\(\int \frac {(a+b x)^2}{x^{5/3}} \, dx\) [665]
Optimal result
Integrand size = 13, antiderivative size = 34 \[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=-\frac {3 a^2}{2 x^{2/3}}+6 a b \sqrt [3]{x}+\frac {3}{4} b^2 x^{4/3}
\]
[Out]
-3/2*a^2/x^(2/3)+6*a*b*x^(1/3)+3/4*b^2*x^(4/3)
Rubi [A] (verified)
Time = 0.00 (sec) , antiderivative size = 34, 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}
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=-\frac {3 a^2}{2 x^{2/3}}+6 a b \sqrt [3]{x}+\frac {3}{4} b^2 x^{4/3}
\]
[In]
Int[(a + b*x)^2/x^(5/3),x]
[Out]
(-3*a^2)/(2*x^(2/3)) + 6*a*b*x^(1/3) + (3*b^2*x^(4/3))/4
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 (\frac {a^2}{x^{5/3}}+\frac {2 a b}{x^{2/3}}+b^2 \sqrt [3]{x}\right ) \, dx \\ & = -\frac {3 a^2}{2 x^{2/3}}+6 a b \sqrt [3]{x}+\frac {3}{4} b^2 x^{4/3} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=-\frac {3 \left (2 a^2-8 a b x-b^2 x^2\right )}{4 x^{2/3}}
\]
[In]
Integrate[(a + b*x)^2/x^(5/3),x]
[Out]
(-3*(2*a^2 - 8*a*b*x - b^2*x^2))/(4*x^(2/3))
Maple [A] (verified)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
| | |
| method | result | size |
| | |
| gosper |
\(-\frac {3 \left (-b^{2} x^{2}-8 a b x +2 a^{2}\right )}{4 x^{\frac {2}{3}}}\) |
\(25\) |
| derivativedivides |
\(-\frac {3 a^{2}}{2 x^{\frac {2}{3}}}+6 a b \,x^{\frac {1}{3}}+\frac {3 b^{2} x^{\frac {4}{3}}}{4}\) |
\(25\) |
| default |
\(-\frac {3 a^{2}}{2 x^{\frac {2}{3}}}+6 a b \,x^{\frac {1}{3}}+\frac {3 b^{2} x^{\frac {4}{3}}}{4}\) |
\(25\) |
| trager |
\(-\frac {3 \left (-b^{2} x^{2}-8 a b x +2 a^{2}\right )}{4 x^{\frac {2}{3}}}\) |
\(25\) |
| risch |
\(-\frac {3 \left (-b^{2} x^{2}-8 a b x +2 a^{2}\right )}{4 x^{\frac {2}{3}}}\) |
\(25\) |
| | |
|
|
|
|
[In]
int((b*x+a)^2/x^(5/3),x,method=_RETURNVERBOSE)
[Out]
-3/4*(-b^2*x^2-8*a*b*x+2*a^2)/x^(2/3)
Fricas [A] (verification not implemented)
none
Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=\frac {3 \, {\left (b^{2} x^{2} + 8 \, a b x - 2 \, a^{2}\right )}}{4 \, x^{\frac {2}{3}}}
\]
[In]
integrate((b*x+a)^2/x^(5/3),x, algorithm="fricas")
[Out]
3/4*(b^2*x^2 + 8*a*b*x - 2*a^2)/x^(2/3)
Sympy [C] (verification not implemented)
Result contains complex when optimal does not.
Time = 1.39 (sec) , antiderivative size = 1957, normalized size of antiderivative = 57.56
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=\text {Too large to display}
\]
[In]
integrate((b*x+a)**2/x**(5/3),x)
[Out]
Piecewise((-27*a**(28/3)*b**(2/3)*(-1 + b*(a/b + x)/a)**(1/3)*exp(2*I*pi/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b
*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) -
27*a**(28/3)*b**(2/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*
exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 72*a**(25/3)*b**(5/3)*(-1 + b*(a/b + x)/a)**(1/3)*(a
/b + x)*exp(2*I*pi/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*e
xp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 81*a**(25/3)*b**(5/3)*(a/b + x)/(-4*a**8*exp(2*I*pi/3
) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp
(2*I*pi/3)) - 60*a**(22/3)*b**(8/3)*(-1 + b*(a/b + x)/a)**(1/3)*(a/b + x)**2*exp(2*I*pi/3)/(-4*a**8*exp(2*I*pi
/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*e
xp(2*I*pi/3)) - 81*a**(22/3)*b**(8/3)*(a/b + x)**2/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3)
- 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 12*a**(19/3)*b**(11/3)*(
-1 + b*(a/b + x)/a)**(1/3)*(a/b + x)**3*exp(2*I*pi/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/
3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 27*a**(19/3)*b**(11/3
)*(a/b + x)**3/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*
pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 3*a**(16/3)*b**(14/3)*(-1 + b*(a/b + x)/a)**(1/3)*(a/b + x)*
*4*exp(2*I*pi/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*
I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)), Abs(b*(a/b + x)/a) > 1), (27*a**(28/3)*b**(2/3)*(1 - b*(a/b
+ x)/a)**(1/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I
*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) - 27*a**(28/3)*b**(2/3)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a
/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) - 72
*a**(25/3)*b**(5/3)*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*
pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 81*a**(25/3)*b**(5
/3)*(a/b + x)/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*p
i/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 60*a**(22/3)*b**(8/3)*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**2
/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5
*b**3*(a/b + x)**3*exp(2*I*pi/3)) - 81*a**(22/3)*b**(8/3)*(a/b + x)**2/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b
+ x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) - 12*a
**(19/3)*b**(11/3)*(1 - b*(a/b + x)/a)**(1/3)*(a/b + x)**3/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*
I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) + 27*a**(19/3)*b**
(11/3)*(a/b + x)**3/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp
(2*I*pi/3) + 4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)) - 3*a**(16/3)*b**(14/3)*(1 - b*(a/b + x)/a)**(1/3)*(a/b +
x)**4/(-4*a**8*exp(2*I*pi/3) + 12*a**7*b*(a/b + x)*exp(2*I*pi/3) - 12*a**6*b**2*(a/b + x)**2*exp(2*I*pi/3) +
4*a**5*b**3*(a/b + x)**3*exp(2*I*pi/3)), True))
Maxima [A] (verification not implemented)
none
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=\frac {3}{4} \, b^{2} x^{\frac {4}{3}} + 6 \, a b x^{\frac {1}{3}} - \frac {3 \, a^{2}}{2 \, x^{\frac {2}{3}}}
\]
[In]
integrate((b*x+a)^2/x^(5/3),x, algorithm="maxima")
[Out]
3/4*b^2*x^(4/3) + 6*a*b*x^(1/3) - 3/2*a^2/x^(2/3)
Giac [A] (verification not implemented)
none
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=\frac {3}{4} \, b^{2} x^{\frac {4}{3}} + 6 \, a b x^{\frac {1}{3}} - \frac {3 \, a^{2}}{2 \, x^{\frac {2}{3}}}
\]
[In]
integrate((b*x+a)^2/x^(5/3),x, algorithm="giac")
[Out]
3/4*b^2*x^(4/3) + 6*a*b*x^(1/3) - 3/2*a^2/x^(2/3)
Mupad [B] (verification not implemented)
Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71
\[
\int \frac {(a+b x)^2}{x^{5/3}} \, dx=\frac {-6\,a^2+24\,a\,b\,x+3\,b^2\,x^2}{4\,x^{2/3}}
\]
[In]
int((a + b*x)^2/x^(5/3),x)
[Out]
(3*b^2*x^2 - 6*a^2 + 24*a*b*x)/(4*x^(2/3))