\(\int \frac {(a+\frac {b}{\sqrt [3]{x}})^3}{x^3} \, dx\) [2419]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {b^3}{3 x^3}-\frac {9 a b^2}{8 x^{8/3}}-\frac {9 a^2 b}{7 x^{7/3}}-\frac {a^3}{2 x^2} \]

[Out]

-1/3*b^3/x^3-9/8*a*b^2/x^(8/3)-9/7*a^2*b/x^(7/3)-1/2*a^3/x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {269, 272, 45} \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {a^3}{2 x^2}-\frac {9 a^2 b}{7 x^{7/3}}-\frac {9 a b^2}{8 x^{8/3}}-\frac {b^3}{3 x^3} \]

[In]

Int[(a + b/x^(1/3))^3/x^3,x]

[Out]

-1/3*b^3/x^3 - (9*a*b^2)/(8*x^(8/3)) - (9*a^2*b)/(7*x^(7/3)) - a^3/(2*x^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 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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=\frac {-56 b^3-189 a b^2 \sqrt [3]{x}-216 a^2 b x^{2/3}-84 a^3 x}{168 x^3} \]

[In]

Integrate[(a + b/x^(1/3))^3/x^3,x]

[Out]

(-56*b^3 - 189*a*b^2*x^(1/3) - 216*a^2*b*x^(2/3) - 84*a^3*x)/(168*x^3)

Maple [A] (verified)

Time = 6.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-\frac {b^{3}}{3 x^{3}}-\frac {9 a \,b^{2}}{8 x^{\frac {8}{3}}}-\frac {9 a^{2} b}{7 x^{\frac {7}{3}}}-\frac {a^{3}}{2 x^{2}}\)	\(36\)
default	\(-\frac {b^{3}}{3 x^{3}}-\frac {9 a \,b^{2}}{8 x^{\frac {8}{3}}}-\frac {9 a^{2} b}{7 x^{\frac {7}{3}}}-\frac {a^{3}}{2 x^{2}}\)	\(36\)
trager	\(\frac {\left (-1+x \right ) \left (3 a^{3} x^{2}+2 b^{3} x^{2}+3 a^{3} x +2 b^{3} x +2 b^{3}\right )}{6 x^{3}}-\frac {9 a \,b^{2}}{8 x^{\frac {8}{3}}}-\frac {9 a^{2} b}{7 x^{\frac {7}{3}}}\)	\(62\)

[In]

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

[Out]

-1/3*b^3/x^3-9/8*a*b^2/x^(8/3)-9/7*a^2*b/x^(7/3)-1/2*a^3/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {84 \, a^{3} x + 216 \, a^{2} b x^{\frac {2}{3}} + 189 \, a b^{2} x^{\frac {1}{3}} + 56 \, b^{3}}{168 \, x^{3}} \]

[In]

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

[Out]

-1/168*(84*a^3*x + 216*a^2*b*x^(2/3) + 189*a*b^2*x^(1/3) + 56*b^3)/x^3

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=- \frac {a^{3}}{2 x^{2}} - \frac {9 a^{2} b}{7 x^{\frac {7}{3}}} - \frac {9 a b^{2}}{8 x^{\frac {8}{3}}} - \frac {b^{3}}{3 x^{3}} \]

[In]

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

[Out]

-a**3/(2*x**2) - 9*a**2*b/(7*x**(7/3)) - 9*a*b**2/(8*x**(8/3)) - b**3/(3*x**3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (35) = 70\).

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.09 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{9}}{3 \, b^{6}} + \frac {15 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{8} a}{8 \, b^{6}} - \frac {30 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{7} a^{2}}{7 \, b^{6}} + \frac {5 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{6} a^{3}}{b^{6}} - \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{5} a^{4}}{b^{6}} + \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{4} a^{5}}{4 \, b^{6}} \]

[In]

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

[Out]

-1/3*(a + b/x^(1/3))^9/b^6 + 15/8*(a + b/x^(1/3))^8*a/b^6 - 30/7*(a + b/x^(1/3))^7*a^2/b^6 + 5*(a + b/x^(1/3))
^6*a^3/b^6 - 3*(a + b/x^(1/3))^5*a^4/b^6 + 3/4*(a + b/x^(1/3))^4*a^5/b^6

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {84 \, a^{3} x + 216 \, a^{2} b x^{\frac {2}{3}} + 189 \, a b^{2} x^{\frac {1}{3}} + 56 \, b^{3}}{168 \, x^{3}} \]

[In]

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

[Out]

-1/168*(84*a^3*x + 216*a^2*b*x^(2/3) + 189*a*b^2*x^(1/3) + 56*b^3)/x^3

Mupad [B] (verification not implemented)

Time = 5.90 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{x^3} \, dx=-\frac {84\,a^3\,x+56\,b^3+189\,a\,b^2\,x^{1/3}+216\,a^2\,b\,x^{2/3}}{168\,x^3} \]

[In]

int((a + b/x^(1/3))^3/x^3,x)

[Out]

-(84*a^3*x + 56*b^3 + 189*a*b^2*x^(1/3) + 216*a^2*b*x^(2/3))/(168*x^3)