3.39 \(\int \sqrt{1-\cot ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ \sin ^{-1}(\cot (x))-\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \cot (x)}{\sqrt{1-\cot ^2(x)}}\right ) \]

[Out]

ArcSin[Cot[x]] - Sqrt[2]*ArcTan[(Sqrt[2]*Cot[x])/Sqrt[1 - Cot[x]^2]]

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Rubi [A]  time = 0.0266071, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3661, 402, 216, 377, 203} \[ \sin ^{-1}(\cot (x))-\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \cot (x)}{\sqrt{1-\cot ^2(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Cot[x]^2],x]

[Out]

ArcSin[Cot[x]] - Sqrt[2]*ArcTan[(Sqrt[2]*Cot[x])/Sqrt[1 - Cot[x]^2]]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

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]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.0689268, size = 62, normalized size = 1.94 \[ \frac{\sin (x) \sqrt{1-\cot ^2(x)} \left (\sqrt{2} \log \left (\sqrt{2} \cos (x)+\sqrt{\cos (2 x)}\right )-\tanh ^{-1}\left (\frac{\cos (x)}{\sqrt{\cos (2 x)}}\right )\right )}{\sqrt{\cos (2 x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Cot[x]^2],x]

[Out]

(Sqrt[1 - Cot[x]^2]*(-ArcTanh[Cos[x]/Sqrt[Cos[2*x]]] + Sqrt[2]*Log[Sqrt[2]*Cos[x] + Sqrt[Cos[2*x]]])*Sin[x])/S
qrt[Cos[2*x]]

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Maple [A]  time = 0.028, size = 34, normalized size = 1.1 \begin{align*} \arcsin \left ( \cot \left ( x \right ) \right ) +\sqrt{2}\arctan \left ({\frac{\sqrt{2}\cot \left ( x \right ) }{-1+ \left ( \cot \left ( x \right ) \right ) ^{2}}\sqrt{1- \left ( \cot \left ( x \right ) \right ) ^{2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cot(x)^2)^(1/2),x)

[Out]

arcsin(cot(x))+2^(1/2)*arctan(2^(1/2)*(1-cot(x)^2)^(1/2)/(-1+cot(x)^2)*cot(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.83894, size = 193, normalized size = 6.03 \begin{align*} \sqrt{2} \arctan \left (\frac{\sqrt{\frac{\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) - 1}} \sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}\right ) - \arctan \left (\frac{\sqrt{2} \sqrt{\frac{\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) - 1}} \sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cot(x)**2)**(1/2),x)

[Out]

Integral(sqrt(1 - cot(x)**2), x)

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Giac [C]  time = 1.3867, size = 230, normalized size = 7.19 \begin{align*} -\frac{1}{2} \,{\left (\pi - \sqrt{2} \pi - 2 \, \sqrt{2} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right ) + 2 \, \arctan \left (-i\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) + \frac{1}{2} \,{\left (\pi \mathrm{sgn}\left (\cos \left (x\right )\right ) - \sqrt{2}{\left (\pi \mathrm{sgn}\left (\cos \left (x\right )\right ) + 2 \, \arctan \left (-\frac{{\left (\frac{{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}^{2}}{\cos \left (x\right )^{2}} - 4\right )} \cos \left (x\right )}{4 \,{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}}\right )\right )} + 2 \, \arctan \left (-\frac{\sqrt{2}{\left (\frac{{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}^{2}}{\cos \left (x\right )^{2}} - 4\right )} \cos \left (x\right )}{4 \,{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}}\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(pi - sqrt(2)*pi - 2*sqrt(2)*arctan(-1/2*I*sqrt(2)) + 2*arctan(-I))*sgn(sin(x)) + 1/2*(pi*sgn(cos(x)) - s
qrt(2)*(pi*sgn(cos(x)) + 2*arctan(-1/4*((sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2))^2/cos(x)^2 - 4)*cos(x)/(sqrt
(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2)))) + 2*arctan(-1/4*sqrt(2)*((sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2))^2/co
s(x)^2 - 4)*cos(x)/(sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2))))*sgn(sin(x))