3.125 \(\int x^3 \sqrt{4-x^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0142777, antiderivative size = 31, normalized size of antiderivative = 1., 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} \[ \frac{1}{5} \left (4-x^2\right )^{5/2}-\frac{4}{3} \left (4-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[4 - x^2],x]

[Out]

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

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0094979, size = 22, normalized size = 0.71 \[ -\frac{1}{15} \left (4-x^2\right )^{3/2} \left (3 x^2+8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[4 - x^2],x]

[Out]

-((4 - x^2)^(3/2)*(8 + 3*x^2))/15

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Maple [A]  time = 0.005, size = 25, normalized size = 0.8 \begin{align*}{\frac{ \left ( -2+x \right ) \left ( 2+x \right ) \left ( 3\,{x}^{2}+8 \right ) }{15}\sqrt{-{x}^{2}+4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(-2+x)*(2+x)*(3*x^2+8)*(-x^2+4)^(1/2)

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Maxima [A]  time = 1.40397, size = 35, normalized size = 1.13 \begin{align*} -\frac{1}{5} \,{\left (-x^{2} + 4\right )}^{\frac{3}{2}} x^{2} - \frac{8}{15} \,{\left (-x^{2} + 4\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(-x^2 + 4)^(3/2)*x^2 - 8/15*(-x^2 + 4)^(3/2)

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Fricas [A]  time = 1.89788, size = 57, normalized size = 1.84 \begin{align*} \frac{1}{15} \,{\left (3 \, x^{4} - 4 \, x^{2} - 32\right )} \sqrt{-x^{2} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*x^4 - 4*x^2 - 32)*sqrt(-x^2 + 4)

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Sympy [A]  time = 0.568048, size = 39, normalized size = 1.26 \begin{align*} \frac{x^{4} \sqrt{4 - x^{2}}}{5} - \frac{4 x^{2} \sqrt{4 - x^{2}}}{15} - \frac{32 \sqrt{4 - x^{2}}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*sqrt(4 - x**2)/5 - 4*x**2*sqrt(4 - x**2)/15 - 32*sqrt(4 - x**2)/15

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Giac [A]  time = 1.05564, size = 41, normalized size = 1.32 \begin{align*} \frac{1}{5} \,{\left (x^{2} - 4\right )}^{2} \sqrt{-x^{2} + 4} - \frac{4}{3} \,{\left (-x^{2} + 4\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^2 - 4)^2*sqrt(-x^2 + 4) - 4/3*(-x^2 + 4)^(3/2)