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]

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

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Rubi [A]  time = 0.03, antiderivative size = 34, 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 = {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 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])

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]]

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*}

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.00 \[ \frac {a^2 x^5}{5}+\frac {3}{8} a b x^{16/3}+\frac {3}{17} b^2 x^{17/3} \]

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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fricas [A]  time = 0.68, size = 24, normalized size = 0.71 \[ \frac {3}{17} \, b^{2} x^{\frac {17}{3}} + \frac {3}{8} \, a b x^{\frac {16}{3}} + \frac {1}{5} \, a^{2} x^{5} \]

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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giac [A]  time = 0.20, size = 24, normalized size = 0.71 \[ \frac {3}{17} \, b^{2} x^{\frac {17}{3}} + \frac {3}{8} \, a b x^{\frac {16}{3}} + \frac {1}{5} \, a^{2} x^{5} \]

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

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maple [A]  time = 0.00, size = 25, normalized size = 0.74 \[ \frac {3 b^{2} x^{\frac {17}{3}}}{17}+\frac {3 a b \,x^{\frac {16}{3}}}{8}+\frac {a^{2} x^{5}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.92, size = 250, normalized size = 7.35 \[ \frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{17}}{17 \, b^{15}} - \frac {21 \, {\left (b x^{\frac {1}{3}} + a\right )}^{16} a}{8 \, b^{15}} + \frac {91 \, {\left (b x^{\frac {1}{3}} + a\right )}^{15} a^{2}}{5 \, b^{15}} - \frac {78 \, {\left (b x^{\frac {1}{3}} + a\right )}^{14} a^{3}}{b^{15}} + \frac {231 \, {\left (b x^{\frac {1}{3}} + a\right )}^{13} a^{4}}{b^{15}} - \frac {1001 \, {\left (b x^{\frac {1}{3}} + a\right )}^{12} a^{5}}{2 \, b^{15}} + \frac {819 \, {\left (b x^{\frac {1}{3}} + a\right )}^{11} a^{6}}{b^{15}} - \frac {5148 \, {\left (b x^{\frac {1}{3}} + a\right )}^{10} a^{7}}{5 \, b^{15}} + \frac {1001 \, {\left (b x^{\frac {1}{3}} + a\right )}^{9} a^{8}}{b^{15}} - \frac {3003 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} a^{9}}{4 \, b^{15}} + \frac {429 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a^{10}}{b^{15}} - \frac {182 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{11}}{b^{15}} + \frac {273 \, {\left (b x^{\frac {1}{3}} + a\right )}^{5} a^{12}}{5 \, b^{15}} - \frac {21 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} a^{13}}{2 \, b^{15}} + \frac {{\left (b x^{\frac {1}{3}} + a\right )}^{3} a^{14}}{b^{15}} \]

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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mupad [B]  time = 0.04, size = 24, normalized size = 0.71 \[ \frac {a^2\,x^5}{5}+\frac {3\,b^2\,x^{17/3}}{17}+\frac {3\,a\,b\,x^{16/3}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^(1/3))^2,x)

[Out]

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

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sympy [A]  time = 2.02, size = 31, normalized size = 0.91 \[ \frac {a^{2} x^{5}}{5} + \frac {3 a b x^{\frac {16}{3}}}{8} + \frac {3 b^{2} x^{\frac {17}{3}}}{17} \]

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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