3.1.51 \(\int \frac {1}{\sqrt {1-\cos ^2(x)}} \, dx\) [51]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3255, 3286, 3855} \begin {gather*} -\frac {\sin (x) \tanh ^{-1}(\cos (x))}{\sqrt {\sin ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 - Cos[x]^2],x]

[Out]

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

Rule 3255

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 3286

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 3855

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*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.87 \begin {gather*} \frac {\left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \sin (x)}{\sqrt {\sin ^2(x)}} \end {gather*}

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 = 0.24, size = 14, normalized size = 0.93

method	result	size
default	\(-\frac {2 \arctanh \left (\cos \left (x \right )\right ) \sin \left (x \right )}{\sqrt {2-2 \cos \left (2 x \right )}}\)	\(14\)
risch	\(\frac {2 \ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {2 \ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (13) = 26\).
time = 0.51, size = 35, normalized size = 2.33 \begin {gather*} \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 {gather*}

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 = 0.40, size = 19, normalized size = 1.27 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {1 - \cos ^{2}{\left (x \right )}}}\, dx \end {gather*}

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 [A]
time = 0.41, size = 21, normalized size = 1.40 \begin {gather*} \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(tan(1/2*x)))/sgn(tan(1/2*x)^3 + tan(1/2*x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.07 \begin {gather*} \int \frac {1}{\sqrt {1-{\cos \left (x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - cos(x)^2)^(1/2),x)

[Out]

int(1/(1 - cos(x)^2)^(1/2), x)

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