\(\int (a+\frac {b}{\sqrt [3]{x}})^3 x^4 \, dx\) [2412]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {b^3 x^4}{4}+\frac {9}{13} a b^2 x^{13/3}+\frac {9}{14} a^2 b x^{14/3}+\frac {a^3 x^5}{5} \]

[Out]

1/4*b^3*x^4+9/13*a*b^2*x^(13/3)+9/14*a^2*b*x^(14/3)+1/5*a^3*x^5

Rubi [A] (verified)

Time = 0.03 (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} \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {a^3 x^5}{5}+\frac {9}{14} a^2 b x^{14/3}+\frac {9}{13} a b^2 x^{13/3}+\frac {b^3 x^4}{4} \]

[In]

Int[(a + b/x^(1/3))^3*x^4,x]

[Out]

(b^3*x^4)/4 + (9*a*b^2*x^(13/3))/13 + (9*a^2*b*x^(14/3))/14 + (a^3*x^5)/5

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 \left (b+a \sqrt [3]{x}\right )^3 x^3 \, dx \\ & = 3 \text {Subst}\left (\int x^{11} (b+a x)^3 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (b^3 x^{11}+3 a b^2 x^{12}+3 a^2 b x^{13}+a^3 x^{14}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {b^3 x^4}{4}+\frac {9}{13} a b^2 x^{13/3}+\frac {9}{14} a^2 b x^{14/3}+\frac {a^3 x^5}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {455 b^3 x^4+1260 a b^2 x^{13/3}+1170 a^2 b x^{14/3}+364 a^3 x^5}{1820} \]

[In]

Integrate[(a + b/x^(1/3))^3*x^4,x]

[Out]

(455*b^3*x^4 + 1260*a*b^2*x^(13/3) + 1170*a^2*b*x^(14/3) + 364*a^3*x^5)/1820

Maple [A] (verified)

Time = 3.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {b^{3} x^{4}}{4}+\frac {9 a \,b^{2} x^{\frac {13}{3}}}{13}+\frac {9 a^{2} b \,x^{\frac {14}{3}}}{14}+\frac {a^{3} x^{5}}{5}\)	\(36\)
default	\(\frac {b^{3} x^{4}}{4}+\frac {9 a \,b^{2} x^{\frac {13}{3}}}{13}+\frac {9 a^{2} b \,x^{\frac {14}{3}}}{14}+\frac {a^{3} x^{5}}{5}\)	\(36\)
trager	\(\frac {\left (4 a^{3} x^{4}+4 a^{3} x^{3}+5 b^{3} x^{3}+4 a^{3} x^{2}+5 b^{3} x^{2}+4 a^{3} x +5 b^{3} x +4 a^{3}+5 b^{3}\right ) \left (-1+x \right )}{20}+\frac {9 a \,b^{2} x^{\frac {13}{3}}}{13}+\frac {9 a^{2} b \,x^{\frac {14}{3}}}{14}\)	\(88\)

[In]

int((a+b/x^(1/3))^3*x^4,x,method=_RETURNVERBOSE)

[Out]

1/4*b^3*x^4+9/13*a*b^2*x^(13/3)+9/14*a^2*b*x^(14/3)+1/5*a^3*x^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {1}{5} \, a^{3} x^{5} + \frac {9}{14} \, a^{2} b x^{\frac {14}{3}} + \frac {9}{13} \, a b^{2} x^{\frac {13}{3}} + \frac {1}{4} \, b^{3} x^{4} \]

[In]

integrate((a+b/x^(1/3))^3*x^4,x, algorithm="fricas")

[Out]

1/5*a^3*x^5 + 9/14*a^2*b*x^(14/3) + 9/13*a*b^2*x^(13/3) + 1/4*b^3*x^4

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {a^{3} x^{5}}{5} + \frac {9 a^{2} b x^{\frac {14}{3}}}{14} + \frac {9 a b^{2} x^{\frac {13}{3}}}{13} + \frac {b^{3} x^{4}}{4} \]

[In]

integrate((a+b/x**(1/3))**3*x**4,x)

[Out]

a**3*x**5/5 + 9*a**2*b*x**(14/3)/14 + 9*a*b**2*x**(13/3)/13 + b**3*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {1}{1820} \, {\left (364 \, a^{3} + \frac {1170 \, a^{2} b}{x^{\frac {1}{3}}} + \frac {1260 \, a b^{2}}{x^{\frac {2}{3}}} + \frac {455 \, b^{3}}{x}\right )} x^{5} \]

[In]

integrate((a+b/x^(1/3))^3*x^4,x, algorithm="maxima")

[Out]

1/1820*(364*a^3 + 1170*a^2*b/x^(1/3) + 1260*a*b^2/x^(2/3) + 455*b^3/x)*x^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {1}{5} \, a^{3} x^{5} + \frac {9}{14} \, a^{2} b x^{\frac {14}{3}} + \frac {9}{13} \, a b^{2} x^{\frac {13}{3}} + \frac {1}{4} \, b^{3} x^{4} \]

[In]

integrate((a+b/x^(1/3))^3*x^4,x, algorithm="giac")

[Out]

1/5*a^3*x^5 + 9/14*a^2*b*x^(14/3) + 9/13*a*b^2*x^(13/3) + 1/4*b^3*x^4

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4 \, dx=\frac {a^3\,x^5}{5}+\frac {b^3\,x^4}{4}+\frac {9\,a\,b^2\,x^{13/3}}{13}+\frac {9\,a^2\,b\,x^{14/3}}{14} \]

[In]

int(x^4*(a + b/x^(1/3))^3,x)

[Out]

(a^3*x^5)/5 + (b^3*x^4)/4 + (9*a*b^2*x^(13/3))/13 + (9*a^2*b*x^(14/3))/14