3.2294 \(\int (a+b \sqrt [3]{x})^2 x^4 \, dx\)

Optimal. Leaf size=34 \[ \frac{a^2 x^5}{5}+\frac{3}{8} a b x^{16/3}+\frac{3}{17} b^2 x^{17/3} \]

[Out]

(a^2*x^5)/5 + (3*a*b*x^(16/3))/8 + (3*b^2*x^(17/3))/17

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Rubi [A]  time = 0.0301086, antiderivative size = 34, 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} \[ \frac{a^2 x^5}{5}+\frac{3}{8} a b x^{16/3}+\frac{3}{17} b^2 x^{17/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^2*x^4,x]

[Out]

(a^2*x^5)/5 + (3*a*b*x^(16/3))/8 + (3*b^2*x^(17/3))/17

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 )^2 x^4 \, dx &=3 \operatorname{Subst}\left (\int x^{14} (a+b x)^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^2 x^{14}+2 a b x^{15}+b^2 x^{16}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^2 x^5}{5}+\frac{3}{8} a b x^{16/3}+\frac{3}{17} b^2 x^{17/3}\\ \end{align*}

Mathematica [A]  time = 0.0245209, size = 34, normalized size = 1. \[ \frac{a^2 x^5}{5}+\frac{3}{8} a b x^{16/3}+\frac{3}{17} b^2 x^{17/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^2*x^4,x]

[Out]

(a^2*x^5)/5 + (3*a*b*x^(16/3))/8 + (3*b^2*x^(17/3))/17

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Maple [A]  time = 0.002, size = 25, normalized size = 0.7 \begin{align*}{\frac{{x}^{5}{a}^{2}}{5}}+{\frac{3\,ab}{8}{x}^{{\frac{16}{3}}}}+{\frac{3\,{b}^{2}}{17}{x}^{{\frac{17}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^2*x^4,x)

[Out]

1/5*x^5*a^2+3/8*a*b*x^(16/3)+3/17*b^2*x^(17/3)

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Maxima [B]  time = 0.957662, size = 338, normalized size = 9.94 \begin{align*} \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{17}}{17 \, b^{15}} - \frac{21 \,{\left (b x^{\frac{1}{3}} + a\right )}^{16} a}{8 \, b^{15}} + \frac{91 \,{\left (b x^{\frac{1}{3}} + a\right )}^{15} a^{2}}{5 \, b^{15}} - \frac{78 \,{\left (b x^{\frac{1}{3}} + a\right )}^{14} a^{3}}{b^{15}} + \frac{231 \,{\left (b x^{\frac{1}{3}} + a\right )}^{13} a^{4}}{b^{15}} - \frac{1001 \,{\left (b x^{\frac{1}{3}} + a\right )}^{12} a^{5}}{2 \, b^{15}} + \frac{819 \,{\left (b x^{\frac{1}{3}} + a\right )}^{11} a^{6}}{b^{15}} - \frac{5148 \,{\left (b x^{\frac{1}{3}} + a\right )}^{10} a^{7}}{5 \, b^{15}} + \frac{1001 \,{\left (b x^{\frac{1}{3}} + a\right )}^{9} a^{8}}{b^{15}} - \frac{3003 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a^{9}}{4 \, b^{15}} + \frac{429 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{10}}{b^{15}} - \frac{182 \,{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{11}}{b^{15}} + \frac{273 \,{\left (b x^{\frac{1}{3}} + a\right )}^{5} a^{12}}{5 \, b^{15}} - \frac{21 \,{\left (b x^{\frac{1}{3}} + a\right )}^{4} a^{13}}{2 \, b^{15}} + \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{3} a^{14}}{b^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^2*x^4,x, algorithm="maxima")

[Out]

3/17*(b*x^(1/3) + a)^17/b^15 - 21/8*(b*x^(1/3) + a)^16*a/b^15 + 91/5*(b*x^(1/3) + a)^15*a^2/b^15 - 78*(b*x^(1/
3) + a)^14*a^3/b^15 + 231*(b*x^(1/3) + a)^13*a^4/b^15 - 1001/2*(b*x^(1/3) + a)^12*a^5/b^15 + 819*(b*x^(1/3) +
a)^11*a^6/b^15 - 5148/5*(b*x^(1/3) + a)^10*a^7/b^15 + 1001*(b*x^(1/3) + a)^9*a^8/b^15 - 3003/4*(b*x^(1/3) + a)
^8*a^9/b^15 + 429*(b*x^(1/3) + a)^7*a^10/b^15 - 182*(b*x^(1/3) + a)^6*a^11/b^15 + 273/5*(b*x^(1/3) + a)^5*a^12
/b^15 - 21/2*(b*x^(1/3) + a)^4*a^13/b^15 + (b*x^(1/3) + a)^3*a^14/b^15

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Fricas [A]  time = 1.42351, size = 70, normalized size = 2.06 \begin{align*} \frac{3}{17} \, b^{2} x^{\frac{17}{3}} + \frac{3}{8} \, a b x^{\frac{16}{3}} + \frac{1}{5} \, a^{2} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^2*x^4,x, algorithm="fricas")

[Out]

3/17*b^2*x^(17/3) + 3/8*a*b*x^(16/3) + 1/5*a^2*x^5

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Sympy [A]  time = 2.98734, size = 31, normalized size = 0.91 \begin{align*} \frac{a^{2} x^{5}}{5} + \frac{3 a b x^{\frac{16}{3}}}{8} + \frac{3 b^{2} x^{\frac{17}{3}}}{17} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**2*x**4,x)

[Out]

a**2*x**5/5 + 3*a*b*x**(16/3)/8 + 3*b**2*x**(17/3)/17

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Giac [A]  time = 1.13648, size = 32, normalized size = 0.94 \begin{align*} \frac{3}{17} \, b^{2} x^{\frac{17}{3}} + \frac{3}{8} \, a b x^{\frac{16}{3}} + \frac{1}{5} \, a^{2} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^2*x^4,x, algorithm="giac")

[Out]

3/17*b^2*x^(17/3) + 3/8*a*b*x^(16/3) + 1/5*a^2*x^5