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

Optimal. Leaf size=33 \[ \tan ^{-1}\left (\frac {\cot (x)}{\sqrt {-\cot ^2(x)-2}}\right )+\tanh ^{-1}\left (\frac {\cot (x)}{\sqrt {-\cot ^2(x)-2}}\right ) \]

[Out]

arctan(cot(x)/(-2-cot(x)^2)^(1/2))+arctanh(cot(x)/(-2-cot(x)^2)^(1/2))

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4128, 402, 217, 203, 377, 206} \[ \tan ^{-1}\left (\frac {\cot (x)}{\sqrt {-\cot ^2(x)-2}}\right )+\tanh ^{-1}\left (\frac {\cot (x)}{\sqrt {-\cot ^2(x)-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 - Csc[x]^2],x]

[Out]

ArcTan[Cot[x]/Sqrt[-2 - Cot[x]^2]] + ArcTanh[Cot[x]/Sqrt[-2 - Cot[x]^2]]

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])

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 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 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

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

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Mathematica [B]  time = 0.04, size = 70, normalized size = 2.12 \[ \frac {\sqrt {2} \sin (x) \sqrt {-\csc ^2(x)-1} \left (\log \left (\sqrt {2} \cos (x)+\sqrt {\cos (2 x)-3}\right )+\tan ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)-3}}\right )\right )}{\sqrt {\cos (2 x)-3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 - Csc[x]^2],x]

[Out]

(Sqrt[2]*Sqrt[-1 - Csc[x]^2]*(ArcTan[(Sqrt[2]*Cos[x])/Sqrt[-3 + Cos[2*x]]] + Log[Sqrt[2]*Cos[x] + Sqrt[-3 + Co
s[2*x]]])*Sin[x])/Sqrt[-3 + Cos[2*x]]

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fricas [C]  time = 0.43, size = 115, normalized size = 3.48 \[ -\frac {1}{2} \, \log \left (-2 \, \sqrt {e^{\left (4 i \, x\right )} - 6 \, e^{\left (2 i \, x\right )} + 1} {\left (e^{\left (2 i \, x\right )} - 1\right )} + 2 \, e^{\left (4 i \, x\right )} - 8 \, e^{\left (2 i \, x\right )} - 2\right ) - i \, \log \left (\sqrt {e^{\left (4 i \, x\right )} - 6 \, e^{\left (2 i \, x\right )} + 1} - e^{\left (2 i \, x\right )} + 2 i + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {e^{\left (4 i \, x\right )} - 6 \, e^{\left (2 i \, x\right )} + 1} - e^{\left (2 i \, x\right )} + 1\right ) + i \, \log \left (\sqrt {e^{\left (4 i \, x\right )} - 6 \, e^{\left (2 i \, x\right )} + 1} - e^{\left (2 i \, x\right )} - 2 i + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-2*sqrt(e^(4*I*x) - 6*e^(2*I*x) + 1)*(e^(2*I*x) - 1) + 2*e^(4*I*x) - 8*e^(2*I*x) - 2) - I*log(sqrt(e^
(4*I*x) - 6*e^(2*I*x) + 1) - e^(2*I*x) + 2*I + 1) + 1/2*log(sqrt(e^(4*I*x) - 6*e^(2*I*x) + 1) - e^(2*I*x) + 1)
+ I*log(sqrt(e^(4*I*x) - 6*e^(2*I*x) + 1) - e^(2*I*x) - 2*I + 1)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: i/2*(2*(2*atan(1/2*(-tan(1/2*x)^2+sqrt(t
an(1/2*x)^4+6*tan(1/2*x)^2+1)-1))-1/2*ln(-tan(1/2*x)^2+sqrt(tan(1/2*x)^4+6*tan(1/2*x)^2+1)+1)+1/2*ln(-tan(1/2*
x)^2+sqrt(tan(1/2*x)^4+6*tan(1/2*x)^2+1)-1)-1/2*ln(-tan(1/2*x)^2+sqrt(tan(1/2*x)^4+6*tan(1/2*x)^2+1)-3))-4*ata
n(i)+ln(2+2*i)+ln(-2+2*i)-ln(2*i))*sign(sin(x))

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maple [B]  time = 1.05, size = 139, normalized size = 4.21 \[ \frac {\sqrt {\frac {\cos ^{2}\relax (x )-2}{\sin \relax (x )^{2}}}\, \left (-1+\cos \relax (x )\right ) \left (\arcsin \left (\frac {\left (2+\cos \relax (x )\right ) \sqrt {2}}{2 \cos \relax (x )+2}\right )-\arctan \left (\frac {\cos ^{2}\relax (x )-3 \cos \relax (x )+2}{\sqrt {\frac {\cos ^{2}\relax (x )-2}{\left (\cos \relax (x )+1\right )^{2}}}\, \sin \relax (x )^{2}}\right )+2 \arctanh \left (\frac {\cos \relax (x ) \sqrt {4}\, \left (-1+\cos \relax (x )\right )}{2 \sin \relax (x )^{2} \sqrt {\frac {\cos ^{2}\relax (x )-2}{\left (\cos \relax (x )+1\right )^{2}}}}\right )\right ) \sqrt {\frac {\cos ^{2}\relax (x )-2}{\left (\cos \relax (x )+1\right )^{2}}}\, \left (\cos \relax (x )+1\right )^{2} \sqrt {4}}{4 \left (\cos ^{2}\relax (x )-2\right ) \sin \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((cos(x)^2-2)/sin(x)^2)^(1/2)*(-1+cos(x))*(arcsin(1/2*(2+cos(x))/(cos(x)+1)*2^(1/2))-arctan((cos(x)^2-3*co
s(x)+2)/((cos(x)^2-2)/(cos(x)+1)^2)^(1/2)/sin(x)^2)+2*arctanh(1/2*cos(x)*4^(1/2)*(-1+cos(x))/sin(x)^2/((cos(x)
^2-2)/(cos(x)+1)^2)^(1/2)))*((cos(x)^2-2)/(cos(x)+1)^2)^(1/2)*(cos(x)+1)^2*4^(1/2)/(cos(x)^2-2)/sin(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-\csc \relax (x)^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-csc(x)^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {-\frac {1}{{\sin \relax (x)}^2}-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- 1/sin(x)^2 - 1)^(1/2),x)

[Out]

int((- 1/sin(x)^2 - 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- \csc ^{2}{\relax (x )} - 1}\, dx \]

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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