3.679 \(\int x^3 (a+b x^2)^{2/3} \, dx\)

Optimal. Leaf size=38 \[ \frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2}-\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2} \]

[Out]

-3/10*a*(b*x^2+a)^(5/3)/b^2+3/16*(b*x^2+a)^(8/3)/b^2

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Rubi [A]  time = 0.02, antiderivative size = 38, 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} \[ \frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2}-\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(2/3),x]

[Out]

(-3*a*(a + b*x^2)^(5/3))/(10*b^2) + (3*(a + b*x^2)^(8/3))/(16*b^2)

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 \left (a+b x^2\right )^{2/3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x)^{2/3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a (a+b x)^{2/3}}{b}+\frac {(a+b x)^{5/3}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2}+\frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.74 \[ \frac {3 \left (a+b x^2\right )^{5/3} \left (5 b x^2-3 a\right )}{80 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(2/3),x]

[Out]

(3*(a + b*x^2)^(5/3)*(-3*a + 5*b*x^2))/(80*b^2)

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fricas [A]  time = 0.89, size = 35, normalized size = 0.92 \[ \frac {3 \, {\left (5 \, b^{2} x^{4} + 2 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{80 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

3/80*(5*b^2*x^4 + 2*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(2/3)/b^2

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giac [A]  time = 1.08, size = 29, normalized size = 0.76 \[ \frac {3 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}} - 8 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a\right )}}{80 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

3/80*(5*(b*x^2 + a)^(8/3) - 8*(b*x^2 + a)^(5/3)*a)/b^2

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maple [A]  time = 0.01, size = 25, normalized size = 0.66 \[ -\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}} \left (-5 b \,x^{2}+3 a \right )}{80 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(2/3),x)

[Out]

-3/80*(b*x^2+a)^(5/3)*(-5*b*x^2+3*a)/b^2

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maxima [A]  time = 1.35, size = 30, normalized size = 0.79 \[ \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}}}{16 \, b^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a}{10 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

3/16*(b*x^2 + a)^(8/3)/b^2 - 3/10*(b*x^2 + a)^(5/3)*a/b^2

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mupad [B]  time = 4.71, size = 33, normalized size = 0.87 \[ {\left (b\,x^2+a\right )}^{2/3}\,\left (\frac {3\,x^4}{16}-\frac {9\,a^2}{80\,b^2}+\frac {3\,a\,x^2}{40\,b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(2/3),x)

[Out]

(a + b*x^2)^(2/3)*((3*x^4)/16 - (9*a^2)/(80*b^2) + (3*a*x^2)/(40*b))

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sympy [A]  time = 0.75, size = 66, normalized size = 1.74 \[ \begin {cases} - \frac {9 a^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}}{80 b^{2}} + \frac {3 a x^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}}{40 b} + \frac {3 x^{4} \left (a + b x^{2}\right )^{\frac {2}{3}}}{16} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**(2/3),x)

[Out]

Piecewise((-9*a**2*(a + b*x**2)**(2/3)/(80*b**2) + 3*a*x**2*(a + b*x**2)**(2/3)/(40*b) + 3*x**4*(a + b*x**2)**
(2/3)/16, Ne(b, 0)), (a**(2/3)*x**4/4, True))

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