Integrand size = 17, antiderivative size = 58 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=-\frac {\csc ^2(a+b x)}{2 b}-\frac {3 \log (\sin (a+b x))}{b}+\frac {3 \sin ^2(a+b x)}{2 b}-\frac {\sin ^4(a+b x)}{4 b} \]
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=-\frac {\csc ^2(a+b x)}{2 b}-\frac {3 \log (\sin (a+b x))}{b}+\frac {3 \sin ^2(a+b x)}{2 b}-\frac {\sin ^4(a+b x)}{4 b} \]
-1/2*Csc[a + b*x]^2/b - (3*Log[Sin[a + b*x]])/b + (3*Sin[a + b*x]^2)/(2*b) - Sin[a + b*x]^4/(4*b)
Time = 0.23 (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.
\(\displaystyle \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right )^4 \tan \left (a+b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right )^4 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^3dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle \frac {\int -\csc ^3(a+b x) \left (1-\sin ^2(a+b x)\right )^3d(-\sin (a+b x))}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \csc ^2(a+b x) (\sin (a+b x)+1)^3d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\csc ^2(a+b x)+3 \csc (a+b x)+\sin (a+b x)+3\right )d\sin ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} \sin ^2(a+b x)-3 \sin (a+b x)+\csc (a+b x)-3 \log \left (\sin ^2(a+b x)\right )}{2 b}\) |
3.2.45.3.1 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 = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {-\frac {\cos ^{8}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{2}-\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{4}-\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{2}-3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(63\) |
default | \(\frac {-\frac {\cos ^{8}\left (b x +a \right )}{2 \sin \left (b x +a \right )^{2}}-\frac {\left (\cos ^{6}\left (b x +a \right )\right )}{2}-\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{4}-\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{2}-3 \ln \left (\sin \left (b x +a \right )\right )}{b}\) | \(63\) |
risch | \(3 i x -\frac {5 \,{\mathrm e}^{2 i \left (b x +a \right )}}{16 b}-\frac {5 \,{\mathrm e}^{-2 i \left (b x +a \right )}}{16 b}+\frac {6 i a}{b}+\frac {2 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {\cos \left (4 b x +4 a \right )}{32 b}\) | \(99\) |
parallelrisch | \(\frac {96 \ln \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-96 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-92 \cos \left (b x +a \right )+28 \cos \left (2 b x +2 a \right )-4 \cos \left (3 b x +3 a \right )+\cos \left (4 b x +4 a \right )+43\right ) \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-4 \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+24\right ) \left (\csc ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}\) | \(113\) |
norman | \(\frac {-\frac {1}{8 b}-\frac {\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}+\frac {10 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {57 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {57 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {3 \ln \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\) | \(133\) |
1/b*(-1/2/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 = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.22 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=-\frac {8 \, \cos \left (b x + a\right )^{6} + 24 \, \cos \left (b x + a\right )^{4} - 51 \, \cos \left (b x + a\right )^{2} + 96 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 3}{32 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]
-1/32*(8*cos(b*x + a)^6 + 24*cos(b*x + a)^4 - 51*cos(b*x + a)^2 + 96*(cos( b*x + a)^2 - 1)*log(1/2*sin(b*x + a)) + 3)/(b*cos(b*x + a)^2 - b)
Leaf count of result is larger than twice the leaf count of optimal. 1484 vs. \(2 (48) = 96\).
Time = 3.46 (sec) , antiderivative size = 1484, normalized size of antiderivative = 25.59 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=\text {Too large to display} \]
Piecewise((24*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**10/(8*b*tan(a /2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32 *b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 96*log(tan(a/2 + b*x/2 )**2 + 1)*tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b *x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a /2 + b*x/2)**2) + 144*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(8* b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)* *6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 96*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan( a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8* b*tan(a/2 + b*x/2)**2) + 24*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)* *2/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b *x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 24*log(ta n(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan( a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8* b*tan(a/2 + b*x/2)**2) - 96*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(8*b *tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)** 6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 144*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b *x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*ta...
Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=-\frac {\sin \left (b x + a\right )^{4} - 6 \, \sin \left (b x + a\right )^{2} + \frac {2}{\sin \left (b x + a\right )^{2}} + 6 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (52) = 104\).
Time = 0.37 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.98 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=\frac {\frac {{\left (\frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} + \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac {2 \, {\left (\frac {76 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {118 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {76 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} - 12 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 24 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{8 \, b} \]
1/8*((12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)*(cos(b*x + a) + 1)/(co s(b*x + a) - 1) + (cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 2*(76*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 118*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^ 2 + 76*(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 - 12*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 24*log(abs(-(cos( b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b
Time = 0.46 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.47 \[ \int \cos ^4(a+b x) \cot ^3(a+b x) \, dx=\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b}-\frac {3\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+\frac {9\,{\mathrm {tan}\left (a+b\,x\right )}^2}{4}+\frac {1}{2}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^6+2\,{\mathrm {tan}\left (a+b\,x\right )}^4+{\mathrm {tan}\left (a+b\,x\right )}^2\right )} \]