\(\int (a+\frac {b}{\sqrt [3]{x}})^2 x^4 \, dx\) [2403]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {3}{13} b^2 x^{13/3}+\frac {3}{7} a b x^{14/3}+\frac {a^2 x^5}{5} \]

[Out]

3/13*b^2*x^(13/3)+3/7*a*b*x^(14/3)+1/5*a^2*x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, 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} \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {a^2 x^5}{5}+\frac {3}{7} a b x^{14/3}+\frac {3}{13} b^2 x^{13/3} \]

[In]

Int[(a + b/x^(1/3))^2*x^4,x]

[Out]

(3*b^2*x^(13/3))/13 + (3*a*b*x^(14/3))/7 + (a^2*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 )^2 x^{10/3} \, dx \\ & = 3 \text {Subst}\left (\int x^{12} (b+a x)^2 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (b^2 x^{12}+2 a b x^{13}+a^2 x^{14}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {3}{13} b^2 x^{13/3}+\frac {3}{7} a b x^{14/3}+\frac {a^2 x^5}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {1}{455} \left (105 b^2+195 a b \sqrt [3]{x}+91 a^2 x^{2/3}\right ) x^{13/3} \]

[In]

Integrate[(a + b/x^(1/3))^2*x^4,x]

[Out]

((105*b^2 + 195*a*b*x^(1/3) + 91*a^2*x^(2/3))*x^(13/3))/455

Maple [A] (verified)

Time = 6.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {3 b^{2} x^{\frac {13}{3}}}{13}+\frac {3 a b \,x^{\frac {14}{3}}}{7}+\frac {x^{5} a^{2}}{5}\)	\(25\)
default	\(\frac {3 b^{2} x^{\frac {13}{3}}}{13}+\frac {3 a b \,x^{\frac {14}{3}}}{7}+\frac {x^{5} a^{2}}{5}\)	\(25\)
trager	\(\frac {a^{2} \left (x^{4}+x^{3}+x^{2}+x +1\right ) \left (-1+x \right )}{5}+\frac {3 b^{2} x^{\frac {13}{3}}}{13}+\frac {3 a b \,x^{\frac {14}{3}}}{7}\)	\(37\)

[In]

int((a+b/x^(1/3))^2*x^4,x,method=_RETURNVERBOSE)

[Out]

3/13*b^2*x^(13/3)+3/7*a*b*x^(14/3)+1/5*x^5*a^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {1}{5} \, a^{2} x^{5} + \frac {3}{7} \, a b x^{\frac {14}{3}} + \frac {3}{13} \, b^{2} x^{\frac {13}{3}} \]

[In]

integrate((a+b/x^(1/3))^2*x^4,x, algorithm="fricas")

[Out]

1/5*a^2*x^5 + 3/7*a*b*x^(14/3) + 3/13*b^2*x^(13/3)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {a^{2} x^{5}}{5} + \frac {3 a b x^{\frac {14}{3}}}{7} + \frac {3 b^{2} x^{\frac {13}{3}}}{13} \]

[In]

integrate((a+b/x**(1/3))**2*x**4,x)

[Out]

a**2*x**5/5 + 3*a*b*x**(14/3)/7 + 3*b**2*x**(13/3)/13

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {1}{455} \, {\left (91 \, a^{2} + \frac {195 \, a b}{x^{\frac {1}{3}}} + \frac {105 \, b^{2}}{x^{\frac {2}{3}}}\right )} x^{5} \]

[In]

integrate((a+b/x^(1/3))^2*x^4,x, algorithm="maxima")

[Out]

1/455*(91*a^2 + 195*a*b/x^(1/3) + 105*b^2/x^(2/3))*x^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {1}{5} \, a^{2} x^{5} + \frac {3}{7} \, a b x^{\frac {14}{3}} + \frac {3}{13} \, b^{2} x^{\frac {13}{3}} \]

[In]

integrate((a+b/x^(1/3))^2*x^4,x, algorithm="giac")

[Out]

1/5*a^2*x^5 + 3/7*a*b*x^(14/3) + 3/13*b^2*x^(13/3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x^4 \, dx=\frac {a^2\,x^5}{5}+\frac {3\,b^2\,x^{13/3}}{13}+\frac {3\,a\,b\,x^{14/3}}{7} \]

[In]

int(x^4*(a + b/x^(1/3))^2,x)

[Out]

(a^2*x^5)/5 + (3*b^2*x^(13/3))/13 + (3*a*b*x^(14/3))/7