3.25.14 \(\int (a+\frac {b}{\sqrt [3]{x}})^3 x^2 \, dx\) [2414]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {269, 272, 45} \begin {gather*} \frac {a^3 x^3}{3}+\frac {9}{8} a^2 b x^{8/3}+\frac {9}{7} a b^2 x^{7/3}+\frac {b^3 x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^3*x^2,x]

[Out]

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

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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.91 \begin {gather*} \frac {1}{168} \left (84 b^3 x^2+216 a b^2 x^{7/3}+189 a^2 b x^{8/3}+56 a^3 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^3*x^2,x]

[Out]

(84*b^3*x^2 + 216*a*b^2*x^(7/3) + 189*a^2*b*x^(8/3) + 56*a^3*x^3)/168

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Maple [A]
time = 0.19, size = 36, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {b^{3} x^{2}}{2}+\frac {9 a \,b^{2} x^{\frac {7}{3}}}{7}+\frac {9 a^{2} b \,x^{\frac {8}{3}}}{8}+\frac {a^{3} x^{3}}{3}\)	\(36\)
default	\(\frac {b^{3} x^{2}}{2}+\frac {9 a \,b^{2} x^{\frac {7}{3}}}{7}+\frac {9 a^{2} b \,x^{\frac {8}{3}}}{8}+\frac {a^{3} x^{3}}{3}\)	\(36\)
trager	\(\frac {\left (2 a^{3} x^{2}+2 a^{3} x +3 b^{3} x +2 a^{3}+3 b^{3}\right ) \left (x -1\right )}{6}+\frac {9 a \,b^{2} x^{\frac {7}{3}}}{7}+\frac {9 a^{2} b \,x^{\frac {8}{3}}}{8}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 37, normalized size = 0.79 \begin {gather*} \frac {1}{168} \, {\left (56 \, a^{3} + \frac {189 \, a^{2} b}{x^{\frac {1}{3}}} + \frac {216 \, a b^{2}}{x^{\frac {2}{3}}} + \frac {84 \, b^{3}}{x}\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(56*a^3 + 189*a^2*b/x^(1/3) + 216*a*b^2/x^(2/3) + 84*b^3/x)*x^3

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Fricas [A]
time = 0.37, size = 35, normalized size = 0.74 \begin {gather*} \frac {1}{3} \, a^{3} x^{3} + \frac {9}{8} \, a^{2} b x^{\frac {8}{3}} + \frac {9}{7} \, a b^{2} x^{\frac {7}{3}} + \frac {1}{2} \, b^{3} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.21, size = 42, normalized size = 0.89 \begin {gather*} \frac {a^{3} x^{3}}{3} + \frac {9 a^{2} b x^{\frac {8}{3}}}{8} + \frac {9 a b^{2} x^{\frac {7}{3}}}{7} + \frac {b^{3} x^{2}}{2} \end {gather*}

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

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Giac [A]
time = 1.13, size = 35, normalized size = 0.74 \begin {gather*} \frac {1}{3} \, a^{3} x^{3} + \frac {9}{8} \, a^{2} b x^{\frac {8}{3}} + \frac {9}{7} \, a b^{2} x^{\frac {7}{3}} + \frac {1}{2} \, b^{3} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 35, normalized size = 0.74 \begin {gather*} \frac {a^3\,x^3}{3}+\frac {b^3\,x^2}{2}+\frac {9\,a\,b^2\,x^{7/3}}{7}+\frac {9\,a^2\,b\,x^{8/3}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/x^(1/3))^3,x)

[Out]

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

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