Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^5(c+d x)}{5 a d} \]
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^2(c+d x) \left (-30+20 \csc (c+d x)+15 \csc ^2(c+d x)-12 \csc ^3(c+d x)\right )}{60 a d} \]
Time = 0.43 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3314, 3042, 25, 3086, 25, 244, 2009, 3087, 15}
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.
\(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cot ^3(c+d x) \csc ^3(c+d x)dx}{a}-\frac {\int \cot ^3(c+d x) \csc ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\int -\csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}+\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\csc ^2(c+d x)-\csc ^4(c+d x)\right )d\csc (c+d x)}{a d}+\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int -\cot ^3(c+d x)d(-\cot (c+d x))}{a d}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\cot ^4(c+d x)}{4 a d}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
3.6.41.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}}{d a}\) | \(49\) |
default | \(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}}{d a}\) | \(49\) |
parallelrisch | \(-\frac {\left (256+1280 \cos \left (2 d x +2 c \right )-75 \sin \left (5 d x +5 c \right )+210 \sin \left (d x +c \right )-585 \sin \left (3 d x +3 c \right )\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{245760 d a}\) | \(74\) |
risch | \(\frac {-\frac {8 i {\mathrm e}^{7 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{8 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{5 i \left (d x +c \right )}}{15}-2 \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {8 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\) | \(103\) |
norman | \(\frac {-\frac {1}{160 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{320 d a}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(166\) |
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]
-1/60*(20*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/((a *d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))
Timed out. \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]
Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \]
Time = 9.93 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-30\,{\sin \left (c+d\,x\right )}^3+20\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-12}{60\,a\,d\,{\sin \left (c+d\,x\right )}^5} \]