3.67 \(\int \frac{\cos (x)}{\sqrt{1+\cos ^2(x)}} \, dx\)

Optimal. Leaf size=9 \[ \sin ^{-1}\left (\frac{\sin (x)}{\sqrt{2}}\right ) \]

[Out]

ArcSin[Sin[x]/Sqrt[2]]

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Rubi [A]  time = 0.0211736, antiderivative size = 9, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3186, 216} \[ \sin ^{-1}\left (\frac{\sin (x)}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/Sqrt[1 + Cos[x]^2],x]

[Out]

ArcSin[Sin[x]/Sqrt[2]]

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]

Rule 216

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

Rubi steps

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

Mathematica [A]  time = 0.0073821, size = 9, normalized size = 1. \[ \sin ^{-1}\left (\frac{\sin (x)}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/Sqrt[1 + Cos[x]^2],x]

[Out]

ArcSin[Sin[x]/Sqrt[2]]

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Maple [B]  time = 0.377, size = 33, normalized size = 3.7 \begin{align*} -{\frac{\arcsin \left ( \left ( \cos \left ( x \right ) \right ) ^{2} \right ) }{2\,\sin \left ( x \right ) }\sqrt{ \left ( 1+ \left ( \cos \left ( x \right ) \right ) ^{2} \right ) \left ( \sin \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{1+ \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(1+cos(x)^2)^(1/2),x)

[Out]

-1/2*((1+cos(x)^2)*sin(x)^2)^(1/2)*arcsin(cos(x)^2)/sin(x)/(1+cos(x)^2)^(1/2)

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Maxima [A]  time = 1.42654, size = 11, normalized size = 1.22 \begin{align*} \arcsin \left (\frac{1}{2} \, \sqrt{2} \sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(1/2*sqrt(2)*sin(x))

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Fricas [B]  time = 1.716, size = 162, normalized size = 18. \begin{align*} \frac{1}{2} \, \arctan \left (\frac{\sqrt{\cos \left (x\right )^{2} + 1} \cos \left (x\right )^{2} \sin \left (x\right ) - \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 1}\right ) + \frac{1}{2} \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(1+cos(x)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33444, size = 11, normalized size = 1.22 \begin{align*} \arcsin \left (\frac{1}{2} \, \sqrt{2} \sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(1/2*sqrt(2)*sin(x))