3.2306 \(\int (a+b \sqrt [3]{x})^3 x \, dx\)

Optimal. Leaf size=47 \[ \frac{9}{7} a^2 b x^{7/3}+\frac{a^3 x^2}{2}+\frac{9}{8} a b^2 x^{8/3}+\frac{b^3 x^3}{3} \]

[Out]

(a^3*x^2)/2 + (9*a^2*b*x^(7/3))/7 + (9*a*b^2*x^(8/3))/8 + (b^3*x^3)/3

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Rubi [A]  time = 0.0255488, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{9}{7} a^2 b x^{7/3}+\frac{a^3 x^2}{2}+\frac{9}{8} a b^2 x^{8/3}+\frac{b^3 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3*x,x]

[Out]

(a^3*x^2)/2 + (9*a^2*b*x^(7/3))/7 + (9*a*b^2*x^(8/3))/8 + (b^3*x^3)/3

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 \, dx &=3 \operatorname{Subst}\left (\int x^5 (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^3 x^5+3 a^2 b x^6+3 a b^2 x^7+b^3 x^8\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^3 x^2}{2}+\frac{9}{7} a^2 b x^{7/3}+\frac{9}{8} a b^2 x^{8/3}+\frac{b^3 x^3}{3}\\ \end{align*}

Mathematica [A]  time = 0.0175139, size = 47, normalized size = 1. \[ \frac{9}{7} a^2 b x^{7/3}+\frac{a^3 x^2}{2}+\frac{9}{8} a b^2 x^{8/3}+\frac{b^3 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3*x,x]

[Out]

(a^3*x^2)/2 + (9*a^2*b*x^(7/3))/7 + (9*a*b^2*x^(8/3))/8 + (b^3*x^3)/3

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Maple [A]  time = 0.003, size = 36, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}{a}^{3}}{2}}+{\frac{9\,b{a}^{2}}{7}{x}^{{\frac{7}{3}}}}+{\frac{9\,{b}^{2}a}{8}{x}^{{\frac{8}{3}}}}+{\frac{{b}^{3}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*a^3+9/7*a^2*b*x^(7/3)+9/8*a*b^2*x^(8/3)+1/3*b^3*x^3

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Maxima [B]  time = 1.00595, size = 132, normalized size = 2.81 \begin{align*} \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{9}}{3 \, b^{6}} - \frac{15 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a}{8 \, b^{6}} + \frac{30 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{2}}{7 \, b^{6}} - \frac{5 \,{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{3}}{b^{6}} + \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{5} a^{4}}{b^{6}} - \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{4} a^{5}}{4 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^(1/3) + a)^9/b^6 - 15/8*(b*x^(1/3) + a)^8*a/b^6 + 30/7*(b*x^(1/3) + a)^7*a^2/b^6 - 5*(b*x^(1/3) + a)^
6*a^3/b^6 + 3*(b*x^(1/3) + a)^5*a^4/b^6 - 3/4*(b*x^(1/3) + a)^4*a^5/b^6

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Fricas [A]  time = 1.44549, size = 90, normalized size = 1.91 \begin{align*} \frac{1}{3} \, b^{3} x^{3} + \frac{9}{8} \, a b^{2} x^{\frac{8}{3}} + \frac{9}{7} \, a^{2} b x^{\frac{7}{3}} + \frac{1}{2} \, a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^3*x^3 + 9/8*a*b^2*x^(8/3) + 9/7*a^2*b*x^(7/3) + 1/2*a^3*x^2

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Sympy [A]  time = 1.81274, size = 42, normalized size = 0.89 \begin{align*} \frac{a^{3} x^{2}}{2} + \frac{9 a^{2} b x^{\frac{7}{3}}}{7} + \frac{9 a b^{2} x^{\frac{8}{3}}}{8} + \frac{b^{3} x^{3}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x**2/2 + 9*a**2*b*x**(7/3)/7 + 9*a*b**2*x**(8/3)/8 + b**3*x**3/3

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Giac [A]  time = 1.10261, size = 47, normalized size = 1. \begin{align*} \frac{1}{3} \, b^{3} x^{3} + \frac{9}{8} \, a b^{2} x^{\frac{8}{3}} + \frac{9}{7} \, a^{2} b x^{\frac{7}{3}} + \frac{1}{2} \, a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^3*x^3 + 9/8*a*b^2*x^(8/3) + 9/7*a^2*b*x^(7/3) + 1/2*a^3*x^2