3.65 \(\int \frac{\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx\)

Optimal. Leaf size=18 \[ -\frac{3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

[Out]

(-3*(a^2 - 4*Sin[x]^2)^(2/3))/8

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Rubi [A]  time = 0.044924, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {12, 261} \[ -\frac{3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/(a^2 - 4*Sin[x]^2)^(1/3),x]

[Out]

(-3*(a^2 - 4*Sin[x]^2)^(2/3))/8

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{2 x}{\sqrt [3]{a^2-4 x^2}} \, dx,x,\sin (x)\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x}{\sqrt [3]{a^2-4 x^2}} \, dx,x,\sin (x)\right )\\ &=-\frac{3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3}\\ \end{align*}

Mathematica [A]  time = 0.0185391, size = 18, normalized size = 1. \[ -\frac{3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]/(a^2 - 4*Sin[x]^2)^(1/3),x]

[Out]

(-3*(a^2 - 4*Sin[x]^2)^(2/3))/8

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Maple [A]  time = 0.02, size = 15, normalized size = 0.8 \begin{align*} -{\frac{3}{8} \left ({a}^{2}-4\, \left ( \sin \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(a^2-4*sin(x)^2)^(1/3),x)

[Out]

-3/8*(a^2-4*sin(x)^2)^(2/3)

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Maxima [A]  time = 0.945078, size = 19, normalized size = 1.06 \begin{align*} -\frac{3}{8} \,{\left (a^{2} - 4 \, \sin \left (x\right )^{2}\right )}^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-4*sin(x)^2)^(1/3),x, algorithm="maxima")

[Out]

-3/8*(a^2 - 4*sin(x)^2)^(2/3)

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Fricas [A]  time = 2.15758, size = 47, normalized size = 2.61 \begin{align*} -\frac{3}{8} \,{\left (a^{2} + 4 \, \cos \left (x\right )^{2} - 4\right )}^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-4*sin(x)^2)^(1/3),x, algorithm="fricas")

[Out]

-3/8*(a^2 + 4*cos(x)^2 - 4)^(2/3)

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Sympy [A]  time = 2.57448, size = 17, normalized size = 0.94 \begin{align*} - \frac{3 \left (a^{2} - 4 \sin ^{2}{\left (x \right )}\right )^{\frac{2}{3}}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a**2-4*sin(x)**2)**(1/3),x)

[Out]

-3*(a**2 - 4*sin(x)**2)**(2/3)/8

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Giac [A]  time = 1.15055, size = 35, normalized size = 1.94 \begin{align*} -\frac{3}{8} \,{\left (a^{2} - \frac{16 \, \tan \left (\frac{1}{2} \, x\right )^{2}}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{2}}\right )}^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-4*sin(x)^2)^(1/3),x, algorithm="giac")

[Out]

-3/8*(a^2 - 16*tan(1/2*x)^2/(tan(1/2*x)^2 + 1)^2)^(2/3)