3.23 \(\int \sqrt{-1+\csc ^2(x)} \, dx\)

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

[Out]

Sqrt[Cot[x]^2]*Log[Sin[x]]*Tan[x]

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Rubi [A]  time = 0.0236293, 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 = {4121, 3658, 3475} \[ \tan (x) \sqrt{\cot ^2(x)} \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Csc[x]^2],x]

[Out]

Sqrt[Cot[x]^2]*Log[Sin[x]]*Tan[x]

Rule 4121

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

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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 3475

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

Rubi steps

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

Mathematica [A]  time = 0.0059525, size = 14, normalized size = 1. \[ \tan (x) \sqrt{\cot ^2(x)} \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + Csc[x]^2],x]

[Out]

Sqrt[Cot[x]^2]*Log[Sin[x]]*Tan[x]

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Maple [B]  time = 0.137, size = 51, normalized size = 3.6 \begin{align*}{\frac{\sqrt{4}\sin \left ( x \right ) }{2\,\cos \left ( x \right ) } \left ( -\ln \left ( 2\, \left ( \cos \left ( x \right ) +1 \right ) ^{-1} \right ) +\ln \left ( -{\frac{-1+\cos \left ( x \right ) }{\sin \left ( x \right ) }} \right ) \right ) \sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}}{ \left ( \cos \left ( x \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*4^(1/2)*(-ln(2/(cos(x)+1))+ln(-(-1+cos(x))/sin(x)))*sin(x)*(-cos(x)^2/(cos(x)^2-1))^(1/2)/cos(x)

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Maxima [A]  time = 1.69768, size = 18, normalized size = 1.29 \begin{align*} -\frac{1}{2} \, \log \left (\tan \left (x\right )^{2} + 1\right ) + \log \left (\tan \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(tan(x)^2 + 1) + log(tan(x))

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Fricas [A]  time = 0.48569, size = 24, normalized size = 1.71 \begin{align*} -\log \left (\frac{1}{2} \, \sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(1/2*sin(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(csc(x)**2 - 1), x)

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Giac [B]  time = 1.39093, size = 59, normalized size = 4.21 \begin{align*} \frac{1}{2} \,{\left (2 \, \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )^{4} - 1\right ) - \log \left (\tan \left (\frac{1}{2} \, x\right )^{2}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )^{4} - 1\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+csc(x)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(2*log(tan(1/2*x)^2 + 1)*sgn(tan(1/2*x)^4 - 1) - log(tan(1/2*x)^2)*sgn(tan(1/2*x)^4 - 1))*sgn(sin(x))