3.185 \(\int \cot ^2(x) \csc ^4(x) \, dx\)

Optimal. Leaf size=17 \[ -\frac {1}{5} \cot ^5(x)-\frac {\cot ^3(x)}{3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2607, 14} \[ -\frac {1}{5} \cot ^5(x)-\frac {\cot ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2*Csc[x]^4,x]

[Out]

-Cot[x]^3/3 - Cot[x]^5/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.59 \[ \frac {2 \cot (x)}{15}-\frac {1}{5} \cot (x) \csc ^4(x)+\frac {1}{15} \cot (x) \csc ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^2*Csc[x]^4,x]

[Out]

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

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fricas [B]  time = 0.41, size = 33, normalized size = 1.94 \[ \frac {2 \, \cos \relax (x)^{5} - 5 \, \cos \relax (x)^{3}}{15 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/sin(x)^6,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 14, normalized size = 0.82 \[ -\frac {5 \, \tan \relax (x)^{2} + 3}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/sin(x)^6,x, algorithm="giac")

[Out]

-1/15*(5*tan(x)^2 + 3)/tan(x)^5

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maple [A]  time = 0.02, size = 22, normalized size = 1.29 \[ -\frac {\cos ^{3}\relax (x )}{5 \sin \relax (x )^{5}}-\frac {2 \left (\cos ^{3}\relax (x )\right )}{15 \sin \relax (x )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/sin(x)^6,x)

[Out]

-1/5*cos(x)^3/sin(x)^5-2/15*cos(x)^3/sin(x)^3

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maxima [A]  time = 0.31, size = 14, normalized size = 0.82 \[ -\frac {5 \, \tan \relax (x)^{2} + 3}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/sin(x)^6,x, algorithm="maxima")

[Out]

-1/15*(5*tan(x)^2 + 3)/tan(x)^5

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mupad [B]  time = 0.08, size = 19, normalized size = 1.12 \[ -{\cos \relax (x)}^3\,\left (\frac {2}{15\,{\sin \relax (x)}^3}+\frac {1}{5\,{\sin \relax (x)}^5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/sin(x)^6,x)

[Out]

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

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sympy [B]  time = 0.07, size = 29, normalized size = 1.71 \[ \frac {2 \cos {\relax (x )}}{15 \sin {\relax (x )}} + \frac {\cos {\relax (x )}}{15 \sin ^{3}{\relax (x )}} - \frac {\cos {\relax (x )}}{5 \sin ^{5}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**2/sin(x)**6,x)

[Out]

2*cos(x)/(15*sin(x)) + cos(x)/(15*sin(x)**3) - cos(x)/(5*sin(x)**5)

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