\(\int x^{5/3} (a+b x)^3 \, dx\) [666]

Optimal result

Integrand size = 13, antiderivative size = 51 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3}{8} a^3 x^{8/3}+\frac {9}{11} a^2 b x^{11/3}+\frac {9}{14} a b^2 x^{14/3}+\frac {3}{17} b^3 x^{17/3} \]

[Out]

3/8*a^3*x^(8/3)+9/11*a^2*b*x^(11/3)+9/14*a*b^2*x^(14/3)+3/17*b^3*x^(17/3)

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} \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3}{8} a^3 x^{8/3}+\frac {9}{11} a^2 b x^{11/3}+\frac {9}{14} a b^2 x^{14/3}+\frac {3}{17} b^3 x^{17/3} \]

[In]

Int[x^(5/3)*(a + b*x)^3,x]

[Out]

(3*a^3*x^(8/3))/8 + (9*a^2*b*x^(11/3))/11 + (9*a*b^2*x^(14/3))/14 + (3*b^3*x^(17/3))/17

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^{5/3}+3 a^2 b x^{8/3}+3 a b^2 x^{11/3}+b^3 x^{14/3}\right ) \, dx \\ & = \frac {3}{8} a^3 x^{8/3}+\frac {9}{11} a^2 b x^{11/3}+\frac {9}{14} a b^2 x^{14/3}+\frac {3}{17} b^3 x^{17/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3 x^{8/3} \left (1309 a^3+2856 a^2 b x+2244 a b^2 x^2+616 b^3 x^3\right )}{10472} \]

[In]

Integrate[x^(5/3)*(a + b*x)^3,x]

[Out]

(3*x^(8/3)*(1309*a^3 + 2856*a^2*b*x + 2244*a*b^2*x^2 + 616*b^3*x^3))/10472

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {3 x^{\frac {8}{3}} \left (616 b^{3} x^{3}+2244 a \,b^{2} x^{2}+2856 a^{2} b x +1309 a^{3}\right )}{10472}\)	\(36\)
derivativedivides	\(\frac {3 a^{3} x^{\frac {8}{3}}}{8}+\frac {9 a^{2} b \,x^{\frac {11}{3}}}{11}+\frac {9 a \,b^{2} x^{\frac {14}{3}}}{14}+\frac {3 b^{3} x^{\frac {17}{3}}}{17}\)	\(36\)
default	\(\frac {3 a^{3} x^{\frac {8}{3}}}{8}+\frac {9 a^{2} b \,x^{\frac {11}{3}}}{11}+\frac {9 a \,b^{2} x^{\frac {14}{3}}}{14}+\frac {3 b^{3} x^{\frac {17}{3}}}{17}\)	\(36\)
trager	\(\frac {3 x^{\frac {8}{3}} \left (616 b^{3} x^{3}+2244 a \,b^{2} x^{2}+2856 a^{2} b x +1309 a^{3}\right )}{10472}\)	\(36\)
risch	\(\frac {3 x^{\frac {8}{3}} \left (616 b^{3} x^{3}+2244 a \,b^{2} x^{2}+2856 a^{2} b x +1309 a^{3}\right )}{10472}\)	\(36\)

[In]

int(x^(5/3)*(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

3/10472*x^(8/3)*(616*b^3*x^3+2244*a*b^2*x^2+2856*a^2*b*x+1309*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3}{10472} \, {\left (616 \, b^{3} x^{5} + 2244 \, a b^{2} x^{4} + 2856 \, a^{2} b x^{3} + 1309 \, a^{3} x^{2}\right )} x^{\frac {2}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^3,x, algorithm="fricas")

[Out]

3/10472*(616*b^3*x^5 + 2244*a*b^2*x^4 + 2856*a^2*b*x^3 + 1309*a^3*x^2)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3 a^{3} x^{\frac {8}{3}}}{8} + \frac {9 a^{2} b x^{\frac {11}{3}}}{11} + \frac {9 a b^{2} x^{\frac {14}{3}}}{14} + \frac {3 b^{3} x^{\frac {17}{3}}}{17} \]

[In]

integrate(x**(5/3)*(b*x+a)**3,x)

[Out]

3*a**3*x**(8/3)/8 + 9*a**2*b*x**(11/3)/11 + 9*a*b**2*x**(14/3)/14 + 3*b**3*x**(17/3)/17

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3}{17} \, b^{3} x^{\frac {17}{3}} + \frac {9}{14} \, a b^{2} x^{\frac {14}{3}} + \frac {9}{11} \, a^{2} b x^{\frac {11}{3}} + \frac {3}{8} \, a^{3} x^{\frac {8}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^3,x, algorithm="maxima")

[Out]

3/17*b^3*x^(17/3) + 9/14*a*b^2*x^(14/3) + 9/11*a^2*b*x^(11/3) + 3/8*a^3*x^(8/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3}{17} \, b^{3} x^{\frac {17}{3}} + \frac {9}{14} \, a b^{2} x^{\frac {14}{3}} + \frac {9}{11} \, a^{2} b x^{\frac {11}{3}} + \frac {3}{8} \, a^{3} x^{\frac {8}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^3,x, algorithm="giac")

[Out]

3/17*b^3*x^(17/3) + 9/14*a*b^2*x^(14/3) + 9/11*a^2*b*x^(11/3) + 3/8*a^3*x^(8/3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{5/3} (a+b x)^3 \, dx=\frac {3\,a^3\,x^{8/3}}{8}+\frac {3\,b^3\,x^{17/3}}{17}+\frac {9\,a^2\,b\,x^{11/3}}{11}+\frac {9\,a\,b^2\,x^{14/3}}{14} \]

[In]

int(x^(5/3)*(a + b*x)^3,x)

[Out]

(3*a^3*x^(8/3))/8 + (3*b^3*x^(17/3))/17 + (9*a^2*b*x^(11/3))/11 + (9*a*b^2*x^(14/3))/14