\(\int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx\) [662]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{2} a^2 x^{2/3}+\frac {6}{5} a b x^{5/3}+\frac {3}{8} b^2 x^{8/3} \]

[Out]

3/2*a^2*x^(2/3)+6/5*a*b*x^(5/3)+3/8*b^2*x^(8/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, 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} \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{2} a^2 x^{2/3}+\frac {6}{5} a b x^{5/3}+\frac {3}{8} b^2 x^{8/3} \]

[In]

Int[(a + b*x)^2/x^(1/3),x]

[Out]

(3*a^2*x^(2/3))/2 + (6*a*b*x^(5/3))/5 + (3*b^2*x^(8/3))/8

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}{\sqrt [3]{x}}+2 a b x^{2/3}+b^2 x^{5/3}\right ) \, dx \\ & = \frac {3}{2} a^2 x^{2/3}+\frac {6}{5} a b x^{5/3}+\frac {3}{8} b^2 x^{8/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{40} x^{2/3} \left (20 a^2+16 a b x+5 b^2 x^2\right ) \]

[In]

Integrate[(a + b*x)^2/x^(1/3),x]

[Out]

(3*x^(2/3)*(20*a^2 + 16*a*b*x + 5*b^2*x^2))/40

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67

method	result	size
trager	\(\left (\frac {3}{8} b^{2} x^{2}+\frac {6}{5} a b x +\frac {3}{2} a^{2}\right ) x^{\frac {2}{3}}\)	\(24\)
gosper	\(\frac {3 x^{\frac {2}{3}} \left (5 b^{2} x^{2}+16 a b x +20 a^{2}\right )}{40}\)	\(25\)
derivativedivides	\(\frac {3 a^{2} x^{\frac {2}{3}}}{2}+\frac {6 a b \,x^{\frac {5}{3}}}{5}+\frac {3 b^{2} x^{\frac {8}{3}}}{8}\)	\(25\)
default	\(\frac {3 a^{2} x^{\frac {2}{3}}}{2}+\frac {6 a b \,x^{\frac {5}{3}}}{5}+\frac {3 b^{2} x^{\frac {8}{3}}}{8}\)	\(25\)
risch	\(\frac {3 x^{\frac {2}{3}} \left (5 b^{2} x^{2}+16 a b x +20 a^{2}\right )}{40}\)	\(25\)

[In]

int((b*x+a)^2/x^(1/3),x,method=_RETURNVERBOSE)

[Out]

(3/8*b^2*x^2+6/5*a*b*x+3/2*a^2)*x^(2/3)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{40} \, {\left (5 \, b^{2} x^{2} + 16 \, a b x + 20 \, a^{2}\right )} x^{\frac {2}{3}} \]

[In]

integrate((b*x+a)^2/x^(1/3),x, algorithm="fricas")

[Out]

3/40*(5*b^2*x^2 + 16*a*b*x + 20*a^2)*x^(2/3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.71 (sec) , antiderivative size = 1765, normalized size of antiderivative = 49.03 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\text {Too large to display} \]

[In]

integrate((b*x+a)**2/x**(1/3),x)

[Out]

Piecewise((-27*a**(32/3)*(-1 + b*(a/b + x)/a)**(2/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a*
*6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 27*a**(32/3)*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 1
20*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 63*a**(29/3)*b
*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a
/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 81*a**(29/3)*b*(a/b + x)*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120
*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 42*a**(26/3)*b**
2*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**2/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3
)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 81*a**(26/3)*b**2*(a/b + x)**2*exp(2*I*pi/3)/(-40*a**8*b**(
2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 18*a**
(23/3)*b**3*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**3/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a*
*6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 27*a**(23/3)*b**3*(a/b + x)**3*exp(2*I*pi/3)/(-40
*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3
) - 27*a**(20/3)*b**4*(-1 + b*(a/b + x)/a)**(2/3)*(a/b + x)**4/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x
) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 15*a**(17/3)*b**5*(-1 + b*(a/b + x)/a)*
*(2/3)*(a/b + x)**5/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**
5*b**(11/3)*(a/b + x)**3), Abs(b*(a/b + x)/a) > 1), (-27*a**(32/3)*(1 - b*(a/b + x)/a)**(2/3)*exp(2*I*pi/3)/(-
40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)*
*3) + 27*a**(32/3)*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x
)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 63*a**(29/3)*b*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)*exp(2*I*pi/3)/(-4
0*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**
3) - 81*a**(29/3)*b*(a/b + x)*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/
3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 42*a**(26/3)*b**2*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)**2*
exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(
11/3)*(a/b + x)**3) + 81*a**(26/3)*b**2*(a/b + x)**2*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b
+ x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 18*a**(23/3)*b**3*(1 - b*(a/b + x)/
a)**(2/3)*(a/b + x)**3*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b
+ x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 27*a**(23/3)*b**3*(a/b + x)**3*exp(2*I*pi/3)/(-40*a**8*b**(2/3) +
120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) - 27*a**(20/3)
*b**4*(1 - b*(a/b + x)/a)**(2/3)*(a/b + x)**4*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) -
120*a**6*b**(8/3)*(a/b + x)**2 + 40*a**5*b**(11/3)*(a/b + x)**3) + 15*a**(17/3)*b**5*(1 - b*(a/b + x)/a)**(2/
3)*(a/b + x)**5*exp(2*I*pi/3)/(-40*a**8*b**(2/3) + 120*a**7*b**(5/3)*(a/b + x) - 120*a**6*b**(8/3)*(a/b + x)**
2 + 40*a**5*b**(11/3)*(a/b + x)**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{8} \, b^{2} x^{\frac {8}{3}} + \frac {6}{5} \, a b x^{\frac {5}{3}} + \frac {3}{2} \, a^{2} x^{\frac {2}{3}} \]

[In]

integrate((b*x+a)^2/x^(1/3),x, algorithm="maxima")

[Out]

3/8*b^2*x^(8/3) + 6/5*a*b*x^(5/3) + 3/2*a^2*x^(2/3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3}{8} \, b^{2} x^{\frac {8}{3}} + \frac {6}{5} \, a b x^{\frac {5}{3}} + \frac {3}{2} \, a^{2} x^{\frac {2}{3}} \]

[In]

integrate((b*x+a)^2/x^(1/3),x, algorithm="giac")

[Out]

3/8*b^2*x^(8/3) + 6/5*a*b*x^(5/3) + 3/2*a^2*x^(2/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2}{\sqrt [3]{x}} \, dx=\frac {3\,x^{2/3}\,\left (20\,a^2+16\,a\,b\,x+5\,b^2\,x^2\right )}{40} \]

[In]

int((a + b*x)^2/x^(1/3),x)

[Out]

(3*x^(2/3)*(20*a^2 + 5*b^2*x^2 + 16*a*b*x))/40