3.314 \(\int \cot ^3(2 x) \csc ^3(2 x) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{6} \csc ^3(2 x)-\frac {1}{10} \csc ^5(2 x) \]

[Out]

1/6*csc(2*x)^3-1/10*csc(2*x)^5

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2606, 14} \[ \frac {1}{6} \csc ^3(2 x)-\frac {1}{10} \csc ^5(2 x) \]

Antiderivative was successfully verified.

[In]

Int[Cot[2*x]^3*Csc[2*x]^3,x]

[Out]

Csc[2*x]^3/6 - Csc[2*x]^5/10

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 2606

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

Rubi steps

\begin {align*} \int \cot ^3(2 x) \csc ^3(2 x) \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\csc (2 x)\right )\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\csc (2 x)\right )\right )\\ &=\frac {1}{6} \csc ^3(2 x)-\frac {1}{10} \csc ^5(2 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.00 \[ \frac {1}{6} \csc ^3(2 x)-\frac {1}{10} \csc ^5(2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[2*x]^3*Csc[2*x]^3,x]

[Out]

Csc[2*x]^3/6 - Csc[2*x]^5/10

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fricas [B]  time = 0.44, size = 36, normalized size = 1.71 \[ -\frac {5 \, \cos \left (2 \, x\right )^{2} - 2}{30 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )} \sin \left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)^3*csc(2*x)^3,x, algorithm="fricas")

[Out]

-1/30*(5*cos(2*x)^2 - 2)/((cos(2*x)^4 - 2*cos(2*x)^2 + 1)*sin(2*x))

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giac [A]  time = 1.02, size = 18, normalized size = 0.86 \[ \frac {5 \, \sin \left (2 \, x\right )^{2} - 3}{30 \, \sin \left (2 \, x\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)^3*csc(2*x)^3,x, algorithm="giac")

[Out]

1/30*(5*sin(2*x)^2 - 3)/sin(2*x)^5

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maple [B]  time = 0.06, size = 58, normalized size = 2.76 \[ \frac {\cos ^{4}\left (2 x \right )}{30 \sin \left (2 x \right )}-\frac {\cos ^{4}\left (2 x \right )}{30 \sin \left (2 x \right )^{3}}+\frac {\left (\cos ^{2}\left (2 x \right )+2\right ) \sin \left (2 x \right )}{30}-\frac {\cos ^{4}\left (2 x \right )}{10 \sin \left (2 x \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(2*x)^3*csc(2*x)^3,x)

[Out]

-1/10/sin(2*x)^5*cos(2*x)^4-1/30/sin(2*x)^3*cos(2*x)^4+1/30/sin(2*x)*cos(2*x)^4+1/30*(2+cos(2*x)^2)*sin(2*x)

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maxima [A]  time = 0.52, size = 18, normalized size = 0.86 \[ \frac {5 \, \sin \left (2 \, x\right )^{2} - 3}{30 \, \sin \left (2 \, x\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)^3*csc(2*x)^3,x, algorithm="maxima")

[Out]

1/30*(5*sin(2*x)^2 - 3)/sin(2*x)^5

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mupad [B]  time = 0.38, size = 18, normalized size = 0.86 \[ \frac {5\,{\sin \left (2\,x\right )}^2-3}{30\,{\sin \left (2\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(2*x)^3/sin(2*x)^3,x)

[Out]

(5*sin(2*x)^2 - 3)/(30*sin(2*x)^5)

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sympy [A]  time = 0.12, size = 19, normalized size = 0.90 \[ - \frac {3 - 5 \sin ^{2}{\left (2 x \right )}}{30 \sin ^{5}{\left (2 x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)**3*csc(2*x)**3,x)

[Out]

-(3 - 5*sin(2*x)**2)/(30*sin(2*x)**5)

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