\(\int x^{2/3} (a+b x)^3 \, dx\) [668]

Optimal result

Integrand size = 13, antiderivative size = 51 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3}{5} a^3 x^{5/3}+\frac {9}{8} a^2 b x^{8/3}+\frac {9}{11} a b^2 x^{11/3}+\frac {3}{14} b^3 x^{14/3} \]

[Out]

3/5*a^3*x^(5/3)+9/8*a^2*b*x^(8/3)+9/11*a*b^2*x^(11/3)+3/14*b^3*x^(14/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3}{5} a^3 x^{5/3}+\frac {9}{8} a^2 b x^{8/3}+\frac {9}{11} a b^2 x^{11/3}+\frac {3}{14} b^3 x^{14/3} \]

[In]

Int[x^(2/3)*(a + b*x)^3,x]

[Out]

(3*a^3*x^(5/3))/5 + (9*a^2*b*x^(8/3))/8 + (9*a*b^2*x^(11/3))/11 + (3*b^3*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])

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 x^{2/3}+3 a^2 b x^{5/3}+3 a b^2 x^{8/3}+b^3 x^{11/3}\right ) \, dx \\ & = \frac {3}{5} a^3 x^{5/3}+\frac {9}{8} a^2 b x^{8/3}+\frac {9}{11} a b^2 x^{11/3}+\frac {3}{14} b^3 x^{14/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3 x^{5/3} \left (616 a^3+1155 a^2 b x+840 a b^2 x^2+220 b^3 x^3\right )}{3080} \]

[In]

Integrate[x^(2/3)*(a + b*x)^3,x]

[Out]

(3*x^(5/3)*(616*a^3 + 1155*a^2*b*x + 840*a*b^2*x^2 + 220*b^3*x^3))/3080

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {3 x^{\frac {5}{3}} \left (220 b^{3} x^{3}+840 a \,b^{2} x^{2}+1155 a^{2} b x +616 a^{3}\right )}{3080}\)	\(36\)
derivativedivides	\(\frac {3 a^{3} x^{\frac {5}{3}}}{5}+\frac {9 a^{2} b \,x^{\frac {8}{3}}}{8}+\frac {9 a \,b^{2} x^{\frac {11}{3}}}{11}+\frac {3 b^{3} x^{\frac {14}{3}}}{14}\)	\(36\)
default	\(\frac {3 a^{3} x^{\frac {5}{3}}}{5}+\frac {9 a^{2} b \,x^{\frac {8}{3}}}{8}+\frac {9 a \,b^{2} x^{\frac {11}{3}}}{11}+\frac {3 b^{3} x^{\frac {14}{3}}}{14}\)	\(36\)
trager	\(\frac {3 x^{\frac {5}{3}} \left (220 b^{3} x^{3}+840 a \,b^{2} x^{2}+1155 a^{2} b x +616 a^{3}\right )}{3080}\)	\(36\)
risch	\(\frac {3 x^{\frac {5}{3}} \left (220 b^{3} x^{3}+840 a \,b^{2} x^{2}+1155 a^{2} b x +616 a^{3}\right )}{3080}\)	\(36\)

[In]

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

[Out]

3/3080*x^(5/3)*(220*b^3*x^3+840*a*b^2*x^2+1155*a^2*b*x+616*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3}{3080} \, {\left (220 \, b^{3} x^{4} + 840 \, a b^{2} x^{3} + 1155 \, a^{2} b x^{2} + 616 \, a^{3} x\right )} x^{\frac {2}{3}} \]

[In]

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

[Out]

3/3080*(220*b^3*x^4 + 840*a*b^2*x^3 + 1155*a^2*b*x^2 + 616*a^3*x)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3 a^{3} x^{\frac {5}{3}}}{5} + \frac {9 a^{2} b x^{\frac {8}{3}}}{8} + \frac {9 a b^{2} x^{\frac {11}{3}}}{11} + \frac {3 b^{3} x^{\frac {14}{3}}}{14} \]

[In]

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

[Out]

3*a**3*x**(5/3)/5 + 9*a**2*b*x**(8/3)/8 + 9*a*b**2*x**(11/3)/11 + 3*b**3*x**(14/3)/14

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3}{14} \, b^{3} x^{\frac {14}{3}} + \frac {9}{11} \, a b^{2} x^{\frac {11}{3}} + \frac {9}{8} \, a^{2} b x^{\frac {8}{3}} + \frac {3}{5} \, a^{3} x^{\frac {5}{3}} \]

[In]

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

[Out]

3/14*b^3*x^(14/3) + 9/11*a*b^2*x^(11/3) + 9/8*a^2*b*x^(8/3) + 3/5*a^3*x^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3}{14} \, b^{3} x^{\frac {14}{3}} + \frac {9}{11} \, a b^{2} x^{\frac {11}{3}} + \frac {9}{8} \, a^{2} b x^{\frac {8}{3}} + \frac {3}{5} \, a^{3} x^{\frac {5}{3}} \]

[In]

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

[Out]

3/14*b^3*x^(14/3) + 9/11*a*b^2*x^(11/3) + 9/8*a^2*b*x^(8/3) + 3/5*a^3*x^(5/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{2/3} (a+b x)^3 \, dx=\frac {3\,a^3\,x^{5/3}}{5}+\frac {3\,b^3\,x^{14/3}}{14}+\frac {9\,a^2\,b\,x^{8/3}}{8}+\frac {9\,a\,b^2\,x^{11/3}}{11} \]

[In]

int(x^(2/3)*(a + b*x)^3,x)

[Out]

(3*a^3*x^(5/3))/5 + (3*b^3*x^(14/3))/14 + (9*a^2*b*x^(8/3))/8 + (9*a*b^2*x^(11/3))/11