3.2315 \(\int (a+b \sqrt [3]{x})^5 x \, dx\)

Optimal. Leaf size=77 \[ \frac{15}{4} a^3 b^2 x^{8/3}+\frac{10}{3} a^2 b^3 x^3+\frac{15}{7} a^4 b x^{7/3}+\frac{a^5 x^2}{2}+\frac{3}{2} a b^4 x^{10/3}+\frac{3}{11} b^5 x^{11/3} \]

[Out]

(a^5*x^2)/2 + (15*a^4*b*x^(7/3))/7 + (15*a^3*b^2*x^(8/3))/4 + (10*a^2*b^3*x^3)/3 + (3*a*b^4*x^(10/3))/2 + (3*b
^5*x^(11/3))/11

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Rubi [A]  time = 0.0408262, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{15}{4} a^3 b^2 x^{8/3}+\frac{10}{3} a^2 b^3 x^3+\frac{15}{7} a^4 b x^{7/3}+\frac{a^5 x^2}{2}+\frac{3}{2} a b^4 x^{10/3}+\frac{3}{11} b^5 x^{11/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^5*x,x]

[Out]

(a^5*x^2)/2 + (15*a^4*b*x^(7/3))/7 + (15*a^3*b^2*x^(8/3))/4 + (10*a^2*b^3*x^3)/3 + (3*a*b^4*x^(10/3))/2 + (3*b
^5*x^(11/3))/11

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 )^5 x \, dx &=3 \operatorname{Subst}\left (\int x^5 (a+b x)^5 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^5 x^5+5 a^4 b x^6+10 a^3 b^2 x^7+10 a^2 b^3 x^8+5 a b^4 x^9+b^5 x^{10}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^5 x^2}{2}+\frac{15}{7} a^4 b x^{7/3}+\frac{15}{4} a^3 b^2 x^{8/3}+\frac{10}{3} a^2 b^3 x^3+\frac{3}{2} a b^4 x^{10/3}+\frac{3}{11} b^5 x^{11/3}\\ \end{align*}

Mathematica [A]  time = 0.0249687, size = 77, normalized size = 1. \[ \frac{15}{4} a^3 b^2 x^{8/3}+\frac{10}{3} a^2 b^3 x^3+\frac{15}{7} a^4 b x^{7/3}+\frac{a^5 x^2}{2}+\frac{3}{2} a b^4 x^{10/3}+\frac{3}{11} b^5 x^{11/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^5*x,x]

[Out]

(a^5*x^2)/2 + (15*a^4*b*x^(7/3))/7 + (15*a^3*b^2*x^(8/3))/4 + (10*a^2*b^3*x^3)/3 + (3*a*b^4*x^(10/3))/2 + (3*b
^5*x^(11/3))/11

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Maple [A]  time = 0.001, size = 58, normalized size = 0.8 \begin{align*}{\frac{{a}^{5}{x}^{2}}{2}}+{\frac{15\,{a}^{4}b}{7}{x}^{{\frac{7}{3}}}}+{\frac{15\,{a}^{3}{b}^{2}}{4}{x}^{{\frac{8}{3}}}}+{\frac{10\,{a}^{2}{b}^{3}{x}^{3}}{3}}+{\frac{3\,a{b}^{4}}{2}{x}^{{\frac{10}{3}}}}+{\frac{3\,{b}^{5}}{11}{x}^{{\frac{11}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^5*x,x)

[Out]

1/2*a^5*x^2+15/7*a^4*b*x^(7/3)+15/4*a^3*b^2*x^(8/3)+10/3*a^2*b^3*x^3+3/2*a*b^4*x^(10/3)+3/11*b^5*x^(11/3)

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Maxima [A]  time = 0.965591, size = 132, normalized size = 1.71 \begin{align*} \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{11}}{11 \, b^{6}} - \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{10} a}{2 \, b^{6}} + \frac{10 \,{\left (b x^{\frac{1}{3}} + a\right )}^{9} a^{2}}{3 \, b^{6}} - \frac{15 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a^{3}}{4 \, b^{6}} + \frac{15 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{4}}{7 \, b^{6}} - \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{5}}{2 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^5*x,x, algorithm="maxima")

[Out]

3/11*(b*x^(1/3) + a)^11/b^6 - 3/2*(b*x^(1/3) + a)^10*a/b^6 + 10/3*(b*x^(1/3) + a)^9*a^2/b^6 - 15/4*(b*x^(1/3)
+ a)^8*a^3/b^6 + 15/7*(b*x^(1/3) + a)^7*a^4/b^6 - 1/2*(b*x^(1/3) + a)^6*a^5/b^6

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Fricas [A]  time = 1.43596, size = 162, normalized size = 2.1 \begin{align*} \frac{10}{3} \, a^{2} b^{3} x^{3} + \frac{1}{2} \, a^{5} x^{2} + \frac{3}{44} \,{\left (4 \, b^{5} x^{3} + 55 \, a^{3} b^{2} x^{2}\right )} x^{\frac{2}{3}} + \frac{3}{14} \,{\left (7 \, a b^{4} x^{3} + 10 \, a^{4} b x^{2}\right )} x^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^5*x,x, algorithm="fricas")

[Out]

10/3*a^2*b^3*x^3 + 1/2*a^5*x^2 + 3/44*(4*b^5*x^3 + 55*a^3*b^2*x^2)*x^(2/3) + 3/14*(7*a*b^4*x^3 + 10*a^4*b*x^2)
*x^(1/3)

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Sympy [A]  time = 1.76514, size = 75, normalized size = 0.97 \begin{align*} \frac{a^{5} x^{2}}{2} + \frac{15 a^{4} b x^{\frac{7}{3}}}{7} + \frac{15 a^{3} b^{2} x^{\frac{8}{3}}}{4} + \frac{10 a^{2} b^{3} x^{3}}{3} + \frac{3 a b^{4} x^{\frac{10}{3}}}{2} + \frac{3 b^{5} x^{\frac{11}{3}}}{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**5*x,x)

[Out]

a**5*x**2/2 + 15*a**4*b*x**(7/3)/7 + 15*a**3*b**2*x**(8/3)/4 + 10*a**2*b**3*x**3/3 + 3*a*b**4*x**(10/3)/2 + 3*
b**5*x**(11/3)/11

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Giac [A]  time = 1.16508, size = 77, normalized size = 1. \begin{align*} \frac{3}{11} \, b^{5} x^{\frac{11}{3}} + \frac{3}{2} \, a b^{4} x^{\frac{10}{3}} + \frac{10}{3} \, a^{2} b^{3} x^{3} + \frac{15}{4} \, a^{3} b^{2} x^{\frac{8}{3}} + \frac{15}{7} \, a^{4} b x^{\frac{7}{3}} + \frac{1}{2} \, a^{5} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^5*x,x, algorithm="giac")

[Out]

3/11*b^5*x^(11/3) + 3/2*a*b^4*x^(10/3) + 10/3*a^2*b^3*x^3 + 15/4*a^3*b^2*x^(8/3) + 15/7*a^4*b*x^(7/3) + 1/2*a^
5*x^2