3.19 \(\int \frac {\cos (x)}{2+2 \sin (x)+\sin ^2(x)} \, dx\)

Optimal. Leaf size=5 \[ \tan ^{-1}(\sin (x)+1) \]

[Out]

arctan(1+sin(x))

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Rubi [A]  time = 0.03, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3258, 617, 204} \[ \tan ^{-1}(\sin (x)+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[1 + Sin[x]]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 5, normalized size = 1.00 \[ \tan ^{-1}(\sin (x)+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[1 + Sin[x]]

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fricas [A]  time = 0.43, size = 5, normalized size = 1.00 \[ \arctan \left (\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sin(x) + 1)

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giac [A]  time = 0.14, size = 5, normalized size = 1.00 \[ \arctan \left (\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sin(x) + 1)

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maple [A]  time = 0.20, size = 6, normalized size = 1.20 \[ \arctan \left (1+\sin \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(1+sin(x))

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maxima [A]  time = 0.43, size = 5, normalized size = 1.00 \[ \arctan \left (\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sin(x) + 1)

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mupad [B]  time = 14.93, size = 5, normalized size = 1.00 \[ \mathrm {atan}\left (\sin \relax (x)+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(sin(x) + 1)

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sympy [A]  time = 0.45, size = 5, normalized size = 1.00 \[ \operatorname {atan}{\left (\sin {\relax (x )} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(sin(x) + 1)

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