\(\int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx\) [11]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=\frac {\arctan \left (\frac {1-\cos (x)}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

1/2*arctan(1/2*(1-cos(x))*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3340, 632, 210} \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=\frac {\arctan \left (\frac {1-\cos (x)}{\sqrt {2}}\right )}{\sqrt {2}} \]

[In]

Int[Sin[x]/(3 - 2*Cos[x] + Cos[x]^2),x]

[Out]

ArcTan[(1 - Cos[x])/Sqrt[2]]/Sqrt[2]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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

Rule 3340

Int[((a_.) + (b_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n_.) + (c_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n2_.))^(p_.)
*sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Module[{g = FreeFactors[Cos[d + e*x], x]}, Dist[-g/e, Subst[Int[(
1 - g^2*x^2)^((m - 1)/2)*(a + b*(f*g*x)^n + c*(f*g*x)^(2*n))^p, x], x, Cos[d + e*x]/g], x]] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] && EqQ[n2, 2*n] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{3-2 x+x^2} \, dx,x,\cos (x)\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,-2+2 \cos (x)\right ) \\ & = \frac {\arctan \left (\frac {1-\cos (x)}{\sqrt {2}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=-\frac {\arctan \left (\frac {-1+\cos (x)}{\sqrt {2}}\right )}{\sqrt {2}} \]

[In]

Integrate[Sin[x]/(3 - 2*Cos[x] + Cos[x]^2),x]

[Out]

-(ArcTan[(-1 + Cos[x])/Sqrt[2]]/Sqrt[2])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 \cos \left (x \right )-2\right ) \sqrt {2}}{4}\right )}{2}\)	\(18\)
default	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 \cos \left (x \right )-2\right ) \sqrt {2}}{4}\right )}{2}\)	\(18\)
risch	\(-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+\left (2 i \sqrt {2}-2\right ) {\mathrm e}^{i x}+1\right )}{4}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+\left (-2 i \sqrt {2}-2\right ) {\mathrm e}^{i x}+1\right )}{4}\)	\(58\)

[In]

int(sin(x)/(3-2*cos(x)+cos(x)^2),x,method=_RETURNVERBOSE)

[Out]

-1/2*2^(1/2)*arctan(1/4*(2*cos(x)-2)*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \cos \left (x\right ) - \frac {1}{2} \, \sqrt {2}\right ) \]

[In]

integrate(sin(x)/(3-2*cos(x)+cos(x)^2),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*cos(x) - 1/2*sqrt(2))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=- \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \cos {\left (x \right )}}{2} - \frac {\sqrt {2}}{2} \right )}}{2} \]

[In]

integrate(sin(x)/(3-2*cos(x)+cos(x)**2),x)

[Out]

-sqrt(2)*atan(sqrt(2)*cos(x)/2 - sqrt(2)/2)/2

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\cos \left (x\right ) - 1\right )}\right ) \]

[In]

integrate(sin(x)/(3-2*cos(x)+cos(x)^2),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(cos(x) - 1))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\cos \left (x\right ) - 1\right )}\right ) \]

[In]

integrate(sin(x)/(3-2*cos(x)+cos(x)^2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(cos(x) - 1))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx=-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (\cos \left (x\right )-1\right )}{2}\right )}{2} \]

[In]

int(sin(x)/(cos(x)^2 - 2*cos(x) + 3),x)

[Out]

-(2^(1/2)*atan((2^(1/2)*(cos(x) - 1))/2))/2