Integrand size = 17, antiderivative size = 58 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3 \csc ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \log (\sin (a+b x))}{b}-\frac {\sin ^2(a+b x)}{2 b} \] Output:
3/2*csc(b*x+a)^2/b-1/4*csc(b*x+a)^4/b+3*ln(sin(b*x+a))/b-1/2*sin(b*x+a)^2/ b
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3 \csc ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \log (\sin (a+b x))}{b}-\frac {\sin ^2(a+b x)}{2 b} \] Input:
Integrate[Cos[a + b*x]^2*Cot[a + b*x]^5,x]
Output:
(3*Csc[a + b*x]^2)/(2*b) - Csc[a + b*x]^4/(4*b) + (3*Log[Sin[a + b*x]])/b - Sin[a + b*x]^2/(2*b)
Time = 0.39 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 25, 3070, 243, 49, 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.
\(\displaystyle \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right )^2 \tan \left (a+b x+\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right )^2 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^5dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle \frac {\int -\csc ^5(a+b x) \left (1-\sin ^2(a+b x)\right )^3d(-\sin (a+b x))}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int -\csc ^3(a+b x) (\sin (a+b x)+1)^3d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\csc ^3(a+b x)-3 \csc ^2(a+b x)-3 \csc (a+b x)-1\right )d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin (a+b x)-\frac {1}{2} \csc ^2(a+b x)-3 \csc (a+b x)+3 \log \left (\sin ^2(a+b x)\right )}{2 b}\) |
Input:
Int[Cos[a + b*x]^2*Cot[a + b*x]^5,x]
Output:
(-3*Csc[a + b*x] - Csc[a + b*x]^2/2 + 3*Log[Sin[a + b*x]^2] + Sin[a + b*x] )/(2*b)
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Time = 3.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.40
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (b x +a \right )^{8}}{4 \sin \left (b x +a \right )^{4}}+\frac {\cos \left (b x +a \right )^{8}}{2 \sin \left (b x +a \right )^{2}}+\frac {\cos \left (b x +a \right )^{6}}{2}+\frac {3 \cos \left (b x +a \right )^{4}}{4}+\frac {3 \cos \left (b x +a \right )^{2}}{2}+3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(81\) |
default | \(\frac {-\frac {\cos \left (b x +a \right )^{8}}{4 \sin \left (b x +a \right )^{4}}+\frac {\cos \left (b x +a \right )^{8}}{2 \sin \left (b x +a \right )^{2}}+\frac {\cos \left (b x +a \right )^{6}}{2}+\frac {3 \cos \left (b x +a \right )^{4}}{4}+\frac {3 \cos \left (b x +a \right )^{2}}{2}+3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(81\) |
risch | \(-3 i x +\frac {{\mathrm e}^{2 i \left (b x +a \right )}}{8 b}+\frac {{\mathrm e}^{-2 i \left (b x +a \right )}}{8 b}-\frac {6 i a}{b}-\frac {2 \left (3 \,{\mathrm e}^{6 i \left (b x +a \right )}-4 \,{\mathrm e}^{4 i \left (b x +a \right )}+3 \,{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\) | \(110\) |
Input:
int(cos(b*x+a)^2*cot(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/4/sin(b*x+a)^4*cos(b*x+a)^8+1/2/sin(b*x+a)^2*cos(b*x+a)^8+1/2*cos( b*x+a)^6+3/4*cos(b*x+a)^4+3/2*cos(b*x+a)^2+3*ln(sin(b*x+a)))
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.55 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{6} - 5 \, \cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 12 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 4}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \] Input:
integrate(cos(b*x+a)^2*cot(b*x+a)^5,x, algorithm="fricas")
Output:
1/4*(2*cos(b*x + a)^6 - 5*cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 12*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*sin(b*x + a)) + 4)/(b*cos(b*x + a)^ 4 - 2*b*cos(b*x + a)^2 + b)
Timed out. \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**2*cot(b*x+a)**5,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{2} - \frac {6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \] Input:
integrate(cos(b*x+a)^2*cot(b*x+a)^5,x, algorithm="maxima")
Output:
-1/4*(2*sin(b*x + a)^2 - (6*sin(b*x + a)^2 - 1)/sin(b*x + a)^4 - 6*log(sin (b*x + a)^2))/b
Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {\cos \left (b x + a\right )^{2}}{2 \, b} + \frac {3 \, \log \left ({\left | \cos \left (b x + a\right )^{2} - 1 \right |}\right )}{2 \, b} - \frac {6 \, \cos \left (b x + a\right )^{2} - 5}{4 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )}^{2} b} \] Input:
integrate(cos(b*x+a)^2*cot(b*x+a)^5,x, algorithm="giac")
Output:
1/2*cos(b*x + a)^2/b + 3/2*log(abs(cos(b*x + a)^2 - 1))/b - 1/4*(6*cos(b*x + a)^2 - 5)/((cos(b*x + a)^2 - 1)^2*b)
Time = 25.56 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^2}{4}-\frac {1}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+{\mathrm {tan}\left (a+b\,x\right )}^4\right )} \] Input:
int(cos(a + b*x)^2*cot(a + b*x)^5,x)
Output:
(3*log(tan(a + b*x)))/b - (3*log(tan(a + b*x)^2 + 1))/(2*b) + ((3*tan(a + b*x)^2)/4 + (3*tan(a + b*x)^4)/2 - 1/4)/(b*(tan(a + b*x)^4 + tan(a + b*x)^ 6))
Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.53 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {-192 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right )^{4}+192 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}-32 \sin \left (b x +a \right )^{6}-81 \sin \left (b x +a \right )^{4}+96 \sin \left (b x +a \right )^{2}-16}{64 \sin \left (b x +a \right )^{4} b} \] Input:
int(cos(b*x+a)^2*cot(b*x+a)^5,x)
Output:
( - 192*log(tan((a + b*x)/2)**2 + 1)*sin(a + b*x)**4 + 192*log(tan((a + b* x)/2))*sin(a + b*x)**4 - 32*sin(a + b*x)**6 - 81*sin(a + b*x)**4 + 96*sin( a + b*x)**2 - 16)/(64*sin(a + b*x)**4*b)