\(\int \cos ^4(2 x) \cot ^5(2 x) \, dx\) [902]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\csc ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+3 \log (\sin (2 x))-\sin ^2(2 x)+\frac {1}{8} \sin ^4(2 x) \]

[Out]

csc(2*x)^2-1/8*csc(2*x)^4+3*ln(sin(2*x))-sin(2*x)^2+1/8*sin(2*x)^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2670, 272, 45} \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\frac {1}{8} \sin ^4(2 x)-\sin ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+\csc ^2(2 x)+3 \log (\sin (2 x)) \]

[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 45

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])

Rule 272

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\left (1-x^2\right )^4}{x^5} \, dx,x,-\sin (2 x)\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {(1-x)^4}{x^3} \, dx,x,\sin ^2(2 x)\right ) \\ & = \frac {1}{4} \text {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] (verified)

Time = 0.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\csc ^2(2 x)-\frac {1}{8} \csc ^4(2 x)+3 \log (\sin (2 x))-\sin ^2(2 x)+\frac {1}{8} \sin ^4(2 x) \]

[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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.64

\[-\frac {\cos \left (2 x \right )^{10}}{8 \sin \left (2 x \right )^{4}}+\frac {3 \cos \left (2 x \right )^{10}}{8 \sin \left (2 x \right )^{2}}+\frac {3 \cos \left (2 x \right )^{8}}{8}+\frac {\cos \left (2 x \right )^{6}}{2}+\frac {3 \cos \left (2 x \right )^{4}}{4}+\frac {3 \cos \left (2 x \right )^{2}}{2}+3 \ln \left (\sin \left (2 x \right )\right )\]

[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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (38) = 76\).

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.88 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\frac {8 \, \cos \left (2 \, x\right )^{8} + 32 \, \cos \left (2 \, x\right )^{6} - 115 \, \cos \left (2 \, x\right )^{4} + 38 \, \cos \left (2 \, x\right )^{2} + 192 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (2 \, x\right )\right ) + 29}{64 \, {\left (\cos \left (2 \, x\right )^{4} - 2 \, \cos \left (2 \, x\right )^{2} + 1\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\frac {8 \sin ^{2}{\left (2 x \right )} - 1}{8 \sin ^{4}{\left (2 x \right )}} + 3 \log {\left (\sin {\left (2 x \right )} \right )} + \frac {\sin ^{4}{\left (2 x \right )}}{8} - \sin ^{2}{\left (2 x \right )} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\frac {1}{8} \, \sin \left (2 \, x\right )^{4} - \sin \left (2 \, x\right )^{2} + \frac {8 \, \sin \left (2 \, x\right )^{2} - 1}{8 \, \sin \left (2 \, x\right )^{4}} + \frac {3}{2} \, \log \left (\sin \left (2 \, x\right )^{2}\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=\frac {1}{8} \, \cos \left (2 \, x\right )^{4} + \frac {3}{4} \, \cos \left (2 \, x\right )^{2} - \frac {8 \, \cos \left (2 \, x\right )^{2} - 7}{8 \, {\left (\cos \left (2 \, x\right )^{2} - 1\right )}^{2}} + \frac {3}{2} \, \log \left (-\cos \left (2 \, x\right )^{2} + 1\right ) \]

[In]

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

[Out]

1/8*cos(2*x)^4 + 3/4*cos(2*x)^2 - 1/8*(8*cos(2*x)^2 - 7)/(cos(2*x)^2 - 1)^2 + 3/2*log(-cos(2*x)^2 + 1)

Mupad [B] (verification not implemented)

Time = 27.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.69 \[ \int \cos ^4(2 x) \cot ^5(2 x) \, dx=3\,\ln \left (\mathrm {tan}\left (2\,x\right )\right )-\frac {3\,\ln \left ({\mathrm {tan}\left (2\,x\right )}^2+1\right )}{2}+\frac {3\,{\mathrm {tan}\left (2\,x\right )}^6+\frac {9\,{\mathrm {tan}\left (2\,x\right )}^4}{2}+{\mathrm {tan}\left (2\,x\right )}^2-\frac {1}{4}}{2\,\left ({\mathrm {tan}\left (2\,x\right )}^8+2\,{\mathrm {tan}\left (2\,x\right )}^6+{\mathrm {tan}\left (2\,x\right )}^4\right )} \]

[In]

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

[Out]

3*log(tan(2*x)) - (3*log(tan(2*x)^2 + 1))/2 + (tan(2*x)^2 + (9*tan(2*x)^4)/2 + 3*tan(2*x)^6 - 1/4)/(2*(tan(2*x
)^4 + 2*tan(2*x)^6 + tan(2*x)^8))