3.18 \(\int \frac{\cos (x)}{10-6 \sin (x)+\sin ^2(x)} \, dx\)

Optimal. Leaf size=9 \[ -\tan ^{-1}(3-\sin (x)) \]

[Out]

-ArcTan[3 - Sin[x]]

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Rubi [A]  time = 0.0285221, antiderivative size = 9, 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 = {3258, 618, 204} \[ -\tan ^{-1}(3-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(10 - 6*Sin[x] + Sin[x]^2),x]

[Out]

-ArcTan[3 - Sin[x]]

Rule 3258

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*sin[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*sin[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_.), x_Symbol] :> Module[{g = FreeFactors[Sin[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, Sin[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{\cos (x)}{10-6 \sin (x)+\sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{10-6 x+x^2} \, dx,x,\sin (x)\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,-6+2 \sin (x)\right )\right )\\ &=-\tan ^{-1}(3-\sin (x))\\ \end{align*}

Mathematica [A]  time = 0.009574, size = 9, normalized size = 1. \[ -\tan ^{-1}(3-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/(10 - 6*Sin[x] + Sin[x]^2),x]

[Out]

-ArcTan[3 - Sin[x]]

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Maple [A]  time = 0.048, size = 6, normalized size = 0.7 \begin{align*} \arctan \left ( -3+\sin \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(10-6*sin(x)+sin(x)^2),x)

[Out]

arctan(-3+sin(x))

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Maxima [A]  time = 1.42398, size = 7, normalized size = 0.78 \begin{align*} \arctan \left (\sin \left (x\right ) - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(10-6*sin(x)+sin(x)^2),x, algorithm="maxima")

[Out]

arctan(sin(x) - 3)

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Fricas [A]  time = 1.39823, size = 27, normalized size = 3. \begin{align*} \arctan \left (\sin \left (x\right ) - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(10-6*sin(x)+sin(x)^2),x, algorithm="fricas")

[Out]

arctan(sin(x) - 3)

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Sympy [A]  time = 0.254743, size = 5, normalized size = 0.56 \begin{align*} \operatorname{atan}{\left (\sin{\left (x \right )} - 3 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(10-6*sin(x)+sin(x)**2),x)

[Out]

atan(sin(x) - 3)

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Giac [A]  time = 1.10479, size = 7, normalized size = 0.78 \begin{align*} \arctan \left (\sin \left (x\right ) - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(10-6*sin(x)+sin(x)^2),x, algorithm="giac")

[Out]

arctan(sin(x) - 3)