\(\int x^{5/3} (a+b x)^2 \, dx\) [658]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{8} a^2 x^{8/3}+\frac {6}{11} a b x^{11/3}+\frac {3}{14} b^2 x^{14/3} \]

[Out]

3/8*a^2*x^(8/3)+6/11*a*b*x^(11/3)+3/14*b^2*x^(14/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, 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} \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{8} a^2 x^{8/3}+\frac {6}{11} a b x^{11/3}+\frac {3}{14} b^2 x^{14/3} \]

[In]

Int[x^(5/3)*(a + b*x)^2,x]

[Out]

(3*a^2*x^(8/3))/8 + (6*a*b*x^(11/3))/11 + (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])

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 x^{5/3}+2 a b x^{8/3}+b^2 x^{11/3}\right ) \, dx \\ & = \frac {3}{8} a^2 x^{8/3}+\frac {6}{11} a b x^{11/3}+\frac {3}{14} b^2 x^{14/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{616} x^{8/3} \left (77 a^2+112 a b x+44 b^2 x^2\right ) \]

[In]

Integrate[x^(5/3)*(a + b*x)^2,x]

[Out]

(3*x^(8/3)*(77*a^2 + 112*a*b*x + 44*b^2*x^2))/616

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {3 x^{\frac {8}{3}} \left (44 b^{2} x^{2}+112 a b x +77 a^{2}\right )}{616}\)	\(25\)
derivativedivides	\(\frac {3 a^{2} x^{\frac {8}{3}}}{8}+\frac {6 a b \,x^{\frac {11}{3}}}{11}+\frac {3 b^{2} x^{\frac {14}{3}}}{14}\)	\(25\)
default	\(\frac {3 a^{2} x^{\frac {8}{3}}}{8}+\frac {6 a b \,x^{\frac {11}{3}}}{11}+\frac {3 b^{2} x^{\frac {14}{3}}}{14}\)	\(25\)
trager	\(\frac {3 x^{\frac {8}{3}} \left (44 b^{2} x^{2}+112 a b x +77 a^{2}\right )}{616}\)	\(25\)
risch	\(\frac {3 x^{\frac {8}{3}} \left (44 b^{2} x^{2}+112 a b x +77 a^{2}\right )}{616}\)	\(25\)

[In]

int(x^(5/3)*(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

3/616*x^(8/3)*(44*b^2*x^2+112*a*b*x+77*a^2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{616} \, {\left (44 \, b^{2} x^{4} + 112 \, a b x^{3} + 77 \, a^{2} x^{2}\right )} x^{\frac {2}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^2,x, algorithm="fricas")

[Out]

3/616*(44*b^2*x^4 + 112*a*b*x^3 + 77*a^2*x^2)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3 a^{2} x^{\frac {8}{3}}}{8} + \frac {6 a b x^{\frac {11}{3}}}{11} + \frac {3 b^{2} x^{\frac {14}{3}}}{14} \]

[In]

integrate(x**(5/3)*(b*x+a)**2,x)

[Out]

3*a**2*x**(8/3)/8 + 6*a*b*x**(11/3)/11 + 3*b**2*x**(14/3)/14

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{14} \, b^{2} x^{\frac {14}{3}} + \frac {6}{11} \, a b x^{\frac {11}{3}} + \frac {3}{8} \, a^{2} x^{\frac {8}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^2,x, algorithm="maxima")

[Out]

3/14*b^2*x^(14/3) + 6/11*a*b*x^(11/3) + 3/8*a^2*x^(8/3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3}{14} \, b^{2} x^{\frac {14}{3}} + \frac {6}{11} \, a b x^{\frac {11}{3}} + \frac {3}{8} \, a^{2} x^{\frac {8}{3}} \]

[In]

integrate(x^(5/3)*(b*x+a)^2,x, algorithm="giac")

[Out]

3/14*b^2*x^(14/3) + 6/11*a*b*x^(11/3) + 3/8*a^2*x^(8/3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{5/3} (a+b x)^2 \, dx=\frac {3\,x^{8/3}\,\left (77\,a^2+112\,a\,b\,x+44\,b^2\,x^2\right )}{616} \]

[In]

int(x^(5/3)*(a + b*x)^2,x)

[Out]

(3*x^(8/3)*(77*a^2 + 44*b^2*x^2 + 112*a*b*x))/616