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

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

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 \[ -\frac {3}{8} \left (a^2-4 \sin ^2(x)\right )^{2/3} \]

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin (2 x)}{\sqrt [3]{a^2-4 \sin ^2(x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.00, size = 15, normalized size = 0.83 \[ -\frac {3}{8} \, {\left (a^{2} + 4 \, \cos \relax (x)^{2} - 4\right )}^{\frac {2}{3}} \]

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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giac [A]  time = 0.92, size = 14, normalized size = 0.78 \[ -\frac {3}{8} \, {\left (a^{2} - 4 \, \sin \relax (x)^{2}\right )}^{\frac {2}{3}} \]

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 - 4*sin(x)^2)^(2/3)

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maple [A]  time = 0.13, size = 15, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.43, size = 14, normalized size = 0.78 \[ -\frac {3}{8} \, {\left (a^{2} - 4 \, \sin \relax (x)^{2}\right )}^{\frac {2}{3}} \]

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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mupad [B]  time = 0.30, size = 14, normalized size = 0.78 \[ -\frac {3\,{\left (a^2-4\,{\sin \relax (x)}^2\right )}^{2/3}}{8} \]

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*(a^2 - 4*sin(x)^2)^(2/3))/8

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sympy [A]  time = 2.29, size = 17, normalized size = 0.94 \[ - \frac {3 \left (a^{2} - 4 \sin ^{2}{\relax (x )}\right )^{\frac {2}{3}}}{8} \]

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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