3.728 \(\int \frac{x^3}{(a+b x^2)^{4/3}} \, dx\)

Optimal. Leaf size=38 \[ \frac{3 a}{2 b^2 \sqrt [3]{a+b x^2}}+\frac{3 \left (a+b x^2\right )^{2/3}}{4 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0234592, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{3 a}{2 b^2 \sqrt [3]{a+b x^2}}+\frac{3 \left (a+b x^2\right )^{2/3}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2)^(4/3),x]

[Out]

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

Mathematica [A]  time = 0.0113592, size = 27, normalized size = 0.71 \[ \frac{3 \left (3 a+b x^2\right )}{4 b^2 \sqrt [3]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^2)^(4/3),x]

[Out]

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

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Maple [A]  time = 0.004, size = 24, normalized size = 0.6 \begin{align*}{\frac{3\,b{x}^{2}+9\,a}{4\,{b}^{2}}{\frac{1}{\sqrt [3]{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.24349, size = 41, normalized size = 1.08 \begin{align*} \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{2}{3}}}{4 \, b^{2}} + \frac{3 \, a}{2 \,{\left (b x^{2} + a\right )}^{\frac{1}{3}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(4/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72117, size = 74, normalized size = 1.95 \begin{align*} \frac{3 \,{\left (b x^{2} + 3 \, a\right )}{\left (b x^{2} + a\right )}^{\frac{2}{3}}}{4 \,{\left (b^{3} x^{2} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(4/3),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.648634, size = 46, normalized size = 1.21 \begin{align*} \begin{cases} \frac{9 a}{4 b^{2} \sqrt [3]{a + b x^{2}}} + \frac{3 x^{2}}{4 b \sqrt [3]{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 a^{\frac{4}{3}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)**(4/3),x)

[Out]

Piecewise((9*a/(4*b**2*(a + b*x**2)**(1/3)) + 3*x**2/(4*b*(a + b*x**2)**(1/3)), Ne(b, 0)), (x**4/(4*a**(4/3)),
True))

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Giac [A]  time = 1.99315, size = 36, normalized size = 0.95 \begin{align*} \frac{3 \,{\left ({\left (b x^{2} + a\right )}^{\frac{2}{3}} + \frac{2 \, a}{{\left (b x^{2} + a\right )}^{\frac{1}{3}}}\right )}}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(4/3),x, algorithm="giac")

[Out]

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