\(\int \cos (2 x) (-1+\csc ^2(2 x))^4 (1-\sin ^2(2 x))^2 \, dx\) [898]

Optimal result

Integrand size = 27, antiderivative size = 63 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=10 \csc (2 x)-\frac {5}{2} \csc ^3(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {15}{2} \sin (2 x)-\sin ^3(2 x)+\frac {1}{10} \sin ^5(2 x) \]

[Out]

10*csc(2*x)-5/2*csc(2*x)^3+3/5*csc(2*x)^5-1/14*csc(2*x)^7+15/2*sin(2*x)-sin(2*x)^3+1/10*sin(2*x)^5

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 63, 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, 4205, 2670, 276} \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{10} \sin ^5(2 x)-\sin ^3(2 x)+\frac {15}{2} \sin (2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {5}{2} \csc ^3(2 x)+10 \csc (2 x) \]

[In]

Int[Cos[2*x]*(-1 + Csc[2*x]^2)^4*(1 - Sin[2*x]^2)^2,x]

[Out]

10*Csc[2*x] - (5*Csc[2*x]^3)/2 + (3*Csc[2*x]^5)/5 - Csc[2*x]^7/14 + (15*Sin[2*x])/2 - Sin[2*x]^3 + Sin[2*x]^5/
10

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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 4205

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[b^p, Int[ActivateTrig[u*tan[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^5(2 x) \left (-1+\csc ^2(2 x)\right )^4 \, dx \\ & = \int \cos ^5(2 x) \cot ^8(2 x) \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\left (1-x^2\right )^6}{x^8} \, dx,x,-\sin (2 x)\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (15+\frac {1}{x^8}-\frac {6}{x^6}+\frac {15}{x^4}-\frac {20}{x^2}-6 x^2+x^4\right ) \, dx,x,-\sin (2 x)\right )\right ) \\ & = 10 \csc (2 x)-\frac {5}{2} \csc ^3(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {15}{2} \sin (2 x)-\sin ^3(2 x)+\frac {1}{10} \sin ^5(2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=10 \csc (2 x)-\frac {5}{2} \csc ^3(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {15}{2} \sin (2 x)-\sin ^3(2 x)+\frac {1}{10} \sin ^5(2 x) \]

[In]

Integrate[Cos[2*x]*(-1 + Csc[2*x]^2)^4*(1 - Sin[2*x]^2)^2,x]

[Out]

10*Csc[2*x] - (5*Csc[2*x]^3)/2 + (3*Csc[2*x]^5)/5 - Csc[2*x]^7/14 + (15*Sin[2*x])/2 - Sin[2*x]^3 + Sin[2*x]^5/
10

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {\sec \left (x \right )^{7} \csc \left (x \right )^{7} \left (6062 \cos \left (16 x \right )+429065 \cos \left (8 x \right )-100940 \cos \left (12 x \right )+7 \cos \left (24 x \right )+196 \cos \left (20 x \right )-952952 \cos \left (4 x \right )+608322\right )}{18350080}\)	\(49\)
derivativedivides	\(\frac {\sin \left (2 x \right )^{5}}{10}-\sin \left (2 x \right )^{3}+\frac {15 \sin \left (2 x \right )}{2}+\frac {10}{\sin \left (2 x \right )}-\frac {5}{2 \sin \left (2 x \right )^{3}}+\frac {3}{5 \sin \left (2 x \right )^{5}}-\frac {1}{14 \sin \left (2 x \right )^{7}}\)	\(56\)
default	\(\frac {\sin \left (2 x \right )^{5}}{10}-\sin \left (2 x \right )^{3}+\frac {15 \sin \left (2 x \right )}{2}+\frac {10}{\sin \left (2 x \right )}-\frac {5}{2 \sin \left (2 x \right )^{3}}+\frac {3}{5 \sin \left (2 x \right )^{5}}-\frac {1}{14 \sin \left (2 x \right )^{7}}\)	\(56\)
risch	\(-\frac {i {\mathrm e}^{10 i x}}{320}-\frac {7 i {\mathrm e}^{6 i x}}{64}-\frac {109 i {\mathrm e}^{2 i x}}{32}+\frac {109 i {\mathrm e}^{-2 i x}}{32}+\frac {7 i {\mathrm e}^{-6 i x}}{64}+\frac {i {\mathrm e}^{-10 i x}}{320}+\frac {4 i \left (175 \,{\mathrm e}^{26 i x}-875 \,{\mathrm e}^{22 i x}+2093 \,{\mathrm e}^{18 i x}-2706 \,{\mathrm e}^{14 i x}+2093 \,{\mathrm e}^{10 i x}-875 \,{\mathrm e}^{6 i x}+175 \,{\mathrm e}^{2 i x}\right )}{35 \left ({\mathrm e}^{4 i x}-1\right )^{7}}\)	\(112\)

[In]

int(cos(2*x)*(-1+csc(2*x)^2)^4*(1-sin(2*x)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/18350080*sec(x)^7*csc(x)^7*(6062*cos(16*x)+429065*cos(8*x)-100940*cos(12*x)+7*cos(24*x)+196*cos(20*x)-952952
*cos(4*x)+608322)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.33 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=-\frac {7 \, \cos \left (2 \, x\right )^{12} + 28 \, \cos \left (2 \, x\right )^{10} + 280 \, \cos \left (2 \, x\right )^{8} - 2240 \, \cos \left (2 \, x\right )^{6} + 4480 \, \cos \left (2 \, x\right )^{4} - 3584 \, \cos \left (2 \, x\right )^{2} + 1024}{70 \, {\left (\cos \left (2 \, x\right )^{6} - 3 \, \cos \left (2 \, x\right )^{4} + 3 \, \cos \left (2 \, x\right )^{2} - 1\right )} \sin \left (2 \, x\right )} \]

[In]

integrate(cos(2*x)*(-1+csc(2*x)^2)^4*(1-sin(2*x)^2)^2,x, algorithm="fricas")

[Out]

-1/70*(7*cos(2*x)^12 + 28*cos(2*x)^10 + 280*cos(2*x)^8 - 2240*cos(2*x)^6 + 4480*cos(2*x)^4 - 3584*cos(2*x)^2 +
1024)/((cos(2*x)^6 - 3*cos(2*x)^4 + 3*cos(2*x)^2 - 1)*sin(2*x))

Sympy [F(-1)]

Timed out. \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(2*x)*(-1+csc(2*x)**2)**4*(1-sin(2*x)**2)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]

[In]

integrate(cos(2*x)*(-1+csc(2*x)^2)^4*(1-sin(2*x)^2)^2,x, algorithm="maxima")

[Out]

1/10*sin(2*x)^5 - sin(2*x)^3 + 1/70*(700*sin(2*x)^6 - 175*sin(2*x)^4 + 42*sin(2*x)^2 - 5)/sin(2*x)^7 + 15/2*si
n(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]

[In]

integrate(cos(2*x)*(-1+csc(2*x)^2)^4*(1-sin(2*x)^2)^2,x, algorithm="giac")

[Out]

1/10*sin(2*x)^5 - sin(2*x)^3 + 1/70*(700*sin(2*x)^6 - 175*sin(2*x)^4 + 42*sin(2*x)^2 - 5)/sin(2*x)^7 + 15/2*si
n(2*x)

Mupad [B] (verification not implemented)

Time = 26.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {\frac {{\sin \left (2\,x\right )}^{12}}{10}-{\sin \left (2\,x\right )}^{10}+\frac {15\,{\sin \left (2\,x\right )}^8}{2}+10\,{\sin \left (2\,x\right )}^6-\frac {5\,{\sin \left (2\,x\right )}^4}{2}+\frac {3\,{\sin \left (2\,x\right )}^2}{5}-\frac {1}{14}}{{\sin \left (2\,x\right )}^7} \]

[In]

int(cos(2*x)*(1/sin(2*x)^2 - 1)^4*(sin(2*x)^2 - 1)^2,x)

[Out]

((3*sin(2*x)^2)/5 - (5*sin(2*x)^4)/2 + 10*sin(2*x)^6 + (15*sin(2*x)^8)/2 - sin(2*x)^10 + sin(2*x)^12/10 - 1/14
)/sin(2*x)^7