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

Optimal. Leaf size=34 \[ 3 a^2 \sqrt [3]{x}+\frac{3}{2} a b x^{4/3}+\frac{3}{7} b^2 x^{7/3} \]

[Out]

3*a^2*x^(1/3) + (3*a*b*x^(4/3))/2 + (3*b^2*x^(7/3))/7

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Rubi [A]  time = 0.0066598, antiderivative size = 34, 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} \[ 3 a^2 \sqrt [3]{x}+\frac{3}{2} a b x^{4/3}+\frac{3}{7} b^2 x^{7/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/x^(2/3),x]

[Out]

3*a^2*x^(1/3) + (3*a*b*x^(4/3))/2 + (3*b^2*x^(7/3))/7

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 \frac{(a+b x)^2}{x^{2/3}} \, dx &=\int \left (\frac{a^2}{x^{2/3}}+2 a b \sqrt [3]{x}+b^2 x^{4/3}\right ) \, dx\\ &=3 a^2 \sqrt [3]{x}+\frac{3}{2} a b x^{4/3}+\frac{3}{7} b^2 x^{7/3}\\ \end{align*}

Mathematica [A]  time = 0.007176, size = 28, normalized size = 0.82 \[ \frac{3}{14} \sqrt [3]{x} \left (14 a^2+7 a b x+2 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/x^(2/3),x]

[Out]

(3*x^(1/3)*(14*a^2 + 7*a*b*x + 2*b^2*x^2))/14

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*}{\frac{6\,{b}^{2}{x}^{2}+21\,abx+42\,{a}^{2}}{14}\sqrt [3]{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*x^(1/3)*(2*b^2*x^2+7*a*b*x+14*a^2)

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Maxima [A]  time = 1.50062, size = 32, normalized size = 0.94 \begin{align*} \frac{3}{7} \, b^{2} x^{\frac{7}{3}} + \frac{3}{2} \, a b x^{\frac{4}{3}} + 3 \, a^{2} x^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*b^2*x^(7/3) + 3/2*a*b*x^(4/3) + 3*a^2*x^(1/3)

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Fricas [A]  time = 1.50627, size = 61, normalized size = 1.79 \begin{align*} \frac{3}{14} \,{\left (2 \, b^{2} x^{2} + 7 \, a b x + 14 \, a^{2}\right )} x^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(2*b^2*x^2 + 7*a*b*x + 14*a^2)*x^(1/3)

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Sympy [C]  time = 2.39756, size = 1742, normalized size = 51.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-27*a**(31/3)*(-1 + b*(a/b + x)/a)**(1/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6
*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 27*a**(31/3)*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a*
*7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 72*a**(28/3)*b*(-1 +
 b*(a/b + x)/a)**(1/3)*(a/b + x)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)*
*2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 81*a**(28/3)*b*(a/b + x)*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4
/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 60*a**(25/3)*b**2*(-1 + b*(a
/b + x)/a)**(1/3)*(a/b + x)**2/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2
 + 14*a**5*b**(10/3)*(a/b + x)**3) + 81*a**(25/3)*b**2*(a/b + x)**2*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b
**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 18*a**(22/3)*b**3*(-1 +
b*(a/b + x)/a)**(1/3)*(a/b + x)**3/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x
)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 27*a**(22/3)*b**3*(a/b + x)**3*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a*
*7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 9*a**(19/3)*b**4*(-1
 + b*(a/b + x)/a)**(1/3)*(a/b + x)**4/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b
+ x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 6*a**(16/3)*b**5*(-1 + b*(a/b + x)/a)**(1/3)*(a/b + x)**5/(-14*a**
8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3), Abs
(b*(a/b + x))/Abs(a) > 1), (-27*a**(31/3)*(1 - b*(a/b + x)/a)**(1/3)*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*
b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 27*a**(31/3)*exp(I*pi/3
)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x
)**3) + 72*a**(28/3)*b*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*
(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 81*a**(28/3)*b*(a/b + x)*exp(I*p
i/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b
+ x)**3) - 60*a**(25/3)*b**2*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**2*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*
b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) + 81*a**(25/3)*b**2*(a/b
+ x)**2*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*
b**(10/3)*(a/b + x)**3) + 18*a**(22/3)*b**3*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**3*exp(I*pi/3)/(-14*a**8*b**(
1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 27*a**(2
2/3)*b**3*(a/b + x)**3*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x
)**2 + 14*a**5*b**(10/3)*(a/b + x)**3) - 9*a**(19/3)*b**4*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**4*exp(I*pi/3)/
(-14*a**8*b**(1/3) + 42*a**7*b**(4/3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)*
*3) + 6*a**(16/3)*b**5*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**5*exp(I*pi/3)/(-14*a**8*b**(1/3) + 42*a**7*b**(4/
3)*(a/b + x) - 42*a**6*b**(7/3)*(a/b + x)**2 + 14*a**5*b**(10/3)*(a/b + x)**3), True))

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Giac [A]  time = 1.0645, size = 32, normalized size = 0.94 \begin{align*} \frac{3}{7} \, b^{2} x^{\frac{7}{3}} + \frac{3}{2} \, a b x^{\frac{4}{3}} + 3 \, a^{2} x^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*b^2*x^(7/3) + 3/2*a*b*x^(4/3) + 3*a^2*x^(1/3)