3.7.64 \(\int \frac {(a+b x)^2}{x^{4/3}} \, dx\)
Optimal. Leaf size=32 \[ -\frac {3 a^2}{\sqrt [3]{x}}+3 a b x^{2/3}+\frac {3}{5} b^2 x^{5/3} \]
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Rubi [A] time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00,
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} \begin {gather*} -\frac {3 a^2}{\sqrt [3]{x}}+3 a b x^{2/3}+\frac {3}{5} b^2 x^{5/3} \end {gather*}
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*}
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Mathematica [A] time = 0.01, size = 27, normalized size = 0.84 \begin {gather*} \frac {3 \left (-5 a^2+5 a b x+b^2 x^2\right )}{5 \sqrt [3]{x}} \end {gather*}
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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IntegrateAlgebraic [A] time = 0.02, size = 27, normalized size = 0.84 \begin {gather*} \frac {3 \left (-5 a^2+5 a b x+b^2 x^2\right )}{5 \sqrt [3]{x}} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(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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fricas [A] time = 1.29, size = 23, normalized size = 0.72 \begin {gather*} \frac {3 \, {\left (b^{2} x^{2} + 5 \, a b x - 5 \, a^{2}\right )}}{5 \, x^{\frac {1}{3}}} \end {gather*}
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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giac [A] time = 1.17, size = 24, normalized size = 0.75 \begin {gather*} \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 {gather*}
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)
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maple [A] time = 0.00, size = 25, normalized size = 0.78 \begin {gather*} -\frac {3 \left (-b^{2} x^{2}-5 a b x +5 a^{2}\right )}{5 x^{\frac {1}{3}}} \end {gather*}
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 = 1.36, size = 24, normalized size = 0.75 \begin {gather*} \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 {gather*}
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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mupad [B] time = 0.04, size = 24, normalized size = 0.75 \begin {gather*} \frac {-15\,a^2+15\,a\,b\,x+3\,b^2\,x^2}{5\,x^{1/3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x)^2/x^(4/3),x)
[Out]
(3*b^2*x^2 - 15*a^2 + 15*a*b*x)/(5*x^(1/3))
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sympy [C] time = 2.09, size = 1826, normalized size = 57.06
result too large to display
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)/a) > 1), (27*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*exp(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*e
xp(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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