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

Optimal. Leaf size=17 \[ -\frac{\sin (x) \tanh ^{-1}(\cos (x))}{\sqrt{-\sin ^2(x)}} \]

[Out]

-((ArcTanh[Cos[x]]*Sin[x])/Sqrt[-Sin[x]^2])

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Rubi [A]  time = 0.0184225, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3176, 3207, 3770} \[ -\frac{\sin (x) \tanh ^{-1}(\cos (x))}{\sqrt{-\sin ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 + Cos[x]^2],x]

[Out]

-((ArcTanh[Cos[x]]*Sin[x])/Sqrt[-Sin[x]^2])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0102452, size = 30, normalized size = 1.76 \[ \frac{\sin (x) \left (\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )\right )}{\sqrt{-\sin ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 + Cos[x]^2],x]

[Out]

((-Log[Cos[x/2]] + Log[Sin[x/2]])*Sin[x])/Sqrt[-Sin[x]^2]

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Maple [B]  time = 0.233, size = 34, normalized size = 2. \begin{align*} -{\frac{\sin \left ( x \right ) }{\cos \left ( x \right ) }\sqrt{- \left ( \cos \left ( x \right ) \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{- \left ( \cos \left ( x \right ) \right ) ^{2}}}} \right ){\frac{1}{\sqrt{- \left ( \sin \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.58404, size = 23, normalized size = 1.35 \begin{align*} -\arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) + \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan2(sin(x), cos(x) + 1) + arctan2(sin(x), cos(x) - 1)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\cos ^{2}{\left (x \right )} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(cos(x)**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\cos \left (x\right )^{2} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(cos(x)^2 - 1), x)