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*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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Rubi [A]  time = 0.123567, antiderivative size = 63, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0306683, size = 63, normalized size = 1. \[ \frac{1}{10} \sin ^5(2 x)-\sin ^3(2 x)+\frac{15}{2} \sin (2 x)-\frac{1}{14} \csc ^7(2 x)+\frac{3}{5} \csc ^5(2 x)-\frac{5}{2} \csc ^3(2 x)+10 \csc (2 x) \]

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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Maple [A]  time = 0.051, size = 56, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sin \left ( 2\,x \right ) \right ) ^{5}}{10}}- \left ( \sin \left ( 2\,x \right ) \right ) ^{3}+{\frac{15\,\sin \left ( 2\,x \right ) }{2}}+10\, \left ( \sin \left ( 2\,x \right ) \right ) ^{-1}-{\frac{5}{2\, \left ( \sin \left ( 2\,x \right ) \right ) ^{3}}}+{\frac{3}{5\, \left ( \sin \left ( 2\,x \right ) \right ) ^{5}}}-{\frac{1}{14\, \left ( \sin \left ( 2\,x \right ) \right ) ^{7}}} \end{align*}

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 = 0.955094, size = 77, normalized size = 1.22 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 2.42875, size = 238, normalized size = 3.78 \begin{align*} -\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 )} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.09792, size = 77, normalized size = 1.22 \begin{align*} \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 ) \end{align*}

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)