3.69 \(\int \frac {\cos (x)}{\sqrt {4-\cos ^2(x)}} \, dx\)

Optimal. Leaf size=9 \[ \sinh ^{-1}\left (\frac {\sin (x)}{\sqrt {3}}\right ) \]

[Out]

arcsinh(1/3*sin(x)*3^(1/2))

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Rubi [A]  time = 0.03, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3186, 215} \[ \sinh ^{-1}\left (\frac {\sin (x)}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/Sqrt[4 - Cos[x]^2],x]

[Out]

ArcSinh[Sin[x]/Sqrt[3]]

Rule 215

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\cos (x)}{\sqrt {4-\cos ^2(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {3+x^2}} \, dx,x,\sin (x)\right )\\ &=\sinh ^{-1}\left (\frac {\sin (x)}{\sqrt {3}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 1.00 \[ \sinh ^{-1}\left (\frac {\sin (x)}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/Sqrt[4 - Cos[x]^2],x]

[Out]

ArcSinh[Sin[x]/Sqrt[3]]

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fricas [B]  time = 1.21, size = 39, normalized size = 4.33 \[ \frac {1}{4} \, \log \left (8 \, \cos \relax (x)^{4} - 4 \, {\left (2 \, \cos \relax (x)^{2} - 5\right )} \sqrt {-\cos \relax (x)^{2} + 4} \sin \relax (x) - 40 \, \cos \relax (x)^{2} + 41\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(4-cos(x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/4*log(8*cos(x)^4 - 4*(2*cos(x)^2 - 5)*sqrt(-cos(x)^2 + 4)*sin(x) - 40*cos(x)^2 + 41)

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giac [A]  time = 1.42, size = 16, normalized size = 1.78 \[ -\log \left (\sqrt {\sin \relax (x)^{2} + 3} - \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(4-cos(x)^2)^(1/2),x, algorithm="giac")

[Out]

-log(sqrt(sin(x)^2 + 3) - sin(x))

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maple [B]  time = 2.15, size = 53, normalized size = 5.89 \[ -\frac {\sqrt {-\left (-4+\cos ^{2}\relax (x )\right ) \left (\sin ^{2}\relax (x )\right )}\, \ln \left (-\left (\sin ^{2}\relax (x )\right )+\sqrt {\sin ^{4}\relax (x )+3 \left (\sin ^{2}\relax (x )\right )}-\frac {3}{2}\right )}{2 \sin \relax (x ) \sqrt {4-\left (\cos ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(4-cos(x)^2)^(1/2),x)

[Out]

-1/2*(-(-4+cos(x)^2)*sin(x)^2)^(1/2)*ln(-sin(x)^2+(sin(x)^4+3*sin(x)^2)^(1/2)-3/2)/sin(x)/(4-cos(x)^2)^(1/2)

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maxima [A]  time = 1.74, size = 8, normalized size = 0.89 \[ \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(4-cos(x)^2)^(1/2),x, algorithm="maxima")

[Out]

arcsinh(1/3*sqrt(3)*sin(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.11 \[ \int \frac {\cos \relax (x)}{\sqrt {4-{\cos \relax (x)}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(4 - cos(x)^2)^(1/2),x)

[Out]

int(cos(x)/(4 - cos(x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\relax (x )}}{\sqrt {- \left (\cos {\relax (x )} - 2\right ) \left (\cos {\relax (x )} + 2\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(4-cos(x)**2)**(1/2),x)

[Out]

Integral(cos(x)/sqrt(-(cos(x) - 2)*(cos(x) + 2)), x)

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