Integrand size = 17, antiderivative size = 69 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \csc ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {6 \log (\sin (a+b x))}{b}-\frac {2 \sin ^2(a+b x)}{b}+\frac {\sin ^4(a+b x)}{4 b} \]
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \csc ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {6 \log (\sin (a+b x))}{b}-\frac {2 \sin ^2(a+b x)}{b}+\frac {\sin ^4(a+b x)}{4 b} \]
(2*Csc[a + b*x]^2)/b - Csc[a + b*x]^4/(4*b) + (6*Log[Sin[a + b*x]])/b - (2 *Sin[a + b*x]^2)/b + Sin[a + b*x]^4/(4*b)
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86, 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.
\(\displaystyle \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right )^4 \tan \left (a+b x+\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right )^4 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^5dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle \frac {\int -\csc ^5(a+b x) \left (1-\sin ^2(a+b x)\right )^4d(-\sin (a+b x))}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int -\csc ^3(a+b x) (\sin (a+b x)+1)^4d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\csc ^3(a+b x)-4 \csc ^2(a+b x)-6 \csc (a+b x)+\sin ^2(a+b x)-4\right )d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} \sin ^2(a+b x)+4 \sin (a+b x)-\frac {1}{2} \csc ^2(a+b x)-4 \csc (a+b x)+6 \log \left (\sin ^2(a+b x)\right )}{2 b}\) |
(-4*Csc[a + b*x] - Csc[a + b*x]^2/2 + 6*Log[Sin[a + b*x]^2] + 4*Sin[a + b* x] + Sin[a + b*x]^2/2)/(2*b)
3.2.71.3.1 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 = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29
method | result | size |
derivativedivides | \(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{10}\left (b x +a \right )\right )}{4 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{8}\left (b x +a \right )\right )}{4}+\cos ^{6}\left (b x +a \right )+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{2}+3 \left (\cos ^{2}\left (b x +a \right )\right )+6 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(89\) |
default | \(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{10}\left (b x +a \right )\right )}{4 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{8}\left (b x +a \right )\right )}{4}+\cos ^{6}\left (b x +a \right )+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{2}+3 \left (\cos ^{2}\left (b x +a \right )\right )+6 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(89\) |
risch | \(-6 i x +\frac {{\mathrm e}^{4 i \left (b x +a \right )}}{64 b}+\frac {7 \,{\mathrm e}^{2 i \left (b x +a \right )}}{16 b}+\frac {7 \,{\mathrm e}^{-2 i \left (b x +a \right )}}{16 b}+\frac {{\mathrm e}^{-4 i \left (b x +a \right )}}{64 b}-\frac {12 i a}{b}-\frac {4 \left (2 \,{\mathrm e}^{6 i \left (b x +a \right )}-3 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\) | \(138\) |
parallelrisch | \(-\frac {3 \left (\left (\frac {3}{4}-\cos \left (2 b x +2 a \right )+\frac {\cos \left (4 b x +4 a \right )}{4}\right ) \ln \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (\cos \left (2 b x +2 a \right )-\frac {\cos \left (4 b x +4 a \right )}{4}-\frac {3}{4}\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\frac {71 \cos \left (2 b x +2 a \right )}{192}+\frac {41 \cos \left (4 b x +4 a \right )}{192}-\frac {\cos \left (6 b x +6 a \right )}{64}-\frac {\cos \left (8 b x +8 a \right )}{1536}+\frac {131}{512}\right ) \left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16 b}\) | \(144\) |
norman | \(\frac {-\frac {1}{64 b}+\frac {3 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {3 \left (\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {\tan ^{16}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}-\frac {93 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {93 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {591 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {6 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {6 \ln \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\) | \(165\) |
1/b*(-1/4/sin(b*x+a)^4*cos(b*x+a)^10+3/4/sin(b*x+a)^2*cos(b*x+a)^10+3/4*co s(b*x+a)^8+cos(b*x+a)^6+3/2*cos(b*x+a)^4+3*cos(b*x+a)^2+6*ln(sin(b*x+a)))
Time = 0.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.45 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {8 \, \cos \left (b x + a\right )^{8} + 32 \, \cos \left (b x + a\right )^{6} - 115 \, \cos \left (b x + a\right )^{4} + 38 \, \cos \left (b x + a\right )^{2} + 192 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 29}{32 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]
1/32*(8*cos(b*x + a)^8 + 32*cos(b*x + a)^6 - 115*cos(b*x + a)^4 + 38*cos(b *x + a)^2 + 192*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*sin(b*x + a)) + 29)/(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. 1664 vs. \(2 (58) = 116\).
Time = 7.68 (sec) , antiderivative size = 1664, normalized size of antiderivative = 24.12 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\text {Too large to display} \]
Piecewise((-384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**12/(64*b*ta n(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)** 8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 1536*log(tan(a /2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**12 + 256* b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2 )**6 + 64*b*tan(a/2 + b*x/2)**4) - 2304*log(tan(a/2 + b*x/2)**2 + 1)*tan(a /2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 3 84*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/ 2)**4) - 1536*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(64*b*tan(a /2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*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)**4/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan (a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*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)** 12/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 153 6*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b *x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 2304*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 3...
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{4} - 8 \, \sin \left (b x + a\right )^{2} + \frac {8 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} + 12 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]
1/4*(sin(b*x + a)^4 - 8*sin(b*x + a)^2 + (8*sin(b*x + a)^2 - 1)/sin(b*x + a)^4 + 12*log(sin(b*x + a)^2))/b
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (65) = 130\).
Time = 0.32 (sec) , antiderivative size = 277, normalized size of antiderivative = 4.01 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {{\left (\frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {288 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {32 \, {\left (\frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {126 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} - 192 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 384 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \]
-1/64*((28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 288*(cos(b*x + a) - 1)^ 2/(cos(b*x + a) + 1)^2 + 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 + 28 *(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a ) + 1)^2 + 32*(84*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 126*(cos(b*x + a ) - 1)^2/(cos(b*x + a) + 1)^2 + 84*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1) ^3 - 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 25)/((cos(b*x + a) - 1 )/(cos(b*x + a) + 1) - 1)^4 - 192*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 384*log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b
Time = 1.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {6\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{b}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^6+\frac {9\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+{\mathrm {tan}\left (a+b\,x\right )}^2-\frac {1}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^8+2\,{\mathrm {tan}\left (a+b\,x\right )}^6+{\mathrm {tan}\left (a+b\,x\right )}^4\right )} \]