Integrand size = 19, antiderivative size = 69 \[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=-\frac {(b \csc (e+f x))^m}{f m}+\frac {2 (b \csc (e+f x))^{2+m}}{b^2 f (2+m)}-\frac {(b \csc (e+f x))^{4+m}}{b^4 f (4+m)} \]
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=-\frac {(b \csc (e+f x))^m \left (8+6 m+m^2-2 m (4+m) \csc ^2(e+f x)+m (2+m) \csc ^4(e+f x)\right )}{f m (2+m) (4+m)} \]
-(((b*Csc[e + f*x])^m*(8 + 6*m + m^2 - 2*m*(4 + m)*Csc[e + f*x]^2 + m*(2 + m)*Csc[e + f*x]^4))/(f*m*(2 + m)*(4 + m)))
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 25, 3086, 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.
\(\displaystyle \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan \left (e+f x-\frac {\pi }{2}\right )^5 \left (-\left (b \sec \left (e+f x-\frac {\pi }{2}\right )\right )^m\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \left (b \sec \left (\frac {1}{2} (2 e-\pi )+f x\right )\right )^m \tan \left (\frac {1}{2} (2 e-\pi )+f x\right )^5dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -\frac {b \int (b \csc (e+f x))^{m-1} \left (1-\csc ^2(e+f x)\right )^2d\csc (e+f x)}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {b \int \left ((b \csc (e+f x))^{m-1}-\frac {2 (b \csc (e+f x))^{m+1}}{b^2}+\frac {(b \csc (e+f x))^{m+3}}{b^4}\right )d\csc (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (\frac {(b \csc (e+f x))^{m+4}}{b^5 (m+4)}-\frac {2 (b \csc (e+f x))^{m+2}}{b^3 (m+2)}+\frac {(b \csc (e+f x))^m}{b m}\right )}{f}\) |
-((b*((b*Csc[e + f*x])^m/(b*m) - (2*(b*Csc[e + f*x])^(2 + m))/(b^3*(2 + m) ) + (b*Csc[e + f*x])^(4 + m)/(b^5*(4 + m))))/f)
3.4.78.3.1 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[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.14 (sec) , antiderivative size = 8846, normalized size of antiderivative = 128.20
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.67 \[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=-\frac {{\left ({\left (m^{2} + 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} - 4 \, {\left (m + 4\right )} \cos \left (f x + e\right )^{2} + 8\right )} \left (\frac {b}{\sin \left (f x + e\right )}\right )^{m}}{{\left (f m^{3} + 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4} + f m^{3} + 6 \, f m^{2} - 2 \, {\left (f m^{3} + 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{2} + 8 \, f m} \]
-((m^2 + 6*m + 8)*cos(f*x + e)^4 - 4*(m + 4)*cos(f*x + e)^2 + 8)*(b/sin(f* x + e))^m/((f*m^3 + 6*f*m^2 + 8*f*m)*cos(f*x + e)^4 + f*m^3 + 6*f*m^2 - 2* (f*m^3 + 6*f*m^2 + 8*f*m)*cos(f*x + e)^2 + 8*f*m)
\[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=\begin {cases} x \left (b \csc {\left (e \right )}\right )^{m} \cot ^{5}{\left (e \right )} & \text {for}\: f = 0 \\\frac {\int \frac {\cot ^{5}{\left (e + f x \right )}}{\csc ^{4}{\left (e + f x \right )}}\, dx}{b^{4}} & \text {for}\: m = -4 \\\frac {\int \frac {\cot ^{5}{\left (e + f x \right )}}{\csc ^{2}{\left (e + f x \right )}}\, dx}{b^{2}} & \text {for}\: m = -2 \\- \frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\log {\left (\tan {\left (e + f x \right )} \right )}}{f} + \frac {1}{2 f \tan ^{2}{\left (e + f x \right )}} - \frac {1}{4 f \tan ^{4}{\left (e + f x \right )}} & \text {for}\: m = 0 \\- \frac {m^{2} \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{4}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} - \frac {2 m \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{4}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} + \frac {4 m \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{2}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} - \frac {8 \left (b \csc {\left (e + f x \right )}\right )^{m}}{f m^{3} + 6 f m^{2} + 8 f m} & \text {otherwise} \end {cases} \]
Piecewise((x*(b*csc(e))**m*cot(e)**5, Eq(f, 0)), (Integral(cot(e + f*x)**5 /csc(e + f*x)**4, x)/b**4, Eq(m, -4)), (Integral(cot(e + f*x)**5/csc(e + f *x)**2, x)/b**2, Eq(m, -2)), (-log(tan(e + f*x)**2 + 1)/(2*f) + log(tan(e + f*x))/f + 1/(2*f*tan(e + f*x)**2) - 1/(4*f*tan(e + f*x)**4), Eq(m, 0)), (-m**2*(b*csc(e + f*x))**m*cot(e + f*x)**4/(f*m**3 + 6*f*m**2 + 8*f*m) - 2 *m*(b*csc(e + f*x))**m*cot(e + f*x)**4/(f*m**3 + 6*f*m**2 + 8*f*m) + 4*m*( b*csc(e + f*x))**m*cot(e + f*x)**2/(f*m**3 + 6*f*m**2 + 8*f*m) - 8*(b*csc( e + f*x))**m/(f*m**3 + 6*f*m**2 + 8*f*m), True))
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=-\frac {\frac {b^{m} \sin \left (f x + e\right )^{-m}}{m} - \frac {2 \, b^{m} \sin \left (f x + e\right )^{-m}}{{\left (m + 2\right )} \sin \left (f x + e\right )^{2}} + \frac {b^{m} \sin \left (f x + e\right )^{-m}}{{\left (m + 4\right )} \sin \left (f x + e\right )^{4}}}{f} \]
-(b^m*sin(f*x + e)^(-m)/m - 2*b^m*sin(f*x + e)^(-m)/((m + 2)*sin(f*x + e)^ 2) + b^m*sin(f*x + e)^(-m)/((m + 4)*sin(f*x + e)^4))/f
\[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{5} \,d x } \]
Time = 8.93 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.22 \[ \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx=-\frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {2\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )}{f\,m}-\frac {\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (6\,m^2+4\,m+48\right )}{f\,m\,\left (m^2+6\,m+8\right )}+\frac {2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (4\,m^2+8\,m-32\right )}{f\,m\,\left (m^2+6\,m+8\right )}\right )}{16\,{\sin \left (e+f\,x\right )}^4} \]
-((b/sin(e + f*x))^m*(sin(4*e + 4*f*x)*1i + 2*sin(2*e + 2*f*x)^2 - 1)*((2* (2*sin(2*e + 2*f*x)^2 - 1)*(sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1 ))/(f*m) - ((sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*(4*m + 6*m^2 + 48))/(f*m*(6*m + m^2 + 8)) + (2*(2*sin(e + f*x)^2 - 1)*(sin(4*e + 4*f*x) *1i - 2*sin(2*e + 2*f*x)^2 + 1)*(8*m + 4*m^2 - 32))/(f*m*(6*m + m^2 + 8))) )/(16*sin(e + f*x)^4)