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

Optimal. Leaf size=36 \[ \frac{3}{5} a^2 x^{5/3}+\frac{3}{4} a b x^{8/3}+\frac{3}{11} b^2 x^{11/3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0068258, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{3}{5} a^2 x^{5/3}+\frac{3}{4} a b x^{8/3}+\frac{3}{11} b^2 x^{11/3} \]

Antiderivative was successfully verified.

[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 43

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*} \int x^{2/3} (a+b x)^2 \, dx &=\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]  time = 0.0074095, size = 28, normalized size = 0.78 \[ \frac{3}{220} x^{5/3} \left (44 a^2+55 a b x+20 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[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]  time = 0.004, size = 25, normalized size = 0.7 \begin{align*}{\frac{60\,{b}^{2}{x}^{2}+165\,abx+132\,{a}^{2}}{220}{x}^{{\frac{5}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.04631, size = 32, normalized size = 0.89 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.51761, size = 70, normalized size = 1.94 \begin{align*} \frac{3}{220} \,{\left (20 \, b^{2} x^{3} + 55 \, a b x^{2} + 44 \, a^{2} x\right )} x^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 1.29595, size = 34, normalized size = 0.94 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Giac [A]  time = 1.06283, size = 32, normalized size = 0.89 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)