3.2278 \(\int \frac{x^2}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=40 \[ \frac{\left (a+b x^{3/2}\right )^{4/3}}{2 b^2}-\frac{2 a \sqrt [3]{a+b x^{3/2}}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0205204, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {266, 43} \[ \frac{\left (a+b x^{3/2}\right )^{4/3}}{2 b^2}-\frac{2 a \sqrt [3]{a+b x^{3/2}}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^(3/2))^(2/3),x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 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 \frac{x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{x}{(a+b x)^{2/3}} \, dx,x,x^{3/2}\right )\\ &=\frac{2}{3} \operatorname{Subst}\left (\int \left (-\frac{a}{b (a+b x)^{2/3}}+\frac{\sqrt [3]{a+b x}}{b}\right ) \, dx,x,x^{3/2}\right )\\ &=-\frac{2 a \sqrt [3]{a+b x^{3/2}}}{b^2}+\frac{\left (a+b x^{3/2}\right )^{4/3}}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0131603, size = 31, normalized size = 0.78 \[ \frac{\left (b x^{3/2}-3 a\right ) \sqrt [3]{a+b x^{3/2}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 30, normalized size = 0.8 \begin{align*} 2\,{\frac{1/4\, \left ( a+b{x}^{3/2} \right ) ^{4/3}-a\sqrt [3]{a+b{x}^{3/2}}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99922, size = 41, normalized size = 1.02 \begin{align*} \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}}}{2 \, b^{2}} - \frac{2 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.52201, size = 66, normalized size = 1.65 \begin{align*} \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}}{\left (b x^{\frac{3}{2}} - 3 \, a\right )}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.09262, size = 49, normalized size = 1.22 \begin{align*} \begin{cases} - \frac{3 a \sqrt [3]{a + b x^{\frac{3}{2}}}}{2 b^{2}} + \frac{x^{\frac{3}{2}} \sqrt [3]{a + b x^{\frac{3}{2}}}}{2 b} & \text{for}\: b \neq 0 \\\frac{x^{3}}{3 a^{\frac{2}{3}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*a*(a + b*x**(3/2))**(1/3)/(2*b**2) + x**(3/2)*(a + b*x**(3/2))**(1/3)/(2*b), Ne(b, 0)), (x**3/(3
*a**(2/3)), True))

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Giac [A]  time = 1.1171, size = 36, normalized size = 0.9 \begin{align*} \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}} - 4 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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