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

Optimal. Leaf size=53 \[ \frac {3 a^2 (a+b x)^{2/3}}{2 b^3}+\frac {3 (a+b x)^{8/3}}{8 b^3}-\frac {6 a (a+b x)^{5/3}}{5 b^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 53, 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 (a+b x)^{2/3}}{2 b^3}+\frac {3 (a+b x)^{8/3}}{8 b^3}-\frac {6 a (a+b x)^{5/3}}{5 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x)^(1/3),x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.66 \begin {gather*} \frac {3 (a+b x)^{2/3} \left (9 a^2-6 a b x+5 b^2 x^2\right )}{40 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x)^(1/3),x]

[Out]

(3*(a + b*x)^(2/3)*(9*a^2 - 6*a*b*x + 5*b^2*x^2))/(40*b^3)

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IntegrateAlgebraic [A]  time = 0.02, size = 45, normalized size = 0.85 \begin {gather*} \frac {3 \left (20 a^2 (a+b x)^{2/3}+5 (a+b x)^{8/3}-16 a (a+b x)^{5/3}\right )}{40 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/(a + b*x)^(1/3),x]

[Out]

(3*(20*a^2*(a + b*x)^(2/3) - 16*a*(a + b*x)^(5/3) + 5*(a + b*x)^(8/3)))/(40*b^3)

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fricas [A]  time = 0.76, size = 31, normalized size = 0.58 \begin {gather*} \frac {3 \, {\left (5 \, b^{2} x^{2} - 6 \, a b x + 9 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{40 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(5*b^2*x^2 - 6*a*b*x + 9*a^2)*(b*x + a)^(2/3)/b^3

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giac [A]  time = 0.91, size = 37, normalized size = 0.70 \begin {gather*} \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {8}{3}} - 16 \, {\left (b x + a\right )}^{\frac {5}{3}} a + 20 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{2}\right )}}{40 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(5*(b*x + a)^(8/3) - 16*(b*x + a)^(5/3)*a + 20*(b*x + a)^(2/3)*a^2)/b^3

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maple [A]  time = 0.00, size = 32, normalized size = 0.60 \begin {gather*} \frac {3 \left (b x +a \right )^{\frac {2}{3}} \left (5 b^{2} x^{2}-6 a b x +9 a^{2}\right )}{40 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(b*x+a)^(2/3)*(5*b^2*x^2-6*a*b*x+9*a^2)/b^3

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maxima [A]  time = 1.38, size = 41, normalized size = 0.77 \begin {gather*} \frac {3 \, {\left (b x + a\right )}^{\frac {8}{3}}}{8 \, b^{3}} - \frac {6 \, {\left (b x + a\right )}^{\frac {5}{3}} a}{5 \, b^{3}} + \frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{2}}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 37, normalized size = 0.70 \begin {gather*} \frac {15\,{\left (a+b\,x\right )}^{8/3}-48\,a\,{\left (a+b\,x\right )}^{5/3}+60\,a^2\,{\left (a+b\,x\right )}^{2/3}}{40\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x)^(1/3),x)

[Out]

(15*(a + b*x)^(8/3) - 48*a*(a + b*x)^(5/3) + 60*a^2*(a + b*x)^(2/3))/(40*b^3)

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sympy [B]  time = 1.77, size = 600, normalized size = 11.32 \begin {gather*} \frac {27 a^{\frac {32}{3}} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} - \frac {27 a^{\frac {32}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} + \frac {63 a^{\frac {29}{3}} b x \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} - \frac {81 a^{\frac {29}{3}} b x}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} + \frac {42 a^{\frac {26}{3}} b^{2} x^{2} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} - \frac {81 a^{\frac {26}{3}} b^{2} x^{2}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} + \frac {18 a^{\frac {23}{3}} b^{3} x^{3} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} - \frac {27 a^{\frac {23}{3}} b^{3} x^{3}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} + \frac {27 a^{\frac {20}{3}} b^{4} x^{4} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} + \frac {15 a^{\frac {17}{3}} b^{5} x^{5} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{8} b^{3} + 120 a^{7} b^{4} x + 120 a^{6} b^{5} x^{2} + 40 a^{5} b^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27*a**(32/3)*(1 + b*x/a)**(2/3)/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 27
*a**(32/3)/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 63*a**(29/3)*b*x*(1 + b
*x/a)**(2/3)/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 81*a**(29/3)*b*x/(40*
a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 42*a**(26/3)*b**2*x**2*(1 + b*x/a)**(2
/3)/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 81*a**(26/3)*b**2*x**2/(40*a**
8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 18*a**(23/3)*b**3*x**3*(1 + b*x/a)**(2/3)
/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) - 27*a**(23/3)*b**3*x**3/(40*a**8*b
**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 27*a**(20/3)*b**4*x**4*(1 + b*x/a)**(2/3)/(4
0*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3) + 15*a**(17/3)*b**5*x**5*(1 + b*x/a)**
(2/3)/(40*a**8*b**3 + 120*a**7*b**4*x + 120*a**6*b**5*x**2 + 40*a**5*b**6*x**3)

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