\(\int (a+b \sqrt [3]{x})^2 x^3 \, dx\) [2295]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {a^2 x^4}{4}+\frac {6}{13} a b x^{13/3}+\frac {3}{14} b^2 x^{14/3} \]

[Out]

1/4*a^2*x^4+6/13*a*b*x^(13/3)+3/14*b^2*x^(14/3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {a^2 x^4}{4}+\frac {6}{13} a b x^{13/3}+\frac {3}{14} b^2 x^{14/3} \]

[In]

Int[(a + b*x^(1/3))^2*x^3,x]

[Out]

(a^2*x^4)/4 + (6*a*b*x^(13/3))/13 + (3*b^2*x^(14/3))/14

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^{11} (a+b x)^2 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (a^2 x^{11}+2 a b x^{12}+b^2 x^{13}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {a^2 x^4}{4}+\frac {6}{13} a b x^{13/3}+\frac {3}{14} b^2 x^{14/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {1}{364} \left (91 a^2+168 a b \sqrt [3]{x}+78 b^2 x^{2/3}\right ) x^4 \]

[In]

Integrate[(a + b*x^(1/3))^2*x^3,x]

[Out]

((91*a^2 + 168*a*b*x^(1/3) + 78*b^2*x^(2/3))*x^4)/364

Maple [A] (verified)

Time = 10.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {a^{2} x^{4}}{4}+\frac {6 a b \,x^{\frac {13}{3}}}{13}+\frac {3 b^{2} x^{\frac {14}{3}}}{14}\)	\(25\)
default	\(\frac {a^{2} x^{4}}{4}+\frac {6 a b \,x^{\frac {13}{3}}}{13}+\frac {3 b^{2} x^{\frac {14}{3}}}{14}\)	\(25\)
trager	\(\frac {a^{2} \left (x^{3}+x^{2}+x +1\right ) \left (-1+x \right )}{4}+\frac {6 a b \,x^{\frac {13}{3}}}{13}+\frac {3 b^{2} x^{\frac {14}{3}}}{14}\)	\(34\)

[In]

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

[Out]

1/4*a^2*x^4+6/13*a*b*x^(13/3)+3/14*b^2*x^(14/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {3}{14} \, b^{2} x^{\frac {14}{3}} + \frac {6}{13} \, a b x^{\frac {13}{3}} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

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

[Out]

3/14*b^2*x^(14/3) + 6/13*a*b*x^(13/3) + 1/4*a^2*x^4

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {a^{2} x^{4}}{4} + \frac {6 a b x^{\frac {13}{3}}}{13} + \frac {3 b^{2} x^{\frac {14}{3}}}{14} \]

[In]

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

[Out]

a**2*x**4/4 + 6*a*b*x**(13/3)/13 + 3*b**2*x**(14/3)/14

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (24) = 48\).

Time = 0.21 (sec) , antiderivative size = 200, normalized size of antiderivative = 5.88 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{14}}{14 \, b^{12}} - \frac {33 \, {\left (b x^{\frac {1}{3}} + a\right )}^{13} a}{13 \, b^{12}} + \frac {55 \, {\left (b x^{\frac {1}{3}} + a\right )}^{12} a^{2}}{4 \, b^{12}} - \frac {45 \, {\left (b x^{\frac {1}{3}} + a\right )}^{11} a^{3}}{b^{12}} + \frac {99 \, {\left (b x^{\frac {1}{3}} + a\right )}^{10} a^{4}}{b^{12}} - \frac {154 \, {\left (b x^{\frac {1}{3}} + a\right )}^{9} a^{5}}{b^{12}} + \frac {693 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} a^{6}}{4 \, b^{12}} - \frac {990 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a^{7}}{7 \, b^{12}} + \frac {165 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{8}}{2 \, b^{12}} - \frac {33 \, {\left (b x^{\frac {1}{3}} + a\right )}^{5} a^{9}}{b^{12}} + \frac {33 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} a^{10}}{4 \, b^{12}} - \frac {{\left (b x^{\frac {1}{3}} + a\right )}^{3} a^{11}}{b^{12}} \]

[In]

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

[Out]

3/14*(b*x^(1/3) + a)^14/b^12 - 33/13*(b*x^(1/3) + a)^13*a/b^12 + 55/4*(b*x^(1/3) + a)^12*a^2/b^12 - 45*(b*x^(1
/3) + a)^11*a^3/b^12 + 99*(b*x^(1/3) + a)^10*a^4/b^12 - 154*(b*x^(1/3) + a)^9*a^5/b^12 + 693/4*(b*x^(1/3) + a)
^8*a^6/b^12 - 990/7*(b*x^(1/3) + a)^7*a^7/b^12 + 165/2*(b*x^(1/3) + a)^6*a^8/b^12 - 33*(b*x^(1/3) + a)^5*a^9/b
^12 + 33/4*(b*x^(1/3) + a)^4*a^10/b^12 - (b*x^(1/3) + a)^3*a^11/b^12

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {3}{14} \, b^{2} x^{\frac {14}{3}} + \frac {6}{13} \, a b x^{\frac {13}{3}} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

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

[Out]

3/14*b^2*x^(14/3) + 6/13*a*b*x^(13/3) + 1/4*a^2*x^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sqrt [3]{x}\right )^2 x^3 \, dx=\frac {a^2\,x^4}{4}+\frac {3\,b^2\,x^{14/3}}{14}+\frac {6\,a\,b\,x^{13/3}}{13} \]

[In]

int(x^3*(a + b*x^(1/3))^2,x)

[Out]

(a^2*x^4)/4 + (3*b^2*x^(14/3))/14 + (6*a*b*x^(13/3))/13