Integrand size = 17, antiderivative size = 58 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3 \csc ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \log (\sin (a+b x))}{b}-\frac {\sin ^2(a+b x)}{2 b} \]
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3 \csc ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {3 \log (\sin (a+b x))}{b}-\frac {\sin ^2(a+b x)}{2 b} \]
(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 = 0.22 (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 |
\(\displaystyle \frac {\sin (a+b x)-\frac {1}{2} \csc ^2(a+b x)-3 \csc (a+b x)+3 \log \left (\sin ^2(a+b x)\right )}{2 b}\) |
3.2.73.3.1 Defintions of rubi rules used
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 = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.40
method | result | size |
derivativedivides | \(\frac {-\frac {\cos ^{8}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {\cos ^{8}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}+\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{2}+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{4}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{2}+3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(81\) |
default | \(\frac {-\frac {\cos ^{8}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {\cos ^{8}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}+\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{2}+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{4}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{2}+3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(81\) |
risch | \(-3 i x +\frac {{\mathrm e}^{2 i \left (b x +a \right )}}{8 b}+\frac {{\mathrm e}^{-2 i \left (b x +a \right )}}{8 b}-\frac {6 i a}{b}-\frac {2 \left (3 \,{\mathrm e}^{6 i \left (b x +a \right )}-4 \,{\mathrm e}^{4 i \left (b x +a \right )}+3 \,{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\) | \(110\) |
parallelrisch | \(\frac {-192 \ln \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+192 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-128 \cos \left (b x +a \right )+16 \cos \left (2 b x +2 a \right )+159\right ) \left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+34 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-\left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+24 \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-105\right ) \left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}\) | \(130\) |
norman | \(\frac {-\frac {1}{64 b}+\frac {9 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}+\frac {9 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}-\frac {\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}-\frac {83 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {3 \ln \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\) | \(133\) |
1/b*(-1/4*cos(b*x+a)^8/sin(b*x+a)^4+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 = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.55 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{6} - 5 \, \cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 12 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 4}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 733 vs. \(2 (48) = 96\).
Time = 3.22 (sec) , antiderivative size = 733, normalized size of antiderivative = 12.64 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\text {Too large to display} \]
Piecewise((-192*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(64*b*tan (a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/ 2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 192*log(ta n(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**8 + 128 *b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 192*log(tan(a/2 + b*x /2))*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2 )**6 + 64*b*tan(a/2 + b*x/2)**4) + 384*log(tan(a/2 + b*x/2))*tan(a/2 + b*x /2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/ 2 + b*x/2)**4) + 192*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a /2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - t an(a/2 + b*x/2)**12/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/ 2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 166*tan(a/ 2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b *tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 1/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4), Ne(b, 0)), (x*cos(a)**7/sin(a)**5, True))
Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{2} - \frac {6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} - 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (52) = 104\).
Time = 0.34 (sec) , antiderivative size = 232, normalized size of antiderivative = 4.00 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {20 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {\frac {18 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {111 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {36 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {72 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}\right )}^{2}} - 96 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 192 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \]
-1/64*(20*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(co s(b*x + a) + 1)^2 + (18*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 111*(cos(b *x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 36*(cos(b*x + a) - 1)^3/(cos(b*x + a ) + 1)^3 - 72*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/((cos(b*x + a ) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2)^2 - 96*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 192*log(abs(-(cos( b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b
Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28 \[ \int \cos ^2(a+b x) \cot ^5(a+b x) \, dx=\frac {3\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^2}{4}-\frac {1}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+{\mathrm {tan}\left (a+b\,x\right )}^4\right )} \]