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

3.9.97.1 Optimal result

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

output

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
 

3.9.97.2 Mathematica [A] (verified)

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

input

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

output

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

3.9.97.3 Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3654, 3042, 4860, 243, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \left (1-\sin ^2(2 x)\right )^2 \cot (2 x) \left (\csc ^2(2 x)-1\right )^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (1-\sin (2 x)^2\right )^2 \cot (2 x) \left (\csc (2 x)^2-1\right )^2dx\)

\(\Big \downarrow \) 3654

\(\displaystyle \int \cos ^4(2 x) \cot (2 x) \left (1-\csc ^2(2 x)\right )^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (2 x)^4 \cot (2 x) \left (1-\csc (2 x)^2\right )^2dx\)

\(\Big \downarrow \) 4860

\(\displaystyle \frac {1}{2} \int \csc ^5(2 x) \left (1-\sin ^2(2 x)\right )^4d\sin (2 x)\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} \int \csc ^3(2 x) \left (1-\sin ^2(2 x)\right )^4d\sin ^2(2 x)\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {1}{4} \int \left (\csc ^3(2 x)-4 \csc ^2(2 x)+6 \csc (2 x)+\sin ^2(2 x)-4\right )d\sin ^2(2 x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \sin ^4(2 x)-4 \sin ^2(2 x)-\frac {1}{2} \csc ^2(2 x)+4 \csc (2 x)+6 \log \left (\sin ^2(2 x)\right )\right )\)

input

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

output

(4*Csc[2*x] - Csc[2*x]^2/2 + 6*Log[Sin[2*x]^2] - 4*Sin[2*x]^2 + Sin[2*x]^4 
/2)/4
 

3.9.97.3.1 Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3654

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ 
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 4860

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto 
rs[Sin[c*(a + b*x)], x]}, Simp[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])
 

3.9.97.4 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	\(-6 i x +\frac {{\mathrm e}^{8 i x}}{128}+\frac {7 \,{\mathrm e}^{4 i x}}{32}+\frac {7 \,{\mathrm e}^{-4 i x}}{32}+\frac {{\mathrm e}^{-8 i x}}{128}-\frac {2 \left (2 \,{\mathrm e}^{12 i x}-3 \,{\mathrm e}^{8 i x}+2 \,{\mathrm e}^{4 i x}\right )}{\left ({\mathrm e}^{4 i x}-1\right )^{4}}+3 \ln \left ({\mathrm e}^{4 i x}-1\right )\)	\(77\)

input

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

output

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
 

3.9.97.5 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 \[ \int \cot (2 x) \left (-1+\csc ^2(2 x)\right )^2 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {8 \, \cos \left (2 \, x\right )^{8} + 32 \, \cos \left (2 \, x\right )^{6} - 115 \, \cos \left (2 \, x\right )^{4} + 38 \, \cos \left (2 \, x\right )^{2} + 192 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (2 \, x\right )\right ) + 29}{64 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )}} \]

input

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

output

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*co 
s(2*x)^2 + 1)
 

3.9.97.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.9.97.7 Maxima [A] (verification not implemented)

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

input

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

output

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)
 

3.9.97.8 Giac [A] (verification not implemented)

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

input

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

output

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)
 

3.9.97.9 Mupad [B] (verification not implemented)

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

input

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

output

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))