3.664 \(\int \frac{(a+b x)^2}{x^{4/3}} \, dx\)

Optimal. Leaf size=32 \[ -\frac{3 a^2}{\sqrt [3]{x}}+3 a b x^{2/3}+\frac{3}{5} b^2 x^{5/3} \]

[Out]

(-3*a^2)/x^(1/3) + 3*a*b*x^(2/3) + (3*b^2*x^(5/3))/5

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Rubi [A]  time = 0.0067156, antiderivative size = 32, 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} \[ -\frac{3 a^2}{\sqrt [3]{x}}+3 a b x^{2/3}+\frac{3}{5} b^2 x^{5/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/x^(4/3),x]

[Out]

(-3*a^2)/x^(1/3) + 3*a*b*x^(2/3) + (3*b^2*x^(5/3))/5

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^{4/3}} \, dx &=\int \left (\frac{a^2}{x^{4/3}}+\frac{2 a b}{\sqrt [3]{x}}+b^2 x^{2/3}\right ) \, dx\\ &=-\frac{3 a^2}{\sqrt [3]{x}}+3 a b x^{2/3}+\frac{3}{5} b^2 x^{5/3}\\ \end{align*}

Mathematica [A]  time = 0.0075086, size = 27, normalized size = 0.84 \[ \frac{3 \left (-5 a^2+5 a b x+b^2 x^2\right )}{5 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/x^(4/3),x]

[Out]

(3*(-5*a^2 + 5*a*b*x + b^2*x^2))/(5*x^(1/3))

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Maple [A]  time = 0.003, size = 25, normalized size = 0.8 \begin{align*} -{\frac{-3\,{b}^{2}{x}^{2}-15\,abx+15\,{a}^{2}}{5}{\frac{1}{\sqrt [3]{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^(4/3),x)

[Out]

-3/5*(-b^2*x^2-5*a*b*x+5*a^2)/x^(1/3)

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Maxima [A]  time = 1.33996, size = 32, normalized size = 1. \begin{align*} \frac{3}{5} \, b^{2} x^{\frac{5}{3}} + 3 \, a b x^{\frac{2}{3}} - \frac{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^(4/3),x, algorithm="maxima")

[Out]

3/5*b^2*x^(5/3) + 3*a*b*x^(2/3) - 3*a^2/x^(1/3)

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Fricas [A]  time = 1.4694, size = 55, normalized size = 1.72 \begin{align*} \frac{3 \,{\left (b^{2} x^{2} + 5 \, a b x - 5 \, a^{2}\right )}}{5 \, x^{\frac{1}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^(4/3),x, algorithm="fricas")

[Out]

3/5*(b^2*x^2 + 5*a*b*x - 5*a^2)/x^(1/3)

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Sympy [C]  time = 2.22666, size = 1828, normalized size = 57.12 \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**(4/3),x)

[Out]

Piecewise((-27*a**(29/3)*b**(1/3)*(-1 + b*(a/b + x)/a)**(2/3)*exp(I*pi/3)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/
b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) - 27*a**(29
/3)*b**(1/3)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) +
5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) + 63*a**(26/3)*b**(4/3)*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)*exp(I*pi/3
)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3
*(a/b + x)**3*exp(I*pi/3)) + 81*a**(26/3)*b**(4/3)*(a/b + x)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*
pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) - 42*a**(23/3)*b**(7/3)*
(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**2*exp(I*pi/3)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) -
15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) - 81*a**(23/3)*b**(7/3)*(a/b + x
)**2/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b
**3*(a/b + x)**3*exp(I*pi/3)) + 3*a**(20/3)*b**(10/3)*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**3*exp(I*pi/3)/(-5
*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b
 + x)**3*exp(I*pi/3)) + 27*a**(20/3)*b**(10/3)*(a/b + x)**3/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*p
i/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) + 3*a**(17/3)*b**(13/3)*(
-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**4*exp(I*pi/3)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 1
5*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)), Abs(b*(a/b + x))/Abs(a) > 1), (2
7*a**(29/3)*b**(1/3)*(1 - b*(a/b + x)/a)**(2/3)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a*
*6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) - 27*a**(29/3)*b**(1/3)/(-5*a**8*exp(
I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*e
xp(I*pi/3)) - 63*a**(26/3)*b**(4/3)*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b
 + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) + 81*a**(26/
3)*b**(4/3)*(a/b + x)/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I
*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) + 42*a**(23/3)*b**(7/3)*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)**2
/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*
(a/b + x)**3*exp(I*pi/3)) - 81*a**(23/3)*b**(7/3)*(a/b + x)**2/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(
I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) - 3*a**(20/3)*b**(10/3
)*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)**3/(-5*a**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**
2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)) + 27*a**(20/3)*b**(10/3)*(a/b + x)**3/(-5*a
**8*exp(I*pi/3) + 15*a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b +
 x)**3*exp(I*pi/3)) - 3*a**(17/3)*b**(13/3)*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)**4/(-5*a**8*exp(I*pi/3) + 15*
a**7*b*(a/b + x)*exp(I*pi/3) - 15*a**6*b**2*(a/b + x)**2*exp(I*pi/3) + 5*a**5*b**3*(a/b + x)**3*exp(I*pi/3)),
True))

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Giac [A]  time = 1.05394, size = 32, normalized size = 1. \begin{align*} \frac{3}{5} \, b^{2} x^{\frac{5}{3}} + 3 \, a b x^{\frac{2}{3}} - \frac{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^(4/3),x, algorithm="giac")

[Out]

3/5*b^2*x^(5/3) + 3*a*b*x^(2/3) - 3*a^2/x^(1/3)