Integrand size = 25, antiderivative size = 127 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {5 a x}{16}-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d} \] Output:
5/16*a*x-a*arctanh(cos(d*x+c))/d+a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d+1/5*a *cos(d*x+c)^5/d+5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x +c)/d+1/6*a*cos(d*x+c)^5*sin(d*x+c)/d
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (300 c+300 d x+1320 \cos (c+d x)+140 \cos (3 (c+d x))+12 \cos (5 (c+d x))-960 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+960 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+225 \sin (2 (c+d x))+45 \sin (4 (c+d x))+5 \sin (6 (c+d x))\right )}{960 d} \] Input:
Integrate[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
(a*(300*c + 300*d*x + 1320*Cos[c + d*x] + 140*Cos[3*(c + d*x)] + 12*Cos[5* (c + d*x)] - 960*Log[Cos[(c + d*x)/2]] + 960*Log[Sin[(c + d*x)/2]] + 225*S in[2*(c + d*x)] + 45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)
Time = 0.54 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3317, 3042, 25, 3072, 254, 2009, 3115, 3042, 3115, 3042, 3115, 24}
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.
| \(\displaystyle \int \cos ^5(c+d x) \cot (c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)}{\sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \cos ^6(c+d x)dx+a \int \cos ^5(c+d x) \cot (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx+a \int -\sin \left (c+d x+\frac {\pi }{2}\right )^5 \tan \left (c+d x+\frac {\pi }{2}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-a \int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \int \frac {\cos ^6(c+d x)}{1-\cos ^2(c+d x)}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \int \left (-\cos ^4(c+d x)-\cos ^2(c+d x)+\frac {1}{1-\cos ^2(c+d x)}-1\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a \left (\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )-\frac {a \left (\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{d}\) |
Input:
Int[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
-((a*(ArcTanh[Cos[c + d*x]] - Cos[c + d*x] - Cos[c + d*x]^3/3 - Cos[c + d* x]^5/5))/d) + a*((Cos[c + d*x]^5*Sin[c + d*x])/(6*d) + (5*((Cos[c + d*x]^3 *Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6 )
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Time = 3.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+a \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(96\) |
| default | \(\frac {a \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+a \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(96\) |
| risch | \(\frac {5 a x}{16}+\frac {11 a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {11 a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {a \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a \cos \left (3 d x +3 c \right )}{48 d}+\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\) | \(146\) |
Input:
int(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/1 6*d*x+5/16*c)+a*(1/5*cos(d*x+c)^5+1/3*cos(d*x+c)^3+cos(d*x+c)+ln(csc(d*x+c )-cot(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {48 \, a \cos \left (d x + c\right )^{5} + 80 \, a \cos \left (d x + c\right )^{3} + 75 \, a d x + 240 \, a \cos \left (d x + c\right ) - 120 \, a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 120 \, a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{5} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/240*(48*a*cos(d*x + c)^5 + 80*a*cos(d*x + c)^3 + 75*a*d*x + 240*a*cos(d* x + c) - 120*a*log(1/2*cos(d*x + c) + 1/2) + 120*a*log(-1/2*cos(d*x + c) + 1/2) + 5*(8*a*cos(d*x + c)^5 + 10*a*cos(d*x + c)^3 + 15*a*cos(d*x + c))*s in(d*x + c))/d
\[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{5}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**5*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**5*cot(c + d*x), x) + Integral(sin(c + d*x)*cos(c + d*x)**5*cot(c + d*x), x))
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {32 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{960 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/960*(32*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log (cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*a - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a)/d
Time = 0.17 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.58 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {75 \, {\left (d x + c\right )} a + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 720 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 450 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3680 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 450 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3360 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1488 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 368 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/240*(75*(d*x + c)*a + 240*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(165*a*ta n(1/2*d*x + 1/2*c)^11 - 720*a*tan(1/2*d*x + 1/2*c)^10 - 25*a*tan(1/2*d*x + 1/2*c)^9 - 2160*a*tan(1/2*d*x + 1/2*c)^8 + 450*a*tan(1/2*d*x + 1/2*c)^7 - 3680*a*tan(1/2*d*x + 1/2*c)^6 - 450*a*tan(1/2*d*x + 1/2*c)^5 - 3360*a*tan (1/2*d*x + 1/2*c)^4 + 25*a*tan(1/2*d*x + 1/2*c)^3 - 1488*a*tan(1/2*d*x + 1 /2*c)^2 - 165*a*tan(1/2*d*x + 1/2*c) - 368*a)/(tan(1/2*d*x + 1/2*c)^2 + 1) ^6)/d
Time = 35.22 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.57 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {92\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {62\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {46\,a}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a\,\mathrm {atan}\left (\frac {25\,a^2}{64\,\left (\frac {5\,a^2}{4}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {5\,a^2}{4}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d} \] Input:
int(cos(c + d*x)^5*cot(c + d*x)*(a + a*sin(c + d*x)),x)
Output:
((46*a)/15 + (11*a*tan(c/2 + (d*x)/2))/8 + (62*a*tan(c/2 + (d*x)/2)^2)/5 - (5*a*tan(c/2 + (d*x)/2)^3)/24 + 28*a*tan(c/2 + (d*x)/2)^4 + (15*a*tan(c/2 + (d*x)/2)^5)/4 + (92*a*tan(c/2 + (d*x)/2)^6)/3 - (15*a*tan(c/2 + (d*x)/2 )^7)/4 + 18*a*tan(c/2 + (d*x)/2)^8 + (5*a*tan(c/2 + (d*x)/2)^9)/24 + 6*a*t an(c/2 + (d*x)/2)^10 - (11*a*tan(c/2 + (d*x)/2)^11)/8)/(d*(6*tan(c/2 + (d* x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a*l og(tan(c/2 + (d*x)/2)))/d + (5*a*atan((25*a^2)/(64*((5*a^2)/4 - (25*a^2*ta n(c/2 + (d*x)/2))/64)) + (5*a^2*tan(c/2 + (d*x)/2))/(4*((5*a^2)/4 - (25*a^ 2*tan(c/2 + (d*x)/2))/64))))/(8*d)
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-130 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-176 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+165 \cos \left (d x +c \right ) \sin \left (d x +c \right )+368 \cos \left (d x +c \right )+240 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 c +75 d x -368\right )}{240 d} \] Input:
int(cos(d*x+c)^5*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
(a*(40*cos(c + d*x)*sin(c + d*x)**5 + 48*cos(c + d*x)*sin(c + d*x)**4 - 13 0*cos(c + d*x)*sin(c + d*x)**3 - 176*cos(c + d*x)*sin(c + d*x)**2 + 165*co s(c + d*x)*sin(c + d*x) + 368*cos(c + d*x) + 240*log(tan((c + d*x)/2)) + 7 5*c + 75*d*x - 368))/(240*d)