\(\int \cot (2 x) (-1+\csc ^2(2 x))^2 (1-\sin ^2(2 x))^2 \, dx\) [897]

Optimal result

Integrand size = 27, antiderivative size = 42 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\csc ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+3 \log (\sin (2 x))-\sin ^2(2 x)+\frac {1}{8} \sin ^4(2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3254, 4445, 272, 45} \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

[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 45

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])

Rule 272

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 3254

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 4445

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])

Rubi steps \begin{align*} \text {integral}& = \int \cos ^4(2 x) \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\left (1-x^2\right )^4}{x^5} \, dx,x,\sin (2 x)\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {(1-x)^4}{x^3} \, dx,x,\sin ^2(2 x)\right ) \\ & = \frac {1}{4} \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\csc ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+3 \log (\sin (2 x))-\sin ^2(2 x)+\frac {1}{8} \sin ^4(2 x) \]

[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

Maple [A] (verified)

Time = 14.66 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\sin \left (2 x \right )^{4}}{8}+\cos \left (2 x \right )^{2}+3 \ln \left (\sin \left (2 x \right )\right )+\frac {1}{\sin \left (2 x \right )^{2}}-\frac {1}{8 \sin \left (2 x \right )^{4}}\)	\(37\)
default	\(\frac {\sin \left (2 x \right )^{4}}{8}+\cos \left (2 x \right )^{2}+3 \ln \left (\sin \left (2 x \right )\right )+\frac {1}{\sin \left (2 x \right )^{2}}-\frac {1}{8 \sin \left (2 x \right )^{4}}\)	\(37\)
risch	\(-6 i x +\frac {{\mathrm e}^{8 i x}}{128}+\frac {7 \,{\mathrm e}^{4 i x}}{32}+\frac {7 \,{\mathrm e}^{-4 i x}}{32}+\frac {{\mathrm e}^{-8 i x}}{128}-\frac {2 \left (2 \,{\mathrm e}^{12 i x}-3 \,{\mathrm e}^{8 i x}+2 \,{\mathrm e}^{4 i x}\right )}{\left ({\mathrm e}^{4 i x}-1\right )^{4}}+3 \ln \left ({\mathrm e}^{4 i x}-1\right )\)	\(77\)

[In]

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

[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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (38) = 76\).

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.88 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {8 \, \cos \left (2 \, x\right )^{8} + 32 \, \cos \left (2 \, x\right )^{6} - 115 \, \cos \left (2 \, x\right )^{4} + 38 \, \cos \left (2 \, x\right )^{2} + 192 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (2 \, x\right )\right ) + 29}{64 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{8} \, \sin \left (2 \, x\right )^{4} - \sin \left (2 \, x\right )^{2} + \frac {8 \, \sin \left (2 \, x\right )^{2} - 1}{8 \, \sin \left (2 \, x\right )^{4}} + \frac {3}{2} \, \log \left (\sin \left (2 \, x\right )^{2}\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{8} \, \cos \left (2 \, x\right )^{4} + \frac {3}{4} \, \cos \left (2 \, x\right )^{2} - \frac {8 \, \cos \left (2 \, x\right )^{2} - 7}{8 \, {\left (\cos \left (2 \, x\right )^{2} - 1\right )}^{2}} + \frac {3}{2} \, \log \left (-\cos \left (2 \, x\right )^{2} + 1\right ) \]

[In]

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

[Out]

1/8*cos(2*x)^4 + 3/4*cos(2*x)^2 - 1/8*(8*cos(2*x)^2 - 7)/(cos(2*x)^2 - 1)^2 + 3/2*log(-cos(2*x)^2 + 1)

Mupad [B] (verification not implemented)

Time = 26.99 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.69 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=3\,\ln \left (\mathrm {tan}\left (2\,x\right )\right )-\frac {3\,\ln \left ({\mathrm {tan}\left (2\,x\right )}^2+1\right )}{2}+\frac {3\,{\mathrm {tan}\left (2\,x\right )}^6+\frac {9\,{\mathrm {tan}\left (2\,x\right )}^4}{2}+{\mathrm {tan}\left (2\,x\right )}^2-\frac {1}{4}}{2\,\left ({\mathrm {tan}\left (2\,x\right )}^8+2\,{\mathrm {tan}\left (2\,x\right )}^6+{\mathrm {tan}\left (2\,x\right )}^4\right )} \]

[In]

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

[Out]

3*log(tan(2*x)) - (3*log(tan(2*x)^2 + 1))/2 + (tan(2*x)^2 + (9*tan(2*x)^4)/2 + 3*tan(2*x)^6 - 1/4)/(2*(tan(2*x
)^4 + 2*tan(2*x)^6 + tan(2*x)^8))