Integrand size = 17, antiderivative size = 68 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\frac {4 \csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {6 \sin (a+b x)}{b}-\frac {4 \sin ^3(a+b x)}{3 b}+\frac {\sin ^5(a+b x)}{5 b} \]
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\frac {4 \csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {6 \sin (a+b x)}{b}-\frac {4 \sin ^3(a+b x)}{3 b}+\frac {\sin ^5(a+b x)}{5 b} \]
(4*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + (6*Sin[a + b*x])/b - (4*Sin[a + b*x]^3)/(3*b) + Sin[a + b*x]^5/(5*b)
Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3070, 244, 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.
\(\displaystyle \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (a+b x+\frac {\pi }{2}\right )^5 \tan \left (a+b x+\frac {\pi }{2}\right )^4dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle -\frac {\int \csc ^4(a+b x) \left (1-\sin ^2(a+b x)\right )^4d(-\sin (a+b x))}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\csc ^4(a+b x)-4 \csc ^2(a+b x)+\sin ^4(a+b x)-4 \sin ^2(a+b x)+6\right )d(-\sin (a+b x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \sin ^5(a+b x)+\frac {4}{3} \sin ^3(a+b x)-6 \sin (a+b x)+\frac {1}{3} \csc ^3(a+b x)-4 \csc (a+b x)}{b}\) |
-((-4*Csc[a + b*x] + Csc[a + b*x]^3/3 - 6*Sin[a + b*x] + (4*Sin[a + b*x]^3 )/3 - Sin[a + b*x]^5/5)/b)
3.2.57.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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 = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {7 \left (\cos ^{10}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {7 \left (\frac {128}{35}+\cos ^{8}\left (b x +a \right )+\frac {8 \left (\cos ^{6}\left (b x +a \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (b x +a \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (b x +a \right )\right )}{35}\right ) \sin \left (b x +a \right )}{3}}{b}\) | \(90\) |
default | \(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {7 \left (\cos ^{10}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {7 \left (\frac {128}{35}+\cos ^{8}\left (b x +a \right )+\frac {8 \left (\cos ^{6}\left (b x +a \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (b x +a \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (b x +a \right )\right )}{35}\right ) \sin \left (b x +a \right )}{3}}{b}\) | \(90\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{11 i \left (b x +a \right )}+56 \,{\mathrm e}^{9 i \left (b x +a \right )}+1044 \,{\mathrm e}^{7 i \left (b x +a \right )}-7524 \cos \left (b x +a \right )-9612 i \sin \left (b x +a \right )-8565 \cos \left (5 b x +5 a \right )-8571 i \sin \left (5 b x +5 a \right )+13706 \cos \left (3 b x +3 a \right )+13594 i \sin \left (3 b x +3 a \right )\right )}{480 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}\) | \(117\) |
parallelrisch | \(\frac {-5 \left (\tan ^{13}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+200 \left (\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2740 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+7800 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+11298 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+7800 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5 \left (\cot ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2740 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+200 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )}{120 b \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{5}}\) | \(135\) |
norman | \(\frac {-\frac {1}{24 b}+\frac {5 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {137 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{6 b}+\frac {65 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {1883 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{20 b}+\frac {65 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {137 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{6 b}+\frac {5 \left (\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {\tan ^{16}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{5} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}\) | \(162\) |
1/b*(-1/3/sin(b*x+a)^3*cos(b*x+a)^10+7/3/sin(b*x+a)*cos(b*x+a)^10+7/3*(128 /35+cos(b*x+a)^8+8/7*cos(b*x+a)^6+48/35*cos(b*x+a)^4+64/35*cos(b*x+a)^2)*s in(b*x+a))
Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{8} + 8 \, \cos \left (b x + a\right )^{6} + 48 \, \cos \left (b x + a\right )^{4} - 192 \, \cos \left (b x + a\right )^{2} + 128}{15 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]
-1/15*(3*cos(b*x + a)^8 + 8*cos(b*x + a)^6 + 48*cos(b*x + a)^4 - 192*cos(b *x + a)^2 + 128)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))
Time = 1.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.54 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\begin {cases} \frac {128 \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {64 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {16 \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} + \frac {8 \cos ^{6}{\left (a + b x \right )}}{3 b \sin {\left (a + b x \right )}} - \frac {\cos ^{8}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{9}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text {otherwise} \end {cases} \]
Piecewise((128*sin(a + b*x)**5/(15*b) + 64*sin(a + b*x)**3*cos(a + b*x)**2 /(3*b) + 16*sin(a + b*x)*cos(a + b*x)**4/b + 8*cos(a + b*x)**6/(3*b*sin(a + b*x)) - cos(a + b*x)**8/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**9/s in(a)**4, True))
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\frac {3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac {5 \, {\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \]
1/15*(3*sin(b*x + a)^5 - 20*sin(b*x + a)^3 + 5*(12*sin(b*x + a)^2 - 1)/sin (b*x + a)^3 + 90*sin(b*x + a))/b
Time = 0.38 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\frac {3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac {5 \, {\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \]
1/15*(3*sin(b*x + a)^5 - 20*sin(b*x + a)^3 + 5*(12*sin(b*x + a)^2 - 1)/sin (b*x + a)^3 + 90*sin(b*x + a))/b
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81 \[ \int \cos ^5(a+b x) \cot ^4(a+b x) \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^8-20\,{\sin \left (a+b\,x\right )}^6+90\,{\sin \left (a+b\,x\right )}^4+60\,{\sin \left (a+b\,x\right )}^2-5}{15\,b\,{\sin \left (a+b\,x\right )}^3} \]