Integrand size = 27, antiderivative size = 117 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {51 \text {arctanh}(\cos (c+d x))}{8 a^3 d}+\frac {7 \cot (c+d x)}{a^3 d}+\frac {\cot ^3(c+d x)}{a^3 d}-\frac {19 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))} \]
-51/8*arctanh(cos(d*x+c))/a^3/d+7*cot(d*x+c)/a^3/d+cot(d*x+c)^3/a^3/d-19/8 *cot(d*x+c)*csc(d*x+c)/a^3/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a^3/d+4*cos(d*x+c )/a^3/d/(1+sin(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(601\) vs. \(2(117)=234\).
Time = 6.89 (sec) , antiderivative size = 601, normalized size of antiderivative = 5.14 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {8 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}{d (a+a \sin (c+d x))^3}+\frac {3 \cot \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{d (a+a \sin (c+d x))^3}-\frac {19 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{32 d (a+a \sin (c+d x))^3}+\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a+a \sin (c+d x))^3}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{64 d (a+a \sin (c+d x))^3}-\frac {51 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a+a \sin (c+d x))^3}+\frac {51 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a+a \sin (c+d x))^3}+\frac {19 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{32 d (a+a \sin (c+d x))^3}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}{64 d (a+a \sin (c+d x))^3}-\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \tan \left (\frac {1}{2} (c+d x)\right )}{d (a+a \sin (c+d x))^3}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \tan \left (\frac {1}{2} (c+d x)\right )}{8 d (a+a \sin (c+d x))^3} \]
(-8*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)/(d*(a + a*Si n[c + d*x])^3) + (3*Cot[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) ^6)/(d*(a + a*Sin[c + d*x])^3) - (19*Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(32*d*(a + a*Sin[c + d*x])^3) + (Cot[(c + d*x)/2]*C sc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(8*d*(a + a*Sin [c + d*x])^3) - (Csc[(c + d*x)/2]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^ 6)/(64*d*(a + a*Sin[c + d*x])^3) - (51*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x )/2] + Sin[(c + d*x)/2])^6)/(8*d*(a + a*Sin[c + d*x])^3) + (51*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(8*d*(a + a*Sin[c + d* x])^3) + (19*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/( 32*d*(a + a*Sin[c + d*x])^3) + (Sec[(c + d*x)/2]^4*(Cos[(c + d*x)/2] + Sin [(c + d*x)/2])^6)/(64*d*(a + a*Sin[c + d*x])^3) - (3*(Cos[(c + d*x)/2] + S in[(c + d*x)/2])^6*Tan[(c + d*x)/2])/(d*(a + a*Sin[c + d*x])^3) - (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*Tan[(c + d*x)/2])/(8*d *(a + a*Sin[c + d*x])^3)
Time = 0.52 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3354, 3042, 3351, 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.
| \(\displaystyle \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^5 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \csc ^5(c+d x) \sec ^2(c+d x) (a-a \sin (c+d x))^3dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3}{\cos (c+d x)^2 \sin (c+d x)^5}dx}{a^6}\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (a \csc ^5(c+d x)-3 a \csc ^4(c+d x)+4 a \csc ^3(c+d x)-4 a \csc ^2(c+d x)+4 a \csc (c+d x)-\frac {4 a}{\sin (c+d x)+1}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {51 a \text {arctanh}(\cos (c+d x))}{8 d}+\frac {a \cot ^3(c+d x)}{d}+\frac {7 a \cot (c+d x)}{d}+\frac {4 a \cos (c+d x)}{d (\sin (c+d x)+1)}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {19 a \cot (c+d x) \csc (c+d x)}{8 d}}{a^4}\) |
((-51*a*ArcTanh[Cos[c + d*x]])/(8*d) + (7*a*Cot[c + d*x])/d + (a*Cot[c + d *x]^3)/d - (19*a*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d) + (4*a*Cos[c + d*x])/(d*(1 + Sin[c + d*x])))/a^4
3.5.39.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 0.57 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {50}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+102 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {128}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{16 d \,a^{3}}\) | \(139\) |
| default | \(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {50}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+102 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {128}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{16 d \,a^{3}}\) | \(139\) |
| parallelrisch | \(\frac {912+408 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-160 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(147\) |
| risch | \(\frac {51 \,{\mathrm e}^{8 i \left (d x +c \right )}-187 \,{\mathrm e}^{6 i \left (d x +c \right )}+51 i {\mathrm e}^{7 i \left (d x +c \right )}+309 \,{\mathrm e}^{4 i \left (d x +c \right )}-171 i {\mathrm e}^{5 i \left (d x +c \right )}-269 \,{\mathrm e}^{2 i \left (d x +c \right )}+133 i {\mathrm e}^{3 i \left (d x +c \right )}+80-29 i {\mathrm e}^{i \left (d x +c \right )}}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d \,a^{3}}-\frac {51 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}+\frac {51 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}\) | \(171\) |
| norman | \(\frac {-\frac {1}{64 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {35 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {5 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {3 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {153 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {931 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {3975 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {5335 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {3431 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {51 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}\) | \(283\) |
1/16/d/a^3*(1/4*tan(1/2*d*x+1/2*c)^4-2*tan(1/2*d*x+1/2*c)^3+10*tan(1/2*d*x +1/2*c)^2-50*tan(1/2*d*x+1/2*c)-1/4/tan(1/2*d*x+1/2*c)^4+2/tan(1/2*d*x+1/2 *c)^3-10/tan(1/2*d*x+1/2*c)^2+50/tan(1/2*d*x+1/2*c)+102*ln(tan(1/2*d*x+1/2 *c))+128/(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (111) = 222\).
Time = 0.28 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.26 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {160 \, \cos \left (d x + c\right )^{5} + 102 \, \cos \left (d x + c\right )^{4} - 298 \, \cos \left (d x + c\right )^{3} - 170 \, \cos \left (d x + c\right )^{2} - 51 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 51 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (80 \, \cos \left (d x + c\right )^{4} + 29 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} - 35 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right ) + 134 \, \cos \left (d x + c\right ) + 64}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right ) + a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]
1/16*(160*cos(d*x + c)^5 + 102*cos(d*x + c)^4 - 298*cos(d*x + c)^3 - 170*c os(d*x + c)^2 - 51*(cos(d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2 *cos(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c) + c os(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 51*(cos(d*x + c)^5 + cos(d* x + c)^4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d *x + c)^2 + 1)*sin(d*x + c) + cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/ 2) - 2*(80*cos(d*x + c)^4 + 29*cos(d*x + c)^3 - 120*cos(d*x + c)^2 - 35*co s(d*x + c) + 32)*sin(d*x + c) + 134*cos(d*x + c) + 64)/(a^3*d*cos(d*x + c) ^5 + a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x + c)^ 2 + a^3*d*cos(d*x + c) + a^3*d + (a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))
Timed out. \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (111) = 222\).
Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.06 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {32 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {160 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {712 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {\frac {200 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{3}} + \frac {408 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{64 \, d} \]
1/64*((7*sin(d*x + c)/(cos(d*x + c) + 1) - 32*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 160*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 712*sin(d*x + c)^4/(co s(d*x + c) + 1)^4 - 1)/(a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin( d*x + c)^5/(cos(d*x + c) + 1)^5) - (200*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^3 + 408*log(sin(d*x + c)/(co s(d*x + c) + 1))/a^3)/d
Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.49 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {408 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {512}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {850 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 200 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} + \frac {a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 200 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{12}}}{64 \, d} \]
1/64*(408*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 512/(a^3*(tan(1/2*d*x + 1/2 *c) + 1)) - (850*tan(1/2*d*x + 1/2*c)^4 - 200*tan(1/2*d*x + 1/2*c)^3 + 40* tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/ 2*c)^4) + (a^9*tan(1/2*d*x + 1/2*c)^4 - 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 40* a^9*tan(1/2*d*x + 1/2*c)^2 - 200*a^9*tan(1/2*d*x + 1/2*c))/a^12)/d
Time = 10.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.50 \[ \int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}+\frac {51\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^3\,d}-\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}-\frac {1}{64}\right )}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]
(5*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - tan(c/2 + (d*x)/2)^3/(8*a^3*d) + tan( c/2 + (d*x)/2)^4/(64*a^3*d) + (51*log(tan(c/2 + (d*x)/2)))/(8*a^3*d) - (25 *tan(c/2 + (d*x)/2))/(8*a^3*d) + (cot(c/2 + (d*x)/2)^4*((7*tan(c/2 + (d*x) /2))/64 - tan(c/2 + (d*x)/2)^2/2 + (5*tan(c/2 + (d*x)/2)^3)/2 + (89*tan(c/ 2 + (d*x)/2)^4)/8 - 1/64))/(a^3*d*(tan(c/2 + (d*x)/2) + 1))