\(\int (a+\frac {b}{\sqrt [3]{x}})^3 x \, dx\) [2415]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=b^3 x+\frac {9}{4} a b^2 x^{4/3}+\frac {9}{5} a^2 b x^{5/3}+\frac {a^3 x^2}{2} \]

[Out]

b^3*x+9/4*a*b^2*x^(4/3)+9/5*a^2*b*x^(5/3)+1/2*a^3*x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 196, 45} \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {a^3 x^2}{2}+\frac {9}{5} a^2 b x^{5/3}+\frac {9}{4} a b^2 x^{4/3}+b^3 x \]

[In]

Int[(a + b/x^(1/3))^3*x,x]

[Out]

b^3*x + (9*a*b^2*x^(4/3))/4 + (9*a^2*b*x^(5/3))/5 + (a^3*x^2)/2

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 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (b+a \sqrt [3]{x}\right )^3 \, dx \\ & = 3 \text {Subst}\left (\int x^2 (b+a x)^3 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (b^3 x^2+3 a b^2 x^3+3 a^2 b x^4+a^3 x^5\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = b^3 x+\frac {9}{4} a b^2 x^{4/3}+\frac {9}{5} a^2 b x^{5/3}+\frac {a^3 x^2}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {1}{20} \left (20 b^3 x+45 a b^2 x^{4/3}+36 a^2 b x^{5/3}+10 a^3 x^2\right ) \]

[In]

Integrate[(a + b/x^(1/3))^3*x,x]

[Out]

(20*b^3*x + 45*a*b^2*x^(4/3) + 36*a^2*b*x^(5/3) + 10*a^3*x^2)/20

Maple [A] (verified)

Time = 6.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(b^{3} x +\frac {9 a \,b^{2} x^{\frac {4}{3}}}{4}+\frac {9 a^{2} b \,x^{\frac {5}{3}}}{5}+\frac {a^{3} x^{2}}{2}\)	\(33\)
default	\(b^{3} x +\frac {9 a \,b^{2} x^{\frac {4}{3}}}{4}+\frac {9 a^{2} b \,x^{\frac {5}{3}}}{5}+\frac {a^{3} x^{2}}{2}\)	\(33\)
trager	\(\frac {\left (-1+x \right ) \left (a^{3} x +a^{3}+2 b^{3}\right )}{2}+\frac {9 a \,b^{2} x^{\frac {4}{3}}}{4}+\frac {9 a^{2} b \,x^{\frac {5}{3}}}{5}\)	\(39\)

[In]

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

[Out]

b^3*x+9/4*a*b^2*x^(4/3)+9/5*a^2*b*x^(5/3)+1/2*a^3*x^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {1}{2} \, a^{3} x^{2} + \frac {9}{5} \, a^{2} b x^{\frac {5}{3}} + \frac {9}{4} \, a b^{2} x^{\frac {4}{3}} + b^{3} x \]

[In]

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

[Out]

1/2*a^3*x^2 + 9/5*a^2*b*x^(5/3) + 9/4*a*b^2*x^(4/3) + b^3*x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {a^{3} x^{2}}{2} + \frac {9 a^{2} b x^{\frac {5}{3}}}{5} + \frac {9 a b^{2} x^{\frac {4}{3}}}{4} + b^{3} x \]

[In]

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

[Out]

a**3*x**2/2 + 9*a**2*b*x**(5/3)/5 + 9*a*b**2*x**(4/3)/4 + b**3*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {1}{20} \, {\left (10 \, a^{3} + \frac {36 \, a^{2} b}{x^{\frac {1}{3}}} + \frac {45 \, a b^{2}}{x^{\frac {2}{3}}} + \frac {20 \, b^{3}}{x}\right )} x^{2} \]

[In]

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

[Out]

1/20*(10*a^3 + 36*a^2*b/x^(1/3) + 45*a*b^2/x^(2/3) + 20*b^3/x)*x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=\frac {1}{2} \, a^{3} x^{2} + \frac {9}{5} \, a^{2} b x^{\frac {5}{3}} + \frac {9}{4} \, a b^{2} x^{\frac {4}{3}} + b^{3} x \]

[In]

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

[Out]

1/2*a^3*x^2 + 9/5*a^2*b*x^(5/3) + 9/4*a*b^2*x^(4/3) + b^3*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x \, dx=b^3\,x+\frac {a^3\,x^2}{2}+\frac {9\,a\,b^2\,x^{4/3}}{4}+\frac {9\,a^2\,b\,x^{5/3}}{5} \]

[In]

int(x*(a + b/x^(1/3))^3,x)

[Out]

b^3*x + (a^3*x^2)/2 + (9*a*b^2*x^(4/3))/4 + (9*a^2*b*x^(5/3))/5