3.2298 \(\int (a+b \sqrt [3]{x})^2 \, dx\)

Optimal. Leaf size=29 \[ a^2 x+\frac{3}{2} a b x^{4/3}+\frac{3}{5} b^2 x^{5/3} \]

[Out]

a^2*x + (3*a*b*x^(4/3))/2 + (3*b^2*x^(5/3))/5

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Rubi [A]  time = 0.0155383, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 43} \[ a^2 x+\frac{3}{2} a b x^{4/3}+\frac{3}{5} b^2 x^{5/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^2,x]

[Out]

a^2*x + (3*a*b*x^(4/3))/2 + (3*b^2*x^(5/3))/5

Rule 190

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]

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

Mathematica [A]  time = 0.0126745, size = 29, normalized size = 1. \[ a^2 x+\frac{3}{2} a b x^{4/3}+\frac{3}{5} b^2 x^{5/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^2,x]

[Out]

a^2*x + (3*a*b*x^(4/3))/2 + (3*b^2*x^(5/3))/5

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Maple [A]  time = 0.001, size = 22, normalized size = 0.8 \begin{align*}{a}^{2}x+{\frac{3\,ab}{2}{x}^{{\frac{4}{3}}}}+{\frac{3\,{b}^{2}}{5}{x}^{{\frac{5}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x+3/2*a*b*x^(4/3)+3/5*b^2*x^(5/3)

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Maxima [A]  time = 1.00659, size = 28, normalized size = 0.97 \begin{align*} \frac{3}{5} \, b^{2} x^{\frac{5}{3}} + \frac{3}{2} \, a b x^{\frac{4}{3}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*b^2*x^(5/3) + 3/2*a*b*x^(4/3) + a^2*x

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Fricas [A]  time = 1.51137, size = 58, normalized size = 2. \begin{align*} \frac{3}{5} \, b^{2} x^{\frac{5}{3}} + \frac{3}{2} \, a b x^{\frac{4}{3}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*b^2*x^(5/3) + 3/2*a*b*x^(4/3) + a^2*x

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Sympy [A]  time = 1.35243, size = 27, normalized size = 0.93 \begin{align*} a^{2} x + \frac{3 a b x^{\frac{4}{3}}}{2} + \frac{3 b^{2} x^{\frac{5}{3}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x + 3*a*b*x**(4/3)/2 + 3*b**2*x**(5/3)/5

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Giac [A]  time = 1.09883, size = 28, normalized size = 0.97 \begin{align*} \frac{3}{5} \, b^{2} x^{\frac{5}{3}} + \frac{3}{2} \, a b x^{\frac{4}{3}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*b^2*x^(5/3) + 3/2*a*b*x^(4/3) + a^2*x