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

Optimal. Leaf size=14 \[ \sqrt{-\sin ^2(x)} (-\cot (x)) \]

[Out]

-(Cot[x]*Sqrt[-Sin[x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0158867, antiderivative size = 14, 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, 2638} \[ \sqrt{-\sin ^2(x)} (-\cot (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Cos[x]^2],x]

[Out]

-(Cot[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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.0046514, size = 14, normalized size = 1. \[ \sqrt{-\sin ^2(x)} (-\cot (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + Cos[x]^2],x]

[Out]

-(Cot[x]*Sqrt[-Sin[x]^2])

________________________________________________________________________________________

Maple [A]  time = 0.231, size = 14, normalized size = 1. \begin{align*}{\cos \left ( x \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+cos(x)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.45688, size = 16, normalized size = 1.14 \begin{align*} -\frac{1}{\sqrt{-\tan \left (x\right )^{2} - 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/sqrt(-tan(x)^2 - 1)

________________________________________________________________________________________

Fricas [A]  time = 1.77924, size = 4, normalized size = 0.29 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(cos(x)**2 - 1), x)

________________________________________________________________________________________

Giac [C]  time = 1.1781, size = 38, normalized size = 2.71 \begin{align*} \frac{2 i \, \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{3} - \tan \left (\frac{1}{2} \, x\right )\right )}{\tan \left (\frac{1}{2} \, x\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*sgn(-tan(1/2*x)^3 - tan(1/2*x))/(tan(1/2*x)^2 + 1)