3.12 \(\int \frac {\sin (x)}{(13-4 \cos (x)+\cos ^2(x))^2} \, dx\)

Optimal. Leaf size=36 \[ \frac {2-\cos (x)}{18 \left (\cos ^2(x)-4 \cos (x)+13\right )}-\frac {1}{54} \tan ^{-1}\left (\frac {1}{3} (\cos (x)-2)\right ) \]

[Out]

-1/54*arctan(-2/3+1/3*cos(x))+1/18*(2-cos(x))/(13-4*cos(x)+cos(x)^2)

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3259, 614, 618, 204} \[ \frac {2-\cos (x)}{18 \left (\cos ^2(x)-4 \cos (x)+13\right )}-\frac {1}{54} \tan ^{-1}\left (\frac {1}{3} (\cos (x)-2)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(13 - 4*Cos[x] + Cos[x]^2)^2,x]

[Out]

-ArcTan[(-2 + Cos[x])/3]/54 + (2 - Cos[x])/(18*(13 - 4*Cos[x] + Cos[x]^2))

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])

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 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]

Rubi steps

\begin {align*} \int \frac {\sin (x)}{\left (13-4 \cos (x)+\cos ^2(x)\right )^2} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{\left (13-4 x+x^2\right )^2} \, dx,x,\cos (x)\right )\\ &=\frac {2-\cos (x)}{18 \left (13-4 \cos (x)+\cos ^2(x)\right )}-\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{13-4 x+x^2} \, dx,x,\cos (x)\right )\\ &=\frac {2-\cos (x)}{18 \left (13-4 \cos (x)+\cos ^2(x)\right )}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{-36-x^2} \, dx,x,-4+2 \cos (x)\right )\\ &=-\frac {1}{54} \tan ^{-1}\left (\frac {1}{3} (-2+\cos (x))\right )+\frac {2-\cos (x)}{18 \left (13-4 \cos (x)+\cos ^2(x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 34, normalized size = 0.94 \[ -\frac {\cos (x)-2}{18 \left (\cos ^2(x)-4 \cos (x)+13\right )}-\frac {1}{54} \tan ^{-1}\left (\frac {1}{3} (\cos (x)-2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(13 - 4*Cos[x] + Cos[x]^2)^2,x]

[Out]

-1/54*ArcTan[(-2 + Cos[x])/3] - (-2 + Cos[x])/(18*(13 - 4*Cos[x] + Cos[x]^2))

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fricas [A]  time = 1.12, size = 38, normalized size = 1.06 \[ -\frac {{\left (\cos \relax (x)^{2} - 4 \, \cos \relax (x) + 13\right )} \arctan \left (\frac {1}{3} \, \cos \relax (x) - \frac {2}{3}\right ) + 3 \, \cos \relax (x) - 6}{54 \, {\left (\cos \relax (x)^{2} - 4 \, \cos \relax (x) + 13\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(13-4*cos(x)+cos(x)^2)^2,x, algorithm="fricas")

[Out]

-1/54*((cos(x)^2 - 4*cos(x) + 13)*arctan(1/3*cos(x) - 2/3) + 3*cos(x) - 6)/(cos(x)^2 - 4*cos(x) + 13)

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giac [A]  time = 0.48, size = 28, normalized size = 0.78 \[ -\frac {\cos \relax (x) - 2}{18 \, {\left (\cos \relax (x)^{2} - 4 \, \cos \relax (x) + 13\right )}} - \frac {1}{54} \, \arctan \left (\frac {1}{3} \, \cos \relax (x) - \frac {2}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(13-4*cos(x)+cos(x)^2)^2,x, algorithm="giac")

[Out]

-1/18*(cos(x) - 2)/(cos(x)^2 - 4*cos(x) + 13) - 1/54*arctan(1/3*cos(x) - 2/3)

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maple [A]  time = 0.08, size = 31, normalized size = 0.86 \[ -\frac {2 \cos \relax (x )-4}{36 \left (13-4 \cos \relax (x )+\cos ^{2}\relax (x )\right )}-\frac {\arctan \left (-\frac {2}{3}+\frac {\cos \relax (x )}{3}\right )}{54} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(13-4*cos(x)+cos(x)^2)^2,x)

[Out]

-1/36*(2*cos(x)-4)/(13-4*cos(x)+cos(x)^2)-1/54*arctan(-2/3+1/3*cos(x))

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maxima [A]  time = 0.90, size = 28, normalized size = 0.78 \[ -\frac {\cos \relax (x) - 2}{18 \, {\left (\cos \relax (x)^{2} - 4 \, \cos \relax (x) + 13\right )}} - \frac {1}{54} \, \arctan \left (\frac {1}{3} \, \cos \relax (x) - \frac {2}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(13-4*cos(x)+cos(x)^2)^2,x, algorithm="maxima")

[Out]

-1/18*(cos(x) - 2)/(cos(x)^2 - 4*cos(x) + 13) - 1/54*arctan(1/3*cos(x) - 2/3)

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mupad [B]  time = 0.06, size = 30, normalized size = 0.83 \[ -\frac {\mathrm {atan}\left (\frac {\cos \relax (x)}{3}-\frac {2}{3}\right )}{54}-\frac {\frac {\cos \relax (x)}{18}-\frac {1}{9}}{{\cos \relax (x)}^2-4\,\cos \relax (x)+13} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(cos(x)^2 - 4*cos(x) + 13)^2,x)

[Out]

- atan(cos(x)/3 - 2/3)/54 - (cos(x)/18 - 1/9)/(cos(x)^2 - 4*cos(x) + 13)

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sympy [B]  time = 1.02, size = 116, normalized size = 3.22 \[ - \frac {\cos ^{2}{\relax (x )} \operatorname {atan}{\left (\frac {\cos {\relax (x )}}{3} - \frac {2}{3} \right )}}{54 \cos ^{2}{\relax (x )} - 216 \cos {\relax (x )} + 702} + \frac {4 \cos {\relax (x )} \operatorname {atan}{\left (\frac {\cos {\relax (x )}}{3} - \frac {2}{3} \right )}}{54 \cos ^{2}{\relax (x )} - 216 \cos {\relax (x )} + 702} - \frac {3 \cos {\relax (x )}}{54 \cos ^{2}{\relax (x )} - 216 \cos {\relax (x )} + 702} - \frac {13 \operatorname {atan}{\left (\frac {\cos {\relax (x )}}{3} - \frac {2}{3} \right )}}{54 \cos ^{2}{\relax (x )} - 216 \cos {\relax (x )} + 702} + \frac {6}{54 \cos ^{2}{\relax (x )} - 216 \cos {\relax (x )} + 702} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(13-4*cos(x)+cos(x)**2)**2,x)

[Out]

-cos(x)**2*atan(cos(x)/3 - 2/3)/(54*cos(x)**2 - 216*cos(x) + 702) + 4*cos(x)*atan(cos(x)/3 - 2/3)/(54*cos(x)**
2 - 216*cos(x) + 702) - 3*cos(x)/(54*cos(x)**2 - 216*cos(x) + 702) - 13*atan(cos(x)/3 - 2/3)/(54*cos(x)**2 - 2
16*cos(x) + 702) + 6/(54*cos(x)**2 - 216*cos(x) + 702)

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