\(\int (a+b \sqrt [3]{x})^2 x \, dx\) [2297]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {a^2 x^2}{2}+\frac {6}{7} a b x^{7/3}+\frac {3}{8} b^2 x^{8/3} \]

[Out]

1/2*a^2*x^2+6/7*a*b*x^(7/3)+3/8*b^2*x^(8/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {a^2 x^2}{2}+\frac {6}{7} a b x^{7/3}+\frac {3}{8} b^2 x^{8/3} \]

[In]

Int[(a + b*x^(1/3))^2*x,x]

[Out]

(a^2*x^2)/2 + (6*a*b*x^(7/3))/7 + (3*b^2*x^(8/3))/8

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {1}{56} \left (28 a^2+48 a b \sqrt [3]{x}+21 b^2 x^{2/3}\right ) x^2 \]

[In]

Integrate[(a + b*x^(1/3))^2*x,x]

[Out]

((28*a^2 + 48*a*b*x^(1/3) + 21*b^2*x^(2/3))*x^2)/56

Maple [A] (verified)

Time = 3.65 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {a^{2} x^{2}}{2}+\frac {6 a b \,x^{\frac {7}{3}}}{7}+\frac {3 b^{2} x^{\frac {8}{3}}}{8}\)	\(25\)
default	\(\frac {a^{2} x^{2}}{2}+\frac {6 a b \,x^{\frac {7}{3}}}{7}+\frac {3 b^{2} x^{\frac {8}{3}}}{8}\)	\(25\)
trager	\(\frac {\left (-1+x \right ) a^{2} \left (1+x \right )}{2}+\frac {6 a b \,x^{\frac {7}{3}}}{7}+\frac {3 b^{2} x^{\frac {8}{3}}}{8}\)	\(28\)

[In]

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

[Out]

1/2*a^2*x^2+6/7*a*b*x^(7/3)+3/8*b^2*x^(8/3)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {3}{8} \, b^{2} x^{\frac {8}{3}} + \frac {6}{7} \, a b x^{\frac {7}{3}} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

3/8*b^2*x^(8/3) + 6/7*a*b*x^(7/3) + 1/2*a^2*x^2

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {a^{2} x^{2}}{2} + \frac {6 a b x^{\frac {7}{3}}}{7} + \frac {3 b^{2} x^{\frac {8}{3}}}{8} \]

[In]

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

[Out]

a**2*x**2/2 + 6*a*b*x**(7/3)/7 + 3*b**2*x**(8/3)/8

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (24) = 48\).

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.88 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8}}{8 \, b^{6}} - \frac {15 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a}{7 \, b^{6}} + \frac {5 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{2}}{b^{6}} - \frac {6 \, {\left (b x^{\frac {1}{3}} + a\right )}^{5} a^{3}}{b^{6}} + \frac {15 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} a^{4}}{4 \, b^{6}} - \frac {{\left (b x^{\frac {1}{3}} + a\right )}^{3} a^{5}}{b^{6}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {3}{8} \, b^{2} x^{\frac {8}{3}} + \frac {6}{7} \, a b x^{\frac {7}{3}} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

3/8*b^2*x^(8/3) + 6/7*a*b*x^(7/3) + 1/2*a^2*x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x \, dx=\frac {a^2\,x^2}{2}+\frac {3\,b^2\,x^{8/3}}{8}+\frac {6\,a\,b\,x^{7/3}}{7} \]

[In]

int(x*(a + b*x^(1/3))^2,x)

[Out]

(a^2*x^2)/2 + (3*b^2*x^(8/3))/8 + (6*a*b*x^(7/3))/7