### 3.11 $$\int \frac{\sin (x)}{3-2 \cos (x)+\cos ^2(x)} \, dx$$

Optimal. Leaf size=19 $\frac{\tan ^{-1}\left (\frac{1-\cos (x)}{\sqrt{2}}\right )}{\sqrt{2}}$

[Out]

ArcTan[(1 - Cos[x])/Sqrt]/Sqrt

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Rubi [A]  time = 0.0363224, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {3259, 618, 204} $\frac{\tan ^{-1}\left (\frac{1-\cos (x)}{\sqrt{2}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

ArcTan[(1 - Cos[x])/Sqrt]/Sqrt

Rule 3259

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]

Rule 618

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 204

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

Rubi steps

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

Mathematica [A]  time = 0.0256319, size = 18, normalized size = 0.95 $-\frac{\tan ^{-1}\left (\frac{\cos (x)-1}{\sqrt{2}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(ArcTan[(-1 + Cos[x])/Sqrt]/Sqrt)

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Maple [A]  time = 0.022, size = 18, normalized size = 1. \begin{align*} -{\frac{\sqrt{2}}{2}\arctan \left ({\frac{ \left ( 2\,\cos \left ( x \right ) -2 \right ) \sqrt{2}}{4}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42794, size = 20, normalized size = 1.05 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\cos \left (x\right ) - 1\right )}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 1.2304, size = 74, normalized size = 3.89 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \cos \left (x\right ) - \frac{1}{2} \, \sqrt{2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [A]  time = 0.5079, size = 26, normalized size = 1.37 \begin{align*} - \frac{\sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} \cos{\left (x \right )}}{2} - \frac{\sqrt{2}}{2} \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.44027, size = 20, normalized size = 1.05 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\cos \left (x\right ) - 1\right )}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))