3.906 \(\int \frac {\sin (2 x)}{\sqrt {9-\cos ^4(x)}} \, dx\)

Optimal. Leaf size=11 \[ -\sin ^{-1}\left (\frac {\cos ^2(x)}{3}\right ) \]

[Out]

-arcsin(1/3*cos(x)^2)

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Rubi [A]  time = 0.05, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {12, 1107, 619, 216} \[ -\sin ^{-1}\left (\frac {\cos ^2(x)}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/Sqrt[9 - Cos[x]^4],x]

[Out]

-ArcSin[Cos[x]^2/3]

Rule 12

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps

\begin {align*} \int \frac {\sin (2 x)}{\sqrt {9-\cos ^4(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {2 x}{\sqrt {8+2 x^2-x^4}} \, dx,x,\sin (x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {8+2 x^2-x^4}} \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {8+2 x-x^2}} \, dx,x,\sin ^2(x)\right )\\ &=-\left (\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{36}}} \, dx,x,2 \cos ^2(x)\right )\right )\\ &=-\sin ^{-1}\left (\frac {\cos ^2(x)}{3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 11, normalized size = 1.00 \[ -\sin ^{-1}\left (\frac {\cos ^2(x)}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]/Sqrt[9 - Cos[x]^4],x]

[Out]

-ArcSin[Cos[x]^2/3]

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fricas [B]  time = 1.31, size = 24, normalized size = 2.18 \[ \arctan \left (\frac {\sqrt {-\cos \relax (x)^{4} + 9} \cos \relax (x)^{2}}{\cos \relax (x)^{4} - 9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(9-cos(x)^4)^(1/2),x, algorithm="fricas")

[Out]

arctan(sqrt(-cos(x)^4 + 9)*cos(x)^2/(cos(x)^4 - 9))

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giac [A]  time = 0.15, size = 9, normalized size = 0.82 \[ -\arcsin \left (\frac {1}{3} \, \cos \relax (x)^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(9-cos(x)^4)^(1/2),x, algorithm="giac")

[Out]

-arcsin(1/3*cos(x)^2)

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maple [A]  time = 0.11, size = 10, normalized size = 0.91 \[ -\arcsin \left (\frac {\left (\cos ^{2}\relax (x )\right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(9-cos(x)^4)^(1/2),x)

[Out]

-arcsin(1/3*cos(x)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (2 \, x\right )}{\sqrt {-\cos \relax (x)^{4} + 9}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(9-cos(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(sin(2*x)/sqrt(-cos(x)^4 + 9), x)

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mupad [B]  time = 3.14, size = 18, normalized size = 1.64 \[ -\mathrm {atan}\left (\frac {{\cos \relax (x)}^2}{\sqrt {9-{\cos \relax (x)}^4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(9 - cos(x)^4)^(1/2),x)

[Out]

-atan(cos(x)^2/(9 - cos(x)^4)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(9-cos(x)**4)**(1/2),x)

[Out]

Timed out

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