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

Optimal. Leaf size=21 \[ -\frac {1}{5} \cot ^5(x)-\frac {2 \cot ^3(x)}{3}-\cot (x) \]

[Out]

-cot(x)-2/3*cot(x)^3-1/5*cot(x)^5

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3175, 3767} \[ -\frac {1}{5} \cot ^5(x)-\frac {2 \cot ^3(x)}{3}-\cot (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[x]^2)^(-3),x]

[Out]

-Cot[x] - (2*Cot[x]^3)/3 - Cot[x]^5/5

Rule 3175

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (1-\cos ^2(x)\right )^3} \, dx &=\int \csc ^6(x) \, dx\\ &=-\operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (x)\right )\\ &=-\cot (x)-\frac {2 \cot ^3(x)}{3}-\frac {\cot ^5(x)}{5}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.29 \[ -\frac {8 \cot (x)}{15}-\frac {1}{5} \cot (x) \csc ^4(x)-\frac {4}{15} \cot (x) \csc ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[x]^2)^(-3),x]

[Out]

(-8*Cot[x])/15 - (4*Cot[x]*Csc[x]^2)/15 - (Cot[x]*Csc[x]^4)/5

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fricas [B]  time = 0.58, size = 37, normalized size = 1.76 \[ -\frac {8 \, \cos \relax (x)^{5} - 20 \, \cos \relax (x)^{3} + 15 \, \cos \relax (x)}{15 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(8*cos(x)^5 - 20*cos(x)^3 + 15*cos(x))/((cos(x)^4 - 2*cos(x)^2 + 1)*sin(x))

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giac [A]  time = 0.59, size = 20, normalized size = 0.95 \[ -\frac {15 \, \tan \relax (x)^{4} + 10 \, \tan \relax (x)^{2} + 3}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*tan(x)^4 + 10*tan(x)^2 + 3)/tan(x)^5

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maple [A]  time = 0.07, size = 20, normalized size = 0.95 \[ -\frac {2}{3 \tan \relax (x )^{3}}-\frac {1}{5 \tan \relax (x )^{5}}-\frac {1}{\tan \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/tan(x)^3-1/5/tan(x)^5-1/tan(x)

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maxima [A]  time = 0.31, size = 20, normalized size = 0.95 \[ -\frac {15 \, \tan \relax (x)^{4} + 10 \, \tan \relax (x)^{2} + 3}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*tan(x)^4 + 10*tan(x)^2 + 3)/tan(x)^5

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mupad [B]  time = 2.24, size = 17, normalized size = 0.81 \[ -\frac {{\mathrm {cot}\relax (x)}^5}{5}-\frac {2\,{\mathrm {cot}\relax (x)}^3}{3}-\mathrm {cot}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cos(x)^2 - 1)^3,x)

[Out]

- cot(x) - (2*cot(x)^3)/3 - cot(x)^5/5

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sympy [B]  time = 3.01, size = 54, normalized size = 2.57 \[ \frac {\tan ^{5}{\left (\frac {x}{2} \right )}}{160} + \frac {5 \tan ^{3}{\left (\frac {x}{2} \right )}}{96} + \frac {5 \tan {\left (\frac {x}{2} \right )}}{16} - \frac {5}{16 \tan {\left (\frac {x}{2} \right )}} - \frac {5}{96 \tan ^{3}{\left (\frac {x}{2} \right )}} - \frac {1}{160 \tan ^{5}{\left (\frac {x}{2} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x/2)**5/160 + 5*tan(x/2)**3/96 + 5*tan(x/2)/16 - 5/(16*tan(x/2)) - 5/(96*tan(x/2)**3) - 1/(160*tan(x/2)**5
)

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