3.5.7 \(\int \frac {x^2}{(a+b x)^{2/3}} \, dx\)

Optimal. Leaf size=51 \[ \frac {3 a^2 \sqrt [3]{a+b x}}{b^3}+\frac {3 (a+b x)^{7/3}}{7 b^3}-\frac {3 a (a+b x)^{4/3}}{2 b^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 51, 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 \sqrt [3]{a+b x}}{b^3}+\frac {3 (a+b x)^{7/3}}{7 b^3}-\frac {3 a (a+b x)^{4/3}}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x)^(2/3),x]

[Out]

(3*a^2*(a + b*x)^(1/3))/b^3 - (3*a*(a + b*x)^(4/3))/(2*b^3) + (3*(a + b*x)^(7/3))/(7*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}{(a+b x)^{2/3}} \, dx &=\int \left (\frac {a^2}{b^2 (a+b x)^{2/3}}-\frac {2 a \sqrt [3]{a+b x}}{b^2}+\frac {(a+b x)^{4/3}}{b^2}\right ) \, dx\\ &=\frac {3 a^2 \sqrt [3]{a+b x}}{b^3}-\frac {3 a (a+b x)^{4/3}}{2 b^3}+\frac {3 (a+b x)^{7/3}}{7 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.69 \begin {gather*} \frac {3 \sqrt [3]{a+b x} \left (9 a^2-3 a b x+2 b^2 x^2\right )}{14 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x)^(2/3),x]

[Out]

(3*(a + b*x)^(1/3)*(9*a^2 - 3*a*b*x + 2*b^2*x^2))/(14*b^3)

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IntegrateAlgebraic [A]  time = 0.02, size = 45, normalized size = 0.88 \begin {gather*} \frac {3 \left (14 a^2 \sqrt [3]{a+b x}+2 (a+b x)^{7/3}-7 a (a+b x)^{4/3}\right )}{14 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(14*a^2*(a + b*x)^(1/3) - 7*a*(a + b*x)^(4/3) + 2*(a + b*x)^(7/3)))/(14*b^3)

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fricas [A]  time = 0.83, size = 31, normalized size = 0.61 \begin {gather*} \frac {3 \, {\left (2 \, b^{2} x^{2} - 3 \, a b x + 9 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {1}{3}}}{14 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(2*b^2*x^2 - 3*a*b*x + 9*a^2)*(b*x + a)^(1/3)/b^3

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giac [A]  time = 0.90, size = 37, normalized size = 0.73 \begin {gather*} \frac {3 \, {\left (2 \, {\left (b x + a\right )}^{\frac {7}{3}} - 7 \, {\left (b x + a\right )}^{\frac {4}{3}} a + 14 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{2}\right )}}{14 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(2*(b*x + a)^(7/3) - 7*(b*x + a)^(4/3)*a + 14*(b*x + a)^(1/3)*a^2)/b^3

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maple [A]  time = 0.00, size = 32, normalized size = 0.63 \begin {gather*} \frac {3 \left (b x +a \right )^{\frac {1}{3}} \left (2 b^{2} x^{2}-3 a b x +9 a^{2}\right )}{14 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(b*x+a)^(1/3)*(2*b^2*x^2-3*a*b*x+9*a^2)/b^3

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maxima [A]  time = 1.31, size = 41, normalized size = 0.80 \begin {gather*} \frac {3 \, {\left (b x + a\right )}^{\frac {7}{3}}}{7 \, b^{3}} - \frac {3 \, {\left (b x + a\right )}^{\frac {4}{3}} a}{2 \, b^{3}} + \frac {3 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{2}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*(b*x + a)^(7/3)/b^3 - 3/2*(b*x + a)^(4/3)*a/b^3 + 3*(b*x + a)^(1/3)*a^2/b^3

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mupad [B]  time = 0.04, size = 37, normalized size = 0.73 \begin {gather*} \frac {6\,{\left (a+b\,x\right )}^{7/3}-21\,a\,{\left (a+b\,x\right )}^{4/3}+42\,a^2\,{\left (a+b\,x\right )}^{1/3}}{14\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x)^(2/3),x)

[Out]

(6*(a + b*x)^(7/3) - 21*a*(a + b*x)^(4/3) + 42*a^2*(a + b*x)^(1/3))/(14*b^3)

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sympy [B]  time = 1.82, size = 600, normalized size = 11.76 \begin {gather*} \frac {27 a^{\frac {31}{3}} \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} - \frac {27 a^{\frac {31}{3}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} + \frac {72 a^{\frac {28}{3}} b x \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} - \frac {81 a^{\frac {28}{3}} b x}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} + \frac {60 a^{\frac {25}{3}} b^{2} x^{2} \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} - \frac {81 a^{\frac {25}{3}} b^{2} x^{2}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} + \frac {18 a^{\frac {22}{3}} b^{3} x^{3} \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} - \frac {27 a^{\frac {22}{3}} b^{3} x^{3}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} + \frac {9 a^{\frac {19}{3}} b^{4} x^{4} \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 a^{5} b^{6} x^{3}} + \frac {6 a^{\frac {16}{3}} b^{5} x^{5} \sqrt [3]{1 + \frac {b x}{a}}}{14 a^{8} b^{3} + 42 a^{7} b^{4} x + 42 a^{6} b^{5} x^{2} + 14 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)**(2/3),x)

[Out]

27*a**(31/3)*(1 + b*x/a)**(1/3)/(14*a**8*b**3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) - 27*a
**(31/3)/(14*a**8*b**3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) + 72*a**(28/3)*b*x*(1 + b*x/a
)**(1/3)/(14*a**8*b**3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) - 81*a**(28/3)*b*x/(14*a**8*b
**3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) + 60*a**(25/3)*b**2*x**2*(1 + b*x/a)**(1/3)/(14*
a**8*b**3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) - 81*a**(25/3)*b**2*x**2/(14*a**8*b**3 + 4
2*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) + 18*a**(22/3)*b**3*x**3*(1 + b*x/a)**(1/3)/(14*a**8*b*
*3 + 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) - 27*a**(22/3)*b**3*x**3/(14*a**8*b**3 + 42*a**7*
b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) + 9*a**(19/3)*b**4*x**4*(1 + b*x/a)**(1/3)/(14*a**8*b**3 + 42*
a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3) + 6*a**(16/3)*b**5*x**5*(1 + b*x/a)**(1/3)/(14*a**8*b**3
+ 42*a**7*b**4*x + 42*a**6*b**5*x**2 + 14*a**5*b**6*x**3)

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