3.2303 \(\int (a+b \sqrt [3]{x})^3 x^4 \, dx\)

Optimal. Leaf size=47 \[ \frac{9}{16} a^2 b x^{16/3}+\frac{a^3 x^5}{5}+\frac{9}{17} a b^2 x^{17/3}+\frac{b^3 x^6}{6} \]

[Out]

(a^3*x^5)/5 + (9*a^2*b*x^(16/3))/16 + (9*a*b^2*x^(17/3))/17 + (b^3*x^6)/6

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Rubi [A]  time = 0.0353205, 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} \[ \frac{9}{16} a^2 b x^{16/3}+\frac{a^3 x^5}{5}+\frac{9}{17} a b^2 x^{17/3}+\frac{b^3 x^6}{6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3*x^4,x]

[Out]

(a^3*x^5)/5 + (9*a^2*b*x^(16/3))/16 + (9*a*b^2*x^(17/3))/17 + (b^3*x^6)/6

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^4 \, dx &=3 \operatorname{Subst}\left (\int x^{14} (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^3 x^{14}+3 a^2 b x^{15}+3 a b^2 x^{16}+b^3 x^{17}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^3 x^5}{5}+\frac{9}{16} a^2 b x^{16/3}+\frac{9}{17} a b^2 x^{17/3}+\frac{b^3 x^6}{6}\\ \end{align*}

Mathematica [A]  time = 0.0267636, size = 47, normalized size = 1. \[ \frac{9}{16} a^2 b x^{16/3}+\frac{a^3 x^5}{5}+\frac{9}{17} a b^2 x^{17/3}+\frac{b^3 x^6}{6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3*x^4,x]

[Out]

(a^3*x^5)/5 + (9*a^2*b*x^(16/3))/16 + (9*a*b^2*x^(17/3))/17 + (b^3*x^6)/6

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Maple [A]  time = 0.002, size = 36, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}{x}^{5}}{5}}+{\frac{9\,b{a}^{2}}{16}{x}^{{\frac{16}{3}}}}+{\frac{9\,{b}^{2}a}{17}{x}^{{\frac{17}{3}}}}+{\frac{{b}^{3}{x}^{6}}{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^3*x^4,x)

[Out]

1/5*a^3*x^5+9/16*a^2*b*x^(16/3)+9/17*a*b^2*x^(17/3)+1/6*b^3*x^6

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Maxima [B]  time = 0.980762, size = 339, normalized size = 7.21 \begin{align*} \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{18}}{6 \, b^{15}} - \frac{42 \,{\left (b x^{\frac{1}{3}} + a\right )}^{17} a}{17 \, b^{15}} + \frac{273 \,{\left (b x^{\frac{1}{3}} + a\right )}^{16} a^{2}}{16 \, b^{15}} - \frac{364 \,{\left (b x^{\frac{1}{3}} + a\right )}^{15} a^{3}}{5 \, b^{15}} + \frac{429 \,{\left (b x^{\frac{1}{3}} + a\right )}^{14} a^{4}}{2 \, b^{15}} - \frac{462 \,{\left (b x^{\frac{1}{3}} + a\right )}^{13} a^{5}}{b^{15}} + \frac{3003 \,{\left (b x^{\frac{1}{3}} + a\right )}^{12} a^{6}}{4 \, b^{15}} - \frac{936 \,{\left (b x^{\frac{1}{3}} + a\right )}^{11} a^{7}}{b^{15}} + \frac{9009 \,{\left (b x^{\frac{1}{3}} + a\right )}^{10} a^{8}}{10 \, b^{15}} - \frac{2002 \,{\left (b x^{\frac{1}{3}} + a\right )}^{9} a^{9}}{3 \, b^{15}} + \frac{3003 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a^{10}}{8 \, b^{15}} - \frac{156 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{11}}{b^{15}} + \frac{91 \,{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{12}}{2 \, b^{15}} - \frac{42 \,{\left (b x^{\frac{1}{3}} + a\right )}^{5} a^{13}}{5 \, b^{15}} + \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{4} a^{14}}{4 \, b^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="maxima")

[Out]

1/6*(b*x^(1/3) + a)^18/b^15 - 42/17*(b*x^(1/3) + a)^17*a/b^15 + 273/16*(b*x^(1/3) + a)^16*a^2/b^15 - 364/5*(b*
x^(1/3) + a)^15*a^3/b^15 + 429/2*(b*x^(1/3) + a)^14*a^4/b^15 - 462*(b*x^(1/3) + a)^13*a^5/b^15 + 3003/4*(b*x^(
1/3) + a)^12*a^6/b^15 - 936*(b*x^(1/3) + a)^11*a^7/b^15 + 9009/10*(b*x^(1/3) + a)^10*a^8/b^15 - 2002/3*(b*x^(1
/3) + a)^9*a^9/b^15 + 3003/8*(b*x^(1/3) + a)^8*a^10/b^15 - 156*(b*x^(1/3) + a)^7*a^11/b^15 + 91/2*(b*x^(1/3) +
a)^6*a^12/b^15 - 42/5*(b*x^(1/3) + a)^5*a^13/b^15 + 3/4*(b*x^(1/3) + a)^4*a^14/b^15

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Fricas [A]  time = 1.45194, size = 96, normalized size = 2.04 \begin{align*} \frac{1}{6} \, b^{3} x^{6} + \frac{9}{17} \, a b^{2} x^{\frac{17}{3}} + \frac{9}{16} \, a^{2} b x^{\frac{16}{3}} + \frac{1}{5} \, a^{3} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 9/17*a*b^2*x^(17/3) + 9/16*a^2*b*x^(16/3) + 1/5*a^3*x^5

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Sympy [A]  time = 2.96906, size = 42, normalized size = 0.89 \begin{align*} \frac{a^{3} x^{5}}{5} + \frac{9 a^{2} b x^{\frac{16}{3}}}{16} + \frac{9 a b^{2} x^{\frac{17}{3}}}{17} + \frac{b^{3} x^{6}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**3*x**4,x)

[Out]

a**3*x**5/5 + 9*a**2*b*x**(16/3)/16 + 9*a*b**2*x**(17/3)/17 + b**3*x**6/6

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Giac [A]  time = 1.10383, size = 47, normalized size = 1. \begin{align*} \frac{1}{6} \, b^{3} x^{6} + \frac{9}{17} \, a b^{2} x^{\frac{17}{3}} + \frac{9}{16} \, a^{2} b x^{\frac{16}{3}} + \frac{1}{5} \, a^{3} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="giac")

[Out]

1/6*b^3*x^6 + 9/17*a*b^2*x^(17/3) + 9/16*a^2*b*x^(16/3) + 1/5*a^3*x^5