3.54 \(\int \frac {1}{(-1+\cos ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=36 \[ \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}+\frac {\sin (x) \tanh ^{-1}(\cos (x))}{2 \sqrt {-\sin ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3176, 3204, 3207, 3770} \[ \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}+\frac {\sin (x) \tanh ^{-1}(\cos (x))}{2 \sqrt {-\sin ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Cos[x]^2)^(-3/2),x]

[Out]

Cot[x]/(2*Sqrt[-Sin[x]^2]) + (ArcTanh[Cos[x]]*Sin[x])/(2*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 3204

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^(p + 1))/(b*f*(
2*p + 1)), x] + Dist[(2*(p + 1))/(b*(2*p + 1)), Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x]
&&  !IntegerQ[p] && LtQ[p, -1]

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 3770

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}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx &=\int \frac {1}{\left (-\sin ^2(x)\right )^{3/2}} \, dx\\ &=\frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {1}{2} \int \frac {1}{\sqrt {-\sin ^2(x)}} \, dx\\ &=\frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {\sin (x) \int \csc (x) \, dx}{2 \sqrt {-\sin ^2(x)}}\\ &=\frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}+\frac {\tanh ^{-1}(\cos (x)) \sin (x)}{2 \sqrt {-\sin ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.47 \[ \frac {\sin (x) \left (\csc ^2\left (\frac {x}{2}\right )-\sec ^2\left (\frac {x}{2}\right )-4 \log \left (\sin \left (\frac {x}{2}\right )\right )+4 \log \left (\cos \left (\frac {x}{2}\right )\right )\right )}{8 \sqrt {-\sin ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Cos[x]^2)^(-3/2),x]

[Out]

((Csc[x/2]^2 + 4*Log[Cos[x/2]] - 4*Log[Sin[x/2]] - Sec[x/2]^2)*Sin[x])/(8*Sqrt[-Sin[x]^2])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+cos(x)^2)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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giac [C]  time = 0.69, size = 90, normalized size = 2.50 \[ -\frac {i \, \tan \left (\frac {1}{2} \, x\right )^{2}}{8 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right )} - \frac {i \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{4 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right )} + \frac {2 i \, \tan \left (\frac {1}{2} \, x\right )^{2} + i}{8 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+cos(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/8*I*tan(1/2*x)^2/sgn(-tan(1/2*x)^3 - tan(1/2*x)) - 1/4*I*log(tan(1/2*x)^2)/sgn(-tan(1/2*x)^3 - tan(1/2*x))
+ 1/8*(2*I*tan(1/2*x)^2 + I)/(sgn(-tan(1/2*x)^3 - tan(1/2*x))*tan(1/2*x)^2)

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maple [A]  time = 1.18, size = 51, normalized size = 1.42 \[ -\frac {\sqrt {-\left (\cos ^{2}\relax (x )\right )}\, \left (-\arctan \left (\frac {1}{\sqrt {-\left (\cos ^{2}\relax (x )\right )}}\right ) \left (\sin ^{2}\relax (x )\right )+\sqrt {-\left (\cos ^{2}\relax (x )\right )}\right )}{2 \sin \relax (x ) \cos \relax (x ) \sqrt {-\left (\sin ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1+cos(x)^2)^(3/2),x)

[Out]

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

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maxima [B]  time = 1.05, size = 284, normalized size = 7.89 \[ \frac {{\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \arctan \left (\sin \relax (x), \cos \relax (x) - 1\right ) + 2 \, {\left (\sin \left (3 \, x\right ) + \sin \relax (x)\right )} \cos \left (4 \, x\right ) - 2 \, {\left (\cos \left (3 \, x\right ) + \cos \relax (x)\right )} \sin \left (4 \, x\right ) - 2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \sin \left (3 \, x\right ) + 4 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \cos \relax (x) \sin \left (2 \, x\right ) - 4 \, \cos \left (2 \, x\right ) \sin \relax (x) + 2 \, \sin \relax (x)}{2 \, {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+cos(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*((2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^
2 + 4*cos(2*x) - 1)*arctan2(sin(x), cos(x) + 1) - (2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - s
in(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*arctan2(sin(x), cos(x) - 1) + 2*(sin(3*x) + s
in(x))*cos(4*x) - 2*(cos(3*x) + cos(x))*sin(4*x) - 2*(2*cos(2*x) - 1)*sin(3*x) + 4*cos(3*x)*sin(2*x) + 4*cos(x
)*sin(2*x) - 4*cos(2*x)*sin(x) + 2*sin(x))/(2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)
^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{{\left ({\cos \relax (x)}^2-1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^2 - 1)^(3/2),x)

[Out]

int(1/(cos(x)^2 - 1)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\cos ^{2}{\relax (x )} - 1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+cos(x)**2)**(3/2),x)

[Out]

Integral((cos(x)**2 - 1)**(-3/2), x)

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