\(\int x^{2/3} (a+b x)^2 \, dx\) [660]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{5} a^2 x^{5/3}+\frac {3}{4} a b x^{8/3}+\frac {3}{11} b^2 x^{11/3} \]

[Out]

3/5*a^2*x^(5/3)+3/4*a*b*x^(8/3)+3/11*b^2*x^(11/3)

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} \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{5} a^2 x^{5/3}+\frac {3}{4} a b x^{8/3}+\frac {3}{11} b^2 x^{11/3} \]

[In]

Int[x^(2/3)*(a + b*x)^2,x]

[Out]

(3*a^2*x^(5/3))/5 + (3*a*b*x^(8/3))/4 + (3*b^2*x^(11/3))/11

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^{2/3}+2 a b x^{5/3}+b^2 x^{8/3}\right ) \, dx \\ & = \frac {3}{5} a^2 x^{5/3}+\frac {3}{4} a b x^{8/3}+\frac {3}{11} b^2 x^{11/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{220} x^{5/3} \left (44 a^2+55 a b x+20 b^2 x^2\right ) \]

[In]

Integrate[x^(2/3)*(a + b*x)^2,x]

[Out]

(3*x^(5/3)*(44*a^2 + 55*a*b*x + 20*b^2*x^2))/220

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {3 x^{\frac {5}{3}} \left (20 b^{2} x^{2}+55 a b x +44 a^{2}\right )}{220}\)	\(25\)
derivativedivides	\(\frac {3 a^{2} x^{\frac {5}{3}}}{5}+\frac {3 a b \,x^{\frac {8}{3}}}{4}+\frac {3 b^{2} x^{\frac {11}{3}}}{11}\)	\(25\)
default	\(\frac {3 a^{2} x^{\frac {5}{3}}}{5}+\frac {3 a b \,x^{\frac {8}{3}}}{4}+\frac {3 b^{2} x^{\frac {11}{3}}}{11}\)	\(25\)
trager	\(\frac {3 x^{\frac {5}{3}} \left (20 b^{2} x^{2}+55 a b x +44 a^{2}\right )}{220}\)	\(25\)
risch	\(\frac {3 x^{\frac {5}{3}} \left (20 b^{2} x^{2}+55 a b x +44 a^{2}\right )}{220}\)	\(25\)

[In]

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

[Out]

3/220*x^(5/3)*(20*b^2*x^2+55*a*b*x+44*a^2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{220} \, {\left (20 \, b^{2} x^{3} + 55 \, a b x^{2} + 44 \, a^{2} x\right )} x^{\frac {2}{3}} \]

[In]

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

[Out]

3/220*(20*b^2*x^3 + 55*a*b*x^2 + 44*a^2*x)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3 a^{2} x^{\frac {5}{3}}}{5} + \frac {3 a b x^{\frac {8}{3}}}{4} + \frac {3 b^{2} x^{\frac {11}{3}}}{11} \]

[In]

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

[Out]

3*a**2*x**(5/3)/5 + 3*a*b*x**(8/3)/4 + 3*b**2*x**(11/3)/11

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{11} \, b^{2} x^{\frac {11}{3}} + \frac {3}{4} \, a b x^{\frac {8}{3}} + \frac {3}{5} \, a^{2} x^{\frac {5}{3}} \]

[In]

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

[Out]

3/11*b^2*x^(11/3) + 3/4*a*b*x^(8/3) + 3/5*a^2*x^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3}{11} \, b^{2} x^{\frac {11}{3}} + \frac {3}{4} \, a b x^{\frac {8}{3}} + \frac {3}{5} \, a^{2} x^{\frac {5}{3}} \]

[In]

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

[Out]

3/11*b^2*x^(11/3) + 3/4*a*b*x^(8/3) + 3/5*a^2*x^(5/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{2/3} (a+b x)^2 \, dx=\frac {3\,x^{5/3}\,\left (44\,a^2+55\,a\,b\,x+20\,b^2\,x^2\right )}{220} \]

[In]

int(x^(2/3)*(a + b*x)^2,x)

[Out]

(3*x^(5/3)*(44*a^2 + 20*b^2*x^2 + 55*a*b*x))/220