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

Optimal. Leaf size=15 \[ -\frac{\sin (x) \tanh ^{-1}(\cos (x))}{\sqrt{\sin ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.0198256, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 = 0.0153835, size = 28, normalized size = 1.87 \[ \frac{\sin (x) \left (\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )\right )}{\sqrt{\sin ^2(x)}} \]

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 [A]  time = 0.336, size = 14, normalized size = 0.9 \begin{align*} -{{\it Artanh} \left ( \cos \left ( x \right ) \right ) \sin \left ( x \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]

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

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Maxima [B]  time = 1.63955, size = 47, normalized size = 3.13 \begin{align*} \frac{1}{2} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - \frac{1}{2} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \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]

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

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Fricas [A]  time = 1.61243, size = 77, normalized size = 5.13 \begin{align*} -\frac{1}{2} \, \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) + \frac{1}{2} \, \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \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]

-1/2*log(1/2*cos(x) + 1/2) + 1/2*log(-1/2*cos(x) + 1/2)

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