3.2305 \(\int (a+b \sqrt [3]{x})^3 x^2 \, dx\)

Optimal. Leaf size=47 \[ \frac{9}{10} a^2 b x^{10/3}+\frac{a^3 x^3}{3}+\frac{9}{11} a b^2 x^{11/3}+\frac{b^3 x^4}{4} \]

[Out]

(a^3*x^3)/3 + (9*a^2*b*x^(10/3))/10 + (9*a*b^2*x^(11/3))/11 + (b^3*x^4)/4

________________________________________________________________________________________

Rubi [A]  time = 0.0300415, antiderivative size = 47, 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} \[ \frac{9}{10} a^2 b x^{10/3}+\frac{a^3 x^3}{3}+\frac{9}{11} a b^2 x^{11/3}+\frac{b^3 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3*x^2,x]

[Out]

(a^3*x^3)/3 + (9*a^2*b*x^(10/3))/10 + (9*a*b^2*x^(11/3))/11 + (b^3*x^4)/4

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 \left (a+b \sqrt [3]{x}\right )^3 x^2 \, dx &=3 \operatorname{Subst}\left (\int x^8 (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^3 x^8+3 a^2 b x^9+3 a b^2 x^{10}+b^3 x^{11}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^3 x^3}{3}+\frac{9}{10} a^2 b x^{10/3}+\frac{9}{11} a b^2 x^{11/3}+\frac{b^3 x^4}{4}\\ \end{align*}

Mathematica [A]  time = 0.021129, size = 41, normalized size = 0.87 \[ \frac{1}{660} x^3 \left (594 a^2 b \sqrt [3]{x}+220 a^3+540 a b^2 x^{2/3}+165 b^3 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3*x^2,x]

[Out]

(x^3*(220*a^3 + 594*a^2*b*x^(1/3) + 540*a*b^2*x^(2/3) + 165*b^3*x))/660

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 36, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}{x}^{3}}{3}}+{\frac{9\,b{a}^{2}}{10}{x}^{{\frac{10}{3}}}}+{\frac{9\,{b}^{2}a}{11}{x}^{{\frac{11}{3}}}}+{\frac{{b}^{3}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^3*x^2,x)

[Out]

1/3*a^3*x^3+9/10*a^2*b*x^(10/3)+9/11*a*b^2*x^(11/3)+1/4*b^3*x^4

________________________________________________________________________________________

Maxima [B]  time = 1.03543, size = 201, normalized size = 4.28 \begin{align*} \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{12}}{4 \, b^{9}} - \frac{24 \,{\left (b x^{\frac{1}{3}} + a\right )}^{11} a}{11 \, b^{9}} + \frac{42 \,{\left (b x^{\frac{1}{3}} + a\right )}^{10} a^{2}}{5 \, b^{9}} - \frac{56 \,{\left (b x^{\frac{1}{3}} + a\right )}^{9} a^{3}}{3 \, b^{9}} + \frac{105 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a^{4}}{4 \, b^{9}} - \frac{24 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{5}}{b^{9}} + \frac{14 \,{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{6}}{b^{9}} - \frac{24 \,{\left (b x^{\frac{1}{3}} + a\right )}^{5} a^{7}}{5 \, b^{9}} + \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{4} a^{8}}{4 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^(1/3) + a)^12/b^9 - 24/11*(b*x^(1/3) + a)^11*a/b^9 + 42/5*(b*x^(1/3) + a)^10*a^2/b^9 - 56/3*(b*x^(1/3
) + a)^9*a^3/b^9 + 105/4*(b*x^(1/3) + a)^8*a^4/b^9 - 24*(b*x^(1/3) + a)^7*a^5/b^9 + 14*(b*x^(1/3) + a)^6*a^6/b
^9 - 24/5*(b*x^(1/3) + a)^5*a^7/b^9 + 3/4*(b*x^(1/3) + a)^4*a^8/b^9

________________________________________________________________________________________

Fricas [A]  time = 1.45876, size = 96, normalized size = 2.04 \begin{align*} \frac{1}{4} \, b^{3} x^{4} + \frac{9}{11} \, a b^{2} x^{\frac{11}{3}} + \frac{9}{10} \, a^{2} b x^{\frac{10}{3}} + \frac{1}{3} \, a^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^3*x^4 + 9/11*a*b^2*x^(11/3) + 9/10*a^2*b*x^(10/3) + 1/3*a^3*x^3

________________________________________________________________________________________

Sympy [A]  time = 2.09301, size = 42, normalized size = 0.89 \begin{align*} \frac{a^{3} x^{3}}{3} + \frac{9 a^{2} b x^{\frac{10}{3}}}{10} + \frac{9 a b^{2} x^{\frac{11}{3}}}{11} + \frac{b^{3} x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**3*x**2,x)

[Out]

a**3*x**3/3 + 9*a**2*b*x**(10/3)/10 + 9*a*b**2*x**(11/3)/11 + b**3*x**4/4

________________________________________________________________________________________

Giac [A]  time = 1.1069, size = 47, normalized size = 1. \begin{align*} \frac{1}{4} \, b^{3} x^{4} + \frac{9}{11} \, a b^{2} x^{\frac{11}{3}} + \frac{9}{10} \, a^{2} b x^{\frac{10}{3}} + \frac{1}{3} \, a^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^3*x^4 + 9/11*a*b^2*x^(11/3) + 9/10*a^2*b*x^(10/3) + 1/3*a^3*x^3