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

Optimal. Leaf size=63 \[ \frac {1}{10} \sin ^5(2 x)-\sin ^3(2 x)+\frac {15}{2} \sin (2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {5}{2} \csc ^3(2 x)+10 \csc (2 x) \]

[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

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Rubi [A]  time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, 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, 4120, 2590, 270} \[ \frac {1}{10} \sin ^5(2 x)-\sin ^3(2 x)+\frac {15}{2} \sin (2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {5}{2} \csc ^3(2 x)+10 \csc (2 x) \]

Antiderivative was successfully verified.

[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 270

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 2590

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 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 4120

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

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Mathematica [A]  time = 0.03, size = 63, normalized size = 1.00 \[ \frac {1}{10} \sin ^5(2 x)-\sin ^3(2 x)+\frac {15}{2} \sin (2 x)-\frac {1}{14} \csc ^7(2 x)+\frac {3}{5} \csc ^5(2 x)-\frac {5}{2} \csc ^3(2 x)+10 \csc (2 x) \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 2.93, size = 84, normalized size = 1.33 \[ -\frac {7 \, \cos \left (2 \, x\right )^{12} + 28 \, \cos \left (2 \, x\right )^{10} + 280 \, \cos \left (2 \, x\right )^{8} - 2240 \, \cos \left (2 \, x\right )^{6} + 4480 \, \cos \left (2 \, x\right )^{4} - 3584 \, \cos \left (2 \, x\right )^{2} + 1024}{70 \, {\left (\cos \left (2 \, x\right )^{6} - 3 \, \cos \left (2 \, x\right )^{4} + 3 \, \cos \left (2 \, x\right )^{2} - 1\right )} \sin \left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.16, size = 57, normalized size = 0.90 \[ \frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.15, size = 56, normalized size = 0.89 \[ \frac {\left (\sin ^{5}\left (2 x \right )\right )}{10}-\left (\sin ^{3}\left (2 x \right )\right )+\frac {15 \sin \left (2 x \right )}{2}+\frac {10}{\sin \left (2 x \right )}-\frac {5}{2 \sin \left (2 x \right )^{3}}+\frac {3}{5 \sin \left (2 x \right )^{5}}-\frac {1}{14 \sin \left (2 x \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 57, normalized size = 0.90 \[ \frac {1}{10} \, \sin \left (2 \, x\right )^{5} - \sin \left (2 \, x\right )^{3} + \frac {700 \, \sin \left (2 \, x\right )^{6} - 175 \, \sin \left (2 \, x\right )^{4} + 42 \, \sin \left (2 \, x\right )^{2} - 5}{70 \, \sin \left (2 \, x\right )^{7}} + \frac {15}{2} \, \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.97, size = 57, normalized size = 0.90 \[ \frac {\frac {{\sin \left (2\,x\right )}^{12}}{10}-{\sin \left (2\,x\right )}^{10}+\frac {15\,{\sin \left (2\,x\right )}^8}{2}+10\,{\sin \left (2\,x\right )}^6-\frac {5\,{\sin \left (2\,x\right )}^4}{2}+\frac {3\,{\sin \left (2\,x\right )}^2}{5}-\frac {1}{14}}{{\sin \left (2\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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