3.897 \(\int \cot (2 x) (-1+\csc ^2(2 x))^2 (1-\sin ^2(2 x))^2 \, dx\)

Optimal. Leaf size=42 \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

[Out]

Csc[2*x]^2 - Csc[2*x]^4/8 + 3*Log[Sin[2*x]] - Sin[2*x]^2 + Sin[2*x]^4/8

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Rubi [A]  time = 0.118309, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3175, 4360, 266, 43} \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

Antiderivative was successfully verified.

[In]

Int[Cot[2*x]*(-1 + Csc[2*x]^2)^2*(1 - Sin[2*x]^2)^2,x]

[Out]

Csc[2*x]^2 - Csc[2*x]^4/8 + 3*Log[Sin[2*x]] - Sin[2*x]^2 + Sin[2*x]^4/8

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 4360

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[1/(b
*c), Subst[Int[SubstFor[1/x, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cot] || EqQ[F, cot])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx &=\int \cos ^4(2 x) \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{x^5} \, dx,x,\sin (2 x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{(1-x)^4}{x^3} \, dx,x,\sin ^2(2 x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (-4+\frac{1}{x^3}-\frac{4}{x^2}+\frac{6}{x}+x\right ) \, dx,x,\sin ^2(2 x)\right )\\ &=\csc ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+3 \log (\sin (2 x))-\sin ^2(2 x)+\frac{1}{8} \sin ^4(2 x)\\ \end{align*}

Mathematica [A]  time = 0.0341055, size = 42, normalized size = 1. \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[2*x]*(-1 + Csc[2*x]^2)^2*(1 - Sin[2*x]^2)^2,x]

[Out]

Csc[2*x]^2 - Csc[2*x]^4/8 + 3*Log[Sin[2*x]] - Sin[2*x]^2 + Sin[2*x]^4/8

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Maple [A]  time = 0.05, size = 37, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sin \left ( 2\,x \right ) \right ) ^{4}}{8}}+ \left ( \cos \left ( 2\,x \right ) \right ) ^{2}+3\,\ln \left ( \sin \left ( 2\,x \right ) \right ) + \left ( \sin \left ( 2\,x \right ) \right ) ^{-2}-{\frac{1}{8\, \left ( \sin \left ( 2\,x \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(2*x)*(-1+csc(2*x)^2)^2*(1-sin(2*x)^2)^2,x)

[Out]

1/8*sin(2*x)^4+cos(2*x)^2+3*ln(sin(2*x))+1/sin(2*x)^2-1/8/sin(2*x)^4

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Maxima [A]  time = 0.939247, size = 59, normalized size = 1.4 \begin{align*} \frac{1}{8} \, \sin \left (2 \, x\right )^{4} - \sin \left (2 \, x\right )^{2} + \frac{8 \, \sin \left (2 \, x\right )^{2} - 1}{8 \, \sin \left (2 \, x\right )^{4}} + \frac{3}{2} \, \log \left (\sin \left (2 \, x\right )^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)*(-1+csc(2*x)^2)^2*(1-sin(2*x)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.51972, size = 220, normalized size = 5.24 \begin{align*} \frac{8 \, \cos \left (2 \, x\right )^{8} + 32 \, \cos \left (2 \, x\right )^{6} - 115 \, \cos \left (2 \, x\right )^{4} + 38 \, \cos \left (2 \, x\right )^{2} + 192 \,{\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \sin \left (2 \, x\right )\right ) + 29}{64 \,{\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)*(-1+csc(2*x)^2)^2*(1-sin(2*x)^2)^2,x, algorithm="fricas")

[Out]

1/64*(8*cos(2*x)^8 + 32*cos(2*x)^6 - 115*cos(2*x)^4 + 38*cos(2*x)^2 + 192*(cos(2*x)^4 - 2*cos(2*x)^2 + 1)*log(
1/2*sin(2*x)) + 29)/(cos(2*x)^4 - 2*cos(2*x)^2 + 1)

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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(cot(2*x)*(-1+csc(2*x)**2)**2*(1-sin(2*x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.09308, size = 70, normalized size = 1.67 \begin{align*} \frac{1}{8} \, \sin \left (2 \, x\right )^{4} - \sin \left (2 \, x\right )^{2} - \frac{18 \, \sin \left (2 \, x\right )^{4} - 8 \, \sin \left (2 \, x\right )^{2} + 1}{8 \, \sin \left (2 \, x\right )^{4}} + \frac{3}{2} \, \log \left (\sin \left (2 \, x\right )^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(2*x)*(-1+csc(2*x)^2)^2*(1-sin(2*x)^2)^2,x, algorithm="giac")

[Out]

1/8*sin(2*x)^4 - sin(2*x)^2 - 1/8*(18*sin(2*x)^4 - 8*sin(2*x)^2 + 1)/sin(2*x)^4 + 3/2*log(sin(2*x)^2)