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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0269628, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, 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}{\sin ^2(x)^{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*}

Mathematica [A]  time = 0.0441808, size = 51, normalized size = 1.59 \[ -\frac{\sin (x) \left (\csc ^2\left (\frac{x}{2}\right )-\sec ^2\left (\frac{x}{2}\right )-4 \log \left (\sin \left (\frac{x}{2}\right )\right )+4 \log \left (\cos \left (\frac{x}{2}\right )\right )\right )}{8 \sqrt{\sin ^2(x)}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.812, size = 37, normalized size = 1.2 \begin{align*} -{\frac{1}{\sin \left ( x \right ) } \left ({\frac{\cos \left ( x \right ) }{2}}+{\frac{ \left ( \ln \left ( 1+\cos \left ( x \right ) \right ) -\ln \left ( \cos \left ( x \right ) -1 \right ) \right ) \left ( \sin \left ( x \right ) \right ) ^{2}}{4}} \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/(1-cos(x)^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.63815, size = 405, normalized size = 12.66 \begin{align*} \frac{4 \,{\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \cos \left (4 \, x\right ) - 4 \,{\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \cos \left (x\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 )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 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 )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) + 4 \,{\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - 8 \, \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - 8 \, \sin \left (2 \, x\right ) \sin \left (x\right ) + 4 \, \cos \left (x\right )}{4 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(cos(3*x) + cos(x))*cos(4*x) - 4*(2*cos(2*x) - 1)*cos(3*x) - 8*cos(2*x)*cos(x) + (2*(2*cos(2*x) - 1)*co
s(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)*log(cos
(x)^2 + sin(x)^2 + 2*cos(x) + 1) - (2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*s
in(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + 4*(sin(3*x) + sin(
x))*sin(4*x) - 8*sin(3*x)*sin(2*x) - 8*sin(2*x)*sin(x) + 4*cos(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)

________________________________________________________________________________________

Fricas [A]  time = 1.60996, size = 150, normalized size = 4.69 \begin{align*} -\frac{{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - 2 \, \cos \left (x\right )}{4 \,{\left (\cos \left (x\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.27541, size = 105, normalized size = 3.28 \begin{align*} \frac{\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{\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 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}{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}} \end{align*}

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