\(\int (a+\frac {b}{\sqrt [3]{x}})^2 \, dx\) [2407]

Optimal result

Integrand size = 11, antiderivative size = 16 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x}{a} \]

[Out]

(a+b/x^(1/3))^3*x/a

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {196, 37} \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=\frac {x \left (a+\frac {b}{\sqrt [3]{x}}\right )^3}{a} \]

[In]

Int[(a + b/x^(1/3))^2,x]

[Out]

((a + b/x^(1/3))^3*x)/a

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rubi steps \begin{align*} \text {integral}& = -\left (3 \text {Subst}\left (\int \frac {(a+b x)^2}{x^4} \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right ) \\ & = \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=3 b^2 \sqrt [3]{x}+3 a b x^{2/3}+a^2 x \]

[In]

Integrate[(a + b/x^(1/3))^2,x]

[Out]

3*b^2*x^(1/3) + 3*a*b*x^(2/3) + a^2*x

Maple [A] (verified)

Time = 3.68 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\left (b +a \,x^{\frac {1}{3}}\right )^{3}}{a}\)	\(14\)
default	\(\frac {\left (b +a \,x^{\frac {1}{3}}\right )^{3}}{a}\)	\(14\)
trager	\(a^{2} \left (-1+x \right )+3 b^{2} x^{\frac {1}{3}}+3 a b \,x^{\frac {2}{3}}\)	\(24\)

[In]

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

[Out]

(b+a*x^(1/3))^3/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=a^{2} x + 3 \, a b x^{\frac {2}{3}} + 3 \, b^{2} x^{\frac {1}{3}} \]

[In]

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

[Out]

a^2*x + 3*a*b*x^(2/3) + 3*b^2*x^(1/3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=a^{2} x + 3 a b x^{\frac {2}{3}} + 3 b^{2} \sqrt [3]{x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=a^{2} x + 3 \, a b x^{\frac {2}{3}} + 3 \, b^{2} x^{\frac {1}{3}} \]

[In]

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

[Out]

a^2*x + 3*a*b*x^(2/3) + 3*b^2*x^(1/3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=a^{2} x + 3 \, a b x^{\frac {2}{3}} + 3 \, b^{2} x^{\frac {1}{3}} \]

[In]

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

[Out]

a^2*x + 3*a*b*x^(2/3) + 3*b^2*x^(1/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 \, dx=a^2\,x+3\,b^2\,x^{1/3}+3\,a\,b\,x^{2/3} \]

[In]

int((a + b/x^(1/3))^2,x)

[Out]

a^2*x + 3*b^2*x^(1/3) + 3*a*b*x^(2/3)