Integrand size = 29, antiderivative size = 112 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d} \]
3/4*arctanh(cos(d*x+c))/a^2/d-2*cot(d*x+c)/a^2/d-cot(d*x+c)^3/a^2/d-1/5*co t(d*x+c)^5/a^2/d+3/4*cot(d*x+c)*csc(d*x+c)/a^2/d+1/2*cot(d*x+c)*csc(d*x+c) ^3/a^2/d
Time = 1.40 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.69 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^5(c+d x) \left (-160 \cos (c+d x)+120 \cos (3 (c+d x))-24 \cos (5 (c+d x))+150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+140 \sin (2 (c+d x))-75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-30 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{320 a^2 d} \]
(Csc[c + d*x]^5*(-160*Cos[c + d*x] + 120*Cos[3*(c + d*x)] - 24*Cos[5*(c + d*x)] + 150*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 150*Log[Sin[(c + d*x)/2]] *Sin[c + d*x] + 140*Sin[2*(c + d*x)] - 75*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 30*Sin[4*(c + d*x)] + 15*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[ 5*(c + d*x)]))/(320*a^2*d)
Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3348, 3042, 3236, 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.
| \(\displaystyle \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^6 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3348 |
| \(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2}{\sin (c+d x)^6}dx}{a^4}\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \frac {\int \left (a^2 \csc ^6(c+d x)-2 a^2 \csc ^5(c+d x)+a^2 \csc ^4(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{4 d}}{a^4}\) |
((3*a^2*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/d - (a^2*Cot[c + d*x]^5)/(5*d) + (3*a^2*Cot[c + d*x]*Csc[c + d* x])/(4*d) + (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d))/a^4
3.5.30.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.30
| method | result | size |
| risch | \(-\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}-40 i {\mathrm e}^{6 i \left (d x +c \right )}-70 \,{\mathrm e}^{7 i \left (d x +c \right )}+200 i {\mathrm e}^{4 i \left (d x +c \right )}-120 i {\mathrm e}^{2 i \left (d x +c \right )}+70 \,{\mathrm e}^{3 i \left (d x +c \right )}+24 i-15 \,{\mathrm e}^{i \left (d x +c \right )}}{10 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}\) | \(146\) |
| parallelrisch | \(\frac {-\left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+5 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-110 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+110 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d \,a^{2}}\) | \(146\) |
| derivativedivides | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {22}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{2}}\) | \(148\) |
| default | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {22}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{2}}\) | \(148\) |
| norman | \(\frac {-\frac {1}{160 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}+\frac {9 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {9 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}-\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}-\frac {45 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {171 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {99 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) | \(283\) |
-1/10*(15*exp(9*I*(d*x+c))-40*I*exp(6*I*(d*x+c))-70*exp(7*I*(d*x+c))+200*I *exp(4*I*(d*x+c))-120*I*exp(2*I*(d*x+c))+70*exp(3*I*(d*x+c))+24*I-15*exp(I *(d*x+c)))/a^2/d/(exp(2*I*(d*x+c))-1)^5-3/4/d/a^2*ln(exp(I*(d*x+c))-1)+3/4 /d/a^2*ln(exp(I*(d*x+c))+1)
Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.60 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {48 \, \cos \left (d x + c\right )^{5} - 120 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 10 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 80 \, \cos \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]
-1/40*(48*cos(d*x + c)^5 - 120*cos(d*x + c)^3 - 15*(cos(d*x + c)^4 - 2*cos (d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 1 0*(3*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 80*cos(d*x + c))/((a^ 2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)*sin(d*x + c))
Timed out. \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (104) = 208\).
Time = 0.33 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.08 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {110 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {110 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{160 \, d} \]
1/160*((110*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5*sin(d*x + c)^4/(co s(d*x + c) + 1)^4 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 - 120*log(sin (d*x + c)/(cos(d*x + c) + 1))/a^2 + (5*sin(d*x + c)/(cos(d*x + c) + 1) - 1 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 110*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)*(cos(d*x + c) + 1)^5/( a^2*sin(d*x + c)^5))/d
Time = 0.37 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.66 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 110 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{160 \, d} \]
-1/160*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (274*tan(1/2*d*x + 1/2*c) ^5 - 110*tan(1/2*d*x + 1/2*c)^4 + 40*tan(1/2*d*x + 1/2*c)^3 - 15*tan(1/2*d *x + 1/2*c)^2 + 5*tan(1/2*d*x + 1/2*c) - 1)/(a^2*tan(1/2*d*x + 1/2*c)^5) - (a^8*tan(1/2*d*x + 1/2*c)^5 - 5*a^8*tan(1/2*d*x + 1/2*c)^4 + 15*a^8*tan(1 /2*d*x + 1/2*c)^3 - 40*a^8*tan(1/2*d*x + 1/2*c)^2 + 110*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d
Time = 10.43 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.58 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-110\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+110\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]
-(cos(c/2 + (d*x)/2)^10 - sin(c/2 + (d*x)/2)^10 + 5*cos(c/2 + (d*x)/2)*sin (c/2 + (d*x)/2)^9 - 5*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 15*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 40*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d* x)/2)^7 - 110*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 110*cos(c/2 + (d *x)/2)^6*sin(c/2 + (d*x)/2)^4 - 40*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) ^3 + 15*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 120*log(sin(c/2 + (d*x )/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)/(160*a ^2*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)