\(\int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx\) [905]

Optimal result

Integrand size = 17, antiderivative size = 14 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \sqrt {9-\sin ^2(x)} \]

[Out]

-2*(9-sin(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 267} \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \sqrt {9-\sin ^2(x)} \]

[In]

Int[Sin[2*x]/Sqrt[9 - Sin[x]^2],x]

[Out]

-2*Sqrt[9 - Sin[x]^2]

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 {9-x^2}} \, dx,x,\sin (x)\right ) \\ & = 2 \text {Subst}\left (\int \frac {x}{\sqrt {9-x^2}} \, dx,x,\sin (x)\right ) \\ & = -2 \sqrt {9-\sin ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \sqrt {9-\sin ^2(x)} \]

[In]

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

[Out]

-2*Sqrt[9 - Sin[x]^2]

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(-2 \sqrt {9-\sin \left (x \right )^{2}}\)	\(13\)
default	\(-2 \sqrt {9-\sin \left (x \right )^{2}}\)	\(13\)

[In]

int(sin(2*x)/(9-sin(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*(9-sin(x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \, \sqrt {\cos \left (x\right )^{2} + 8} \]

[In]

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

[Out]

-2*sqrt(cos(x)^2 + 8)

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=- 2 \sqrt {9 - \sin ^{2}{\left (x \right )}} \]

[In]

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

[Out]

-2*sqrt(9 - sin(x)**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \, \sqrt {-\sin \left (x\right )^{2} + 9} \]

[In]

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

[Out]

-2*sqrt(-sin(x)^2 + 9)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2 \, \sqrt {-\sin \left (x\right )^{2} + 9} \]

[In]

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

[Out]

-2*sqrt(-sin(x)^2 + 9)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {\sin (2 x)}{\sqrt {9-\sin ^2(x)}} \, dx=-2\,\sqrt {{\cos \left (x\right )}^2+8} \]

[In]

int(sin(2*x)/(9 - sin(x)^2)^(1/2),x)

[Out]

-2*(cos(x)^2 + 8)^(1/2)