Integrand size = 29, antiderivative size = 118 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {x}{a^2}+\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}+\frac {\cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{2 a^2 d} \] Output:
x/a^2+3/4*arctanh(cos(d*x+c))/a^2/d+cot(d*x+c)/a^2/d-1/3*cot(d*x+c)^3/a^2/ d-1/5*cot(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)^3* csc(d*x+c)/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(254\) vs. \(2(118)=236\).
Time = 2.84 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^5(c+d x) \left (-40 \cos (c+d x)-220 \cos (3 (c+d x))+68 \cos (5 (c+d x))+600 c \sin (c+d x)+600 d x \sin (c+d x)+450 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-450 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-60 \sin (2 (c+d x))-300 c \sin (3 (c+d x))-300 d x \sin (3 (c+d x))-225 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+225 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+150 \sin (4 (c+d x))+60 c \sin (5 (c+d x))+60 d x \sin (5 (c+d x))+45 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-45 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^2,x]
Output:
(Csc[c + d*x]^5*(-40*Cos[c + d*x] - 220*Cos[3*(c + d*x)] + 68*Cos[5*(c + d *x)] + 600*c*Sin[c + d*x] + 600*d*x*Sin[c + d*x] + 450*Log[Cos[(c + d*x)/2 ]]*Sin[c + d*x] - 450*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 60*Sin[2*(c + d *x)] - 300*c*Sin[3*(c + d*x)] - 300*d*x*Sin[3*(c + d*x)] - 225*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 225*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 150*Sin[4*(c + d*x)] + 60*c*Sin[5*(c + d*x)] + 60*d*x*Sin[5*(c + d*x)] + 45*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 45*Log[Sin[(c + d*x)/2]]*Sin[5 *(c + d*x)]))/(960*a^2*d)
Time = 0.58 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3352, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^6 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \cot ^4(c+d x) \csc ^2(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x)^4 (a-a \sin (c+d x))^2}{\sin (c+d x)^6}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (a^2 \cot ^4(c+d x)+a^2 \csc ^2(c+d x) \cot ^4(c+d x)-2 a^2 \csc (c+d x) \cot ^4(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3 a^2 \text {arctanh}(\cos (c+d x))}{4 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^3(c+d x) \csc (c+d x)}{2 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+a^2 x}{a^4}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*x + (3*a^2*ArcTanh[Cos[c + d*x]])/(4*d) + (a^2*Cot[c + d*x])/d - (a^2 *Cot[c + d*x]^3)/(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]^3*Csc[c + d*x])/(2*d))/a^4
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 = 1.06 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.36
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {18}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2}}\) | \(160\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {18}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2}}\) | \(160\) |
| risch | \(\frac {x}{a^{2}}+\frac {60 i {\mathrm e}^{8 i \left (d x +c \right )}+75 \,{\mathrm e}^{9 i \left (d x +c \right )}-360 i {\mathrm e}^{6 i \left (d x +c \right )}-30 \,{\mathrm e}^{7 i \left (d x +c \right )}+320 i {\mathrm e}^{4 i \left (d x +c \right )}-280 i {\mathrm e}^{2 i \left (d x +c \right )}+30 \,{\mathrm e}^{3 i \left (d x +c \right )}+68 i-75 \,{\mathrm e}^{i \left (d x +c \right )}}{30 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}\) | \(163\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/32/d/a^2*(1/5*tan(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^4+1/3*tan(1/2*d*x+ 1/2*c)^3+8*tan(1/2*d*x+1/2*c)^2-18*tan(1/2*d*x+1/2*c)-1/5/tan(1/2*d*x+1/2* c)^5-1/3/tan(1/2*d*x+1/2*c)^3+1/tan(1/2*d*x+1/2*c)^4-8/tan(1/2*d*x+1/2*c)^ 2+18/tan(1/2*d*x+1/2*c)-24*ln(tan(1/2*d*x+1/2*c))+64*arctan(tan(1/2*d*x+1/ 2*c)))
Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {136 \, \cos \left (d x + c\right )^{5} - 280 \, \cos \left (d x + c\right )^{3} + 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (4 \, d x \cos \left (d x + c\right )^{4} - 8 \, d x \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right )^{3} + 4 \, d x - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 120 \, \cos \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/120*(136*cos(d*x + c)^5 - 280*cos(d*x + c)^3 + 45*(cos(d*x + c)^4 - 2*co s(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 45*(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) + 30*(4*d*x*cos(d*x + c)^4 - 8*d*x*cos(d*x + c)^2 + 5*cos(d*x + c)^3 + 4*d*x - 3*cos(d*x + c))*sin(d*x + c) + 120*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))
\[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{6}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**6/(a+a*sin(d*x+c))**2,x)
Output:
Integral(cos(c + d*x)**2*cot(c + d*x)**6/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (108) = 216\).
Time = 0.11 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {270 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {960 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {270 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{480 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/480*((270*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 - 960*arc tan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 360*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - (15*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^2/(co s(d*x + c) + 1)^2 - 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 270*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3)*(cos(d*x + c) + 1)^5/(a^2*sin(d*x + c)^5) )/d
Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {480 \, {\left (d x + c\right )}}{a^{2}} - \frac {360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 270 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 270 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{480 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/480*(480*(d*x + c)/a^2 - 360*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (822*t an(1/2*d*x + 1/2*c)^5 + 270*tan(1/2*d*x + 1/2*c)^4 - 120*tan(1/2*d*x + 1/2 *c)^3 - 5*tan(1/2*d*x + 1/2*c)^2 + 15*tan(1/2*d*x + 1/2*c) - 3)/(a^2*tan(1 /2*d*x + 1/2*c)^5) + (3*a^8*tan(1/2*d*x + 1/2*c)^5 - 15*a^8*tan(1/2*d*x + 1/2*c)^4 + 5*a^8*tan(1/2*d*x + 1/2*c)^3 + 120*a^8*tan(1/2*d*x + 1/2*c)^2 - 270*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d
Time = 32.47 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.09 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+960\,\mathrm {atan}\left (\frac {4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:
int((cos(c + d*x)^2*cot(c + d*x)^6)/(a + a*sin(c + d*x))^2,x)
Output:
-(3*cos(c/2 + (d*x)/2)^10 - 3*sin(c/2 + (d*x)/2)^10 + 15*cos(c/2 + (d*x)/2 )*sin(c/2 + (d*x)/2)^9 - 15*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 5*co s(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 120*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^7 + 270*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 270*cos(c/ 2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + ( d*x)/2)^3 + 5*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 960*atan((4*cos( c/2 + (d*x)/2) - 3*sin(c/2 + (d*x)/2))/(3*cos(c/2 + (d*x)/2) + 4*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5 + 360*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)/(4 80*a^2*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)
Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {68 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right )-12 \cos \left (d x +c \right )-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+60 \sin \left (d x +c \right )^{5} d x}{60 \sin \left (d x +c \right )^{5} a^{2} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^2,x)
Output:
(68*cos(c + d*x)*sin(c + d*x)**4 - 75*cos(c + d*x)*sin(c + d*x)**3 + 4*cos (c + d*x)*sin(c + d*x)**2 + 30*cos(c + d*x)*sin(c + d*x) - 12*cos(c + d*x) - 45*log(tan((c + d*x)/2))*sin(c + d*x)**5 + 60*sin(c + d*x)**5*d*x)/(60* sin(c + d*x)**5*a**2*d)