Integrand size = 27, antiderivative size = 63 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=10 \csc (2 x)-\frac {5}{2} \csc ^3(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {15}{2} \sin (2 x)-\sin ^3(2 x)+\frac {1}{10} \sin ^5(2 x) \]
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)-si n(2*x)^3+1/10*sin(2*x)^5
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=10 \csc (2 x)-\frac {5}{2} \csc ^3(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {15}{2} \sin (2 x)-\sin ^3(2 x)+\frac {1}{10} \sin ^5(2 x) \]
10*Csc[2*x] - (5*Csc[2*x]^3)/2 + (3*Csc[2*x]^5)/5 - Csc[2*x]^7/14 + (15*Si n[2*x])/2 - Sin[2*x]^3 + Sin[2*x]^5/10
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3654, 3042, 4608, 3042, 3070, 244, 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.
\(\displaystyle \int \left (1-\sin ^2(2 x)\right )^2 \cos (2 x) \left (\csc ^2(2 x)-1\right )^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (1-\sin (2 x)^2\right )^2 \cos (2 x) \left (\csc (2 x)^2-1\right )^4dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \int \cos ^5(2 x) \left (1-\csc ^2(2 x)\right )^4dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (2 x+\frac {\pi }{2}\right )^5 \left (1-\sec \left (2 x+\frac {\pi }{2}\right )^2\right )^4dx\) |
\(\Big \downarrow \) 4608 |
\(\displaystyle \int \cos ^5(2 x) \cot ^8(2 x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (2 x+\frac {\pi }{2}\right )^5 \tan \left (2 x+\frac {\pi }{2}\right )^8dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle -\frac {1}{2} \int \csc ^8(2 x) \left (1-\sin ^2(2 x)\right )^6d(-\sin (2 x))\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {1}{2} \int \left (\csc ^8(2 x)-6 \csc ^6(2 x)+15 \csc ^4(2 x)-20 \csc ^2(2 x)+\sin ^4(2 x)-6 \sin ^2(2 x)+15\right )d(-\sin (2 x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \sin ^5(2 x)-2 \sin ^3(2 x)+15 \sin (2 x)-\frac {1}{7} \csc ^7(2 x)+\frac {6}{5} \csc ^5(2 x)-5 \csc ^3(2 x)+20 \csc (2 x)\right )\) |
(20*Csc[2*x] - 5*Csc[2*x]^3 + (6*Csc[2*x]^5)/5 - Csc[2*x]^7/7 + 15*Sin[2*x ] - 2*Sin[2*x]^3 + Sin[2*x]^5/5)/2
3.9.98.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-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]
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]
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ 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]
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 | \(-\frac {i {\mathrm e}^{10 i x}}{320}-\frac {7 i {\mathrm e}^{6 i x}}{64}-\frac {109 i {\mathrm e}^{2 i x}}{32}+\frac {109 i {\mathrm e}^{-2 i x}}{32}+\frac {7 i {\mathrm e}^{-6 i x}}{64}+\frac {i {\mathrm e}^{-10 i x}}{320}+\frac {4 i \left (175 \,{\mathrm e}^{26 i x}-875 \,{\mathrm e}^{22 i x}+2093 \,{\mathrm e}^{18 i x}-2706 \,{\mathrm e}^{14 i x}+2093 \,{\mathrm e}^{10 i x}-875 \,{\mathrm e}^{6 i x}+175 \,{\mathrm e}^{2 i x}\right )}{35 \left ({\mathrm e}^{4 i x}-1\right )^{7}}\) | \(112\) |
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)
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.33 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=-\frac {7 \, \cos \left (2 \, x\right )^{12} + 28 \, \cos \left (2 \, x\right )^{10} + 280 \, \cos \left (2 \, x\right )^{8} - 2240 \, \cos \left (2 \, x\right )^{6} + 4480 \, \cos \left (2 \, x\right )^{4} - 3584 \, \cos \left (2 \, x\right )^{2} + 1024}{70 \, {\left (\cos \left (2 \, x\right )^{6} - 3 \, \cos \left (2 \, x\right )^{4} + 3 \, \cos \left (2 \, x\right )^{2} - 1\right )} \sin \left (2 \, x\right )} \]
-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))
Timed out. \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]
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*sin(2*x)
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]
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*sin(2*x)
Time = 26.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \cos (2 x) \left (-1+\csc ^2(2 x)\right )^4 \left (1-\sin ^2(2 x)\right )^2 \, dx=\frac {\frac {{\sin \left (2\,x\right )}^{12}}{10}-{\sin \left (2\,x\right )}^{10}+\frac {15\,{\sin \left (2\,x\right )}^8}{2}+10\,{\sin \left (2\,x\right )}^6-\frac {5\,{\sin \left (2\,x\right )}^4}{2}+\frac {3\,{\sin \left (2\,x\right )}^2}{5}-\frac {1}{14}}{{\sin \left (2\,x\right )}^7} \]