\(\int x^3 \sqrt {4-9 x^2} \, dx\) [140]

Optimal result

Integrand size = 15, antiderivative size = 31 \[ \int x^3 \sqrt {4-9 x^2} \, dx=-\frac {4}{243} \left (4-9 x^2\right )^{3/2}+\frac {1}{405} \left (4-9 x^2\right )^{5/2} \]

[Out]

-4/243*(-9*x^2+4)^(3/2)+1/405*(-9*x^2+4)^(5/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int x^3 \sqrt {4-9 x^2} \, dx=\frac {1}{405} \left (4-9 x^2\right )^{5/2}-\frac {4}{243} \left (4-9 x^2\right )^{3/2} \]

[In]

Int[x^3*Sqrt[4 - 9*x^2],x]

[Out]

(-4*(4 - 9*x^2)^(3/2))/243 + (4 - 9*x^2)^(5/2)/405

Rule 45

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 272

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \sqrt {4-9 x} x \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {4}{9} \sqrt {4-9 x}-\frac {1}{9} (4-9 x)^{3/2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {4}{243} \left (4-9 x^2\right )^{3/2}+\frac {1}{405} \left (4-9 x^2\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int x^3 \sqrt {4-9 x^2} \, dx=\frac {\left (-8-27 x^2\right ) \left (4-9 x^2\right )^{3/2}}{1215} \]

[In]

Integrate[x^3*Sqrt[4 - 9*x^2],x]

[Out]

((-8 - 27*x^2)*(4 - 9*x^2)^(3/2))/1215

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {\left (27 x^{2}+8\right ) \left (-9 x^{2}+4\right )^{\frac {3}{2}}}{1215}\)	\(19\)
trager	\(\left (\frac {1}{5} x^{4}-\frac {4}{135} x^{2}-\frac {32}{1215}\right ) \sqrt {-9 x^{2}+4}\)	\(23\)
default	\(-\frac {x^{2} \left (-9 x^{2}+4\right )^{\frac {3}{2}}}{45}-\frac {8 \left (-9 x^{2}+4\right )^{\frac {3}{2}}}{1215}\)	\(27\)
gosper	\(\frac {\left (-2+3 x \right ) \left (2+3 x \right ) \left (27 x^{2}+8\right ) \sqrt {-9 x^{2}+4}}{1215}\)	\(29\)
risch	\(-\frac {\left (243 x^{4}-36 x^{2}-32\right ) \left (9 x^{2}-4\right )}{1215 \sqrt {-9 x^{2}+4}}\)	\(31\)
meijerg	\(-\frac {8 \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (1-\frac {9 x^{2}}{4}\right )^{\frac {3}{2}} \left (\frac {27 x^{2}}{4}+2\right )}{15}\right )}{81 \sqrt {\pi }}\)	\(33\)

[In]

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

[Out]

-1/1215*(27*x^2+8)*(-9*x^2+4)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int x^3 \sqrt {4-9 x^2} \, dx=\frac {1}{1215} \, {\left (243 \, x^{4} - 36 \, x^{2} - 32\right )} \sqrt {-9 \, x^{2} + 4} \]

[In]

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

[Out]

1/1215*(243*x^4 - 36*x^2 - 32)*sqrt(-9*x^2 + 4)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int x^3 \sqrt {4-9 x^2} \, dx=\frac {x^{4} \sqrt {4 - 9 x^{2}}}{5} - \frac {4 x^{2} \sqrt {4 - 9 x^{2}}}{135} - \frac {32 \sqrt {4 - 9 x^{2}}}{1215} \]

[In]

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

[Out]

x**4*sqrt(4 - 9*x**2)/5 - 4*x**2*sqrt(4 - 9*x**2)/135 - 32*sqrt(4 - 9*x**2)/1215

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int x^3 \sqrt {4-9 x^2} \, dx=-\frac {1}{45} \, {\left (-9 \, x^{2} + 4\right )}^{\frac {3}{2}} x^{2} - \frac {8}{1215} \, {\left (-9 \, x^{2} + 4\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

-1/45*(-9*x^2 + 4)^(3/2)*x^2 - 8/1215*(-9*x^2 + 4)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int x^3 \sqrt {4-9 x^2} \, dx=\frac {1}{405} \, {\left (9 \, x^{2} - 4\right )}^{2} \sqrt {-9 \, x^{2} + 4} - \frac {4}{243} \, {\left (-9 \, x^{2} + 4\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

1/405*(9*x^2 - 4)^2*sqrt(-9*x^2 + 4) - 4/243*(-9*x^2 + 4)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int x^3 \sqrt {4-9 x^2} \, dx=-\frac {\sqrt {\frac {4}{9}-x^2}\,\left (-\frac {9\,x^4}{5}+\frac {4\,x^2}{15}+\frac {32}{135}\right )}{3} \]

[In]

int(x^3*(4 - 9*x^2)^(1/2),x)

[Out]

-((4/9 - x^2)^(1/2)*((4*x^2)/15 - (9*x^4)/5 + 32/135))/3