3.24.5 \(\int (a+b \sqrt [3]{x})^3 x^2 \, dx\) [2305]

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

[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

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Rubi [A]
time = 0.02, antiderivative size = 47, 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 = {272, 45} \begin {gather*} \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 {gather*}

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 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*} \int \left (a+b \sqrt [3]{x}\right )^3 x^2 \, dx &=3 \text {Subst}\left (\int x^8 (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {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*}

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Mathematica [A]
time = 0.02, size = 41, normalized size = 0.87 \begin {gather*} \frac {1}{660} x^3 \left (220 a^3+594 a^2 b \sqrt [3]{x}+540 a b^2 x^{2/3}+165 b^3 x\right ) \end {gather*}

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

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Maple [A]
time = 0.18, size = 36, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {a^{3} x^{3}}{3}+\frac {9 a^{2} b \,x^{\frac {10}{3}}}{10}+\frac {9 a \,b^{2} x^{\frac {11}{3}}}{11}+\frac {b^{3} x^{4}}{4}\)	\(36\)
default	\(\frac {a^{3} x^{3}}{3}+\frac {9 a^{2} b \,x^{\frac {10}{3}}}{10}+\frac {9 a \,b^{2} x^{\frac {11}{3}}}{11}+\frac {b^{3} x^{4}}{4}\)	\(36\)
trager	\(\frac {\left (3 b^{3} x^{3}+4 a^{3} x^{2}+3 b^{3} x^{2}+4 a^{3} x +3 b^{3} x +4 a^{3}+3 b^{3}\right ) \left (x -1\right )}{12}+\frac {9 a^{2} b \,x^{\frac {10}{3}}}{10}+\frac {9 a \,b^{2} x^{\frac {11}{3}}}{11}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (35) = 70\).
time = 0.30, size = 149, normalized size = 3.17 \begin {gather*} \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 {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]

1/4*(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

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Fricas [A]
time = 0.35, size = 35, normalized size = 0.74 \begin {gather*} \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 {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]

1/4*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

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Sympy [A]
time = 0.87, size = 42, normalized size = 0.89 \begin {gather*} \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 {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**(10/3)/10 + 9*a*b**2*x**(11/3)/11 + b**3*x**4/4

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Giac [A]
time = 1.81, size = 35, normalized size = 0.74 \begin {gather*} \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 {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]

1/4*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

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Mupad [B]
time = 0.04, size = 35, normalized size = 0.74 \begin {gather*} \frac {a^3\,x^3}{3}+\frac {b^3\,x^4}{4}+\frac {9\,a^2\,b\,x^{10/3}}{10}+\frac {9\,a\,b^2\,x^{11/3}}{11} \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^4)/4 + (9*a^2*b*x^(10/3))/10 + (9*a*b^2*x^(11/3))/11

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