out]

-x**2*sqrt(9 - 4*x**2)/12 - 3*sqrt(9 - 4*x**2)/8

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