\(\int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx\) [65]

Optimal result

Integrand size = 19, antiderivative size = 18 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 267} \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

[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 267

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*} \text {integral}& = \text {Subst}\left (\int \frac {2 x}{\sqrt [3]{a^2-4 x^2}} \, dx,x,\sin (x)\right ) \\ & = 2 \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

[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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {3 \left (a^{2}-4 \left (\sin ^{2}\left (x \right )\right )\right )^{\frac {2}{3}}}{8}\)	\(15\)
default	\(-\frac {3 \left (a^{2}-4 \left (\sin ^{2}\left (x \right )\right )\right )^{\frac {2}{3}}}{8}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \, {\left (a^{2} + 4 \, \cos \left (x\right )^{2} - 4\right )}^{\frac {2}{3}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=- \frac {3 \left (a^{2} - 4 \sin ^{2}{\left (x \right )}\right )^{\frac {2}{3}}}{8} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \, {\left (a^{2} - 4 \, \sin \left (x\right )^{2}\right )}^{\frac {2}{3}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3}{8} \, {\left (a^{2} - 4 \, \sin \left (x\right )^{2}\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

-3/8*(a^2 - 4*sin(x)^2)^(2/3)

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx=-\frac {3\,{\left (a^2-4\,{\sin \left (x\right )}^2\right )}^{2/3}}{8} \]

[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