3.902 \(\int \cos ^4(2 x) \cot ^5(2 x) \, dx\)

Optimal. Leaf size=42 \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

[Out]

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

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Rubi [A]  time = 0.0406121, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2590, 266, 43} \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

Antiderivative was successfully verified.

[In]

Int[Cos[2*x]^4*Cot[2*x]^5,x]

[Out]

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos ^4(2 x) \cot ^5(2 x) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{x^5} \, dx,x,-\sin (2 x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{(1-x)^4}{x^3} \, dx,x,\sin ^2(2 x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (-4+\frac{1}{x^3}-\frac{4}{x^2}+\frac{6}{x}+x\right ) \, dx,x,\sin ^2(2 x)\right )\\ &=\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)\\ \end{align*}

Mathematica [A]  time = 0.0279163, size = 42, normalized size = 1. \[ \frac{1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac{1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[2*x]^4*Cot[2*x]^5,x]

[Out]

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

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Maple [A]  time = 0.027, size = 69, normalized size = 1.6 \begin{align*} -{\frac{ \left ( \cos \left ( 2\,x \right ) \right ) ^{10}}{8\, \left ( \sin \left ( 2\,x \right ) \right ) ^{4}}}+{\frac{3\, \left ( \cos \left ( 2\,x \right ) \right ) ^{10}}{8\, \left ( \sin \left ( 2\,x \right ) \right ) ^{2}}}+{\frac{3\, \left ( \cos \left ( 2\,x \right ) \right ) ^{8}}{8}}+{\frac{ \left ( \cos \left ( 2\,x \right ) \right ) ^{6}}{2}}+{\frac{3\, \left ( \cos \left ( 2\,x \right ) \right ) ^{4}}{4}}+{\frac{3\, \left ( \cos \left ( 2\,x \right ) \right ) ^{2}}{2}}+3\,\ln \left ( \sin \left ( 2\,x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)^4*cot(2*x)^5,x)

[Out]

-1/8/sin(2*x)^4*cos(2*x)^10+3/8/sin(2*x)^2*cos(2*x)^10+3/8*cos(2*x)^8+1/2*cos(2*x)^6+3/4*cos(2*x)^4+3/2*cos(2*
x)^2+3*ln(sin(2*x))

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Maxima [A]  time = 0.95173, size = 59, normalized size = 1.4 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*cot(2*x)^5,x, algorithm="maxima")

[Out]

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)

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Fricas [B]  time = 2.23845, size = 220, normalized size = 5.24 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*cot(2*x)^5,x, algorithm="fricas")

[Out]

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*cos(2*x)^2 + 1)

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Sympy [A]  time = 0.108527, size = 41, normalized size = 0.98 \begin{align*} \frac{8 \sin ^{2}{\left (2 x \right )} - 1}{8 \sin ^{4}{\left (2 x \right )}} + 3 \log{\left (\sin{\left (2 x \right )} \right )} + \frac{\sin ^{4}{\left (2 x \right )}}{8} - \sin ^{2}{\left (2 x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)**4*cot(2*x)**5,x)

[Out]

(8*sin(2*x)**2 - 1)/(8*sin(2*x)**4) + 3*log(sin(2*x)) + sin(2*x)**4/8 - sin(2*x)**2

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Giac [B]  time = 1.14504, size = 300, normalized size = 7.14 \begin{align*} -\frac{{\left (\frac{28 \,{\left (\cos \left (2 \, x\right ) - 1\right )}}{\cos \left (2 \, x\right ) + 1} + \frac{288 \,{\left (\cos \left (2 \, x\right ) - 1\right )}^{2}}{{\left (\cos \left (2 \, x\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (2 \, x\right ) + 1\right )}^{2}}{128 \,{\left (\cos \left (2 \, x\right ) - 1\right )}^{2}} - \frac{7 \,{\left (\cos \left (2 \, x\right ) - 1\right )}}{32 \,{\left (\cos \left (2 \, x\right ) + 1\right )}} - \frac{{\left (\cos \left (2 \, x\right ) - 1\right )}^{2}}{128 \,{\left (\cos \left (2 \, x\right ) + 1\right )}^{2}} - \frac{\frac{84 \,{\left (\cos \left (2 \, x\right ) - 1\right )}}{\cos \left (2 \, x\right ) + 1} - \frac{126 \,{\left (\cos \left (2 \, x\right ) - 1\right )}^{2}}{{\left (\cos \left (2 \, x\right ) + 1\right )}^{2}} + \frac{84 \,{\left (\cos \left (2 \, x\right ) - 1\right )}^{3}}{{\left (\cos \left (2 \, x\right ) + 1\right )}^{3}} - \frac{25 \,{\left (\cos \left (2 \, x\right ) - 1\right )}^{4}}{{\left (\cos \left (2 \, x\right ) + 1\right )}^{4}} - 25}{4 \,{\left (\frac{\cos \left (2 \, x\right ) - 1}{\cos \left (2 \, x\right ) + 1} - 1\right )}^{4}} - 3 \, \log \left (-\frac{\cos \left (2 \, x\right ) - 1}{\cos \left (2 \, x\right ) + 1} + 1\right ) + \frac{3}{2} \, \log \left (-\frac{\cos \left (2 \, x\right ) - 1}{\cos \left (2 \, x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*cot(2*x)^5,x, algorithm="giac")

[Out]

-1/128*(28*(cos(2*x) - 1)/(cos(2*x) + 1) + 288*(cos(2*x) - 1)^2/(cos(2*x) + 1)^2 + 1)*(cos(2*x) + 1)^2/(cos(2*
x) - 1)^2 - 7/32*(cos(2*x) - 1)/(cos(2*x) + 1) - 1/128*(cos(2*x) - 1)^2/(cos(2*x) + 1)^2 - 1/4*(84*(cos(2*x) -
1)/(cos(2*x) + 1) - 126*(cos(2*x) - 1)^2/(cos(2*x) + 1)^2 + 84*(cos(2*x) - 1)^3/(cos(2*x) + 1)^3 - 25*(cos(2*
x) - 1)^4/(cos(2*x) + 1)^4 - 25)/((cos(2*x) - 1)/(cos(2*x) + 1) - 1)^4 - 3*log(-(cos(2*x) - 1)/(cos(2*x) + 1)
+ 1) + 3/2*log(-(cos(2*x) - 1)/(cos(2*x) + 1))