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

Optimal. Leaf size=31 \[ \frac {1}{80} \left (4 x^2-9\right )^{5/2}+\frac {3}{16} \left (4 x^2-9\right )^{3/2} \]

[Out]

3/16*(4*x^2-9)^(3/2)+1/80*(4*x^2-9)^(5/2)

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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} \[ \frac {1}{80} \left (4 x^2-9\right )^{5/2}+\frac {3}{16} \left (4 x^2-9\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[-9 + 4*x^2],x]

[Out]

(3*(-9 + 4*x^2)^(3/2))/16 + (-9 + 4*x^2)^(5/2)/80

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 x^3 \sqrt {-9+4 x^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int 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} (-9+4 x)^{3/2}\right ) \, dx,x,x^2\right )\\ &=\frac {3}{16} \left (-9+4 x^2\right )^{3/2}+\frac {1}{80} \left (-9+4 x^2\right )^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.71 \[ \frac {1}{40} \left (2 x^2+3\right ) \left (4 x^2-9\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[-9 + 4*x^2],x]

[Out]

((3 + 2*x^2)*(-9 + 4*x^2)^(3/2))/40

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fricas [A]  time = 0.88, size = 23, normalized size = 0.74 \[ \frac {1}{40} \, {\left (8 \, x^{4} - 6 \, x^{2} - 27\right )} \sqrt {4 \, x^{2} - 9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*x^4 - 6*x^2 - 27)*sqrt(4*x^2 - 9)

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giac [A]  time = 1.07, size = 23, normalized size = 0.74 \[ \frac {1}{80} \, {\left (4 \, x^{2} - 9\right )}^{\frac {5}{2}} + \frac {3}{16} \, {\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(4*x^2 - 9)^(5/2) + 3/16*(4*x^2 - 9)^(3/2)

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maple [A]  time = 0.00, size = 29, normalized size = 0.94 \[ \frac {\left (2 x -3\right ) \left (2 x +3\right ) \left (2 x^{2}+3\right ) \sqrt {4 x^{2}-9}}{40} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(2*x-3)*(2*x+3)*(2*x^2+3)*(4*x^2-9)^(1/2)

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maxima [A]  time = 2.91, size = 26, normalized size = 0.84 \[ \frac {1}{20} \, {\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}} x^{2} + \frac {3}{40} \, {\left (4 \, x^{2} - 9\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*x^2 - 9)^(3/2)*x^2 + 3/40*(4*x^2 - 9)^(3/2)

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mupad [B]  time = 5.51, size = 23, normalized size = 0.74 \[ -\sqrt {4\,x^2-9}\,\left (-\frac {x^4}{5}+\frac {3\,x^2}{20}+\frac {27}{40}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(4*x^2 - 9)^(1/2),x)

[Out]

-(4*x^2 - 9)^(1/2)*((3*x^2)/20 - x^4/5 + 27/40)

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sympy [A]  time = 0.66, size = 44, normalized size = 1.42 \[ \frac {x^{4} \sqrt {4 x^{2} - 9}}{5} - \frac {3 x^{2} \sqrt {4 x^{2} - 9}}{20} - \frac {27 \sqrt {4 x^{2} - 9}}{40} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*sqrt(4*x**2 - 9)/5 - 3*x**2*sqrt(4*x**2 - 9)/20 - 27*sqrt(4*x**2 - 9)/40

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