Integrand size = 29, antiderivative size = 158 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {25 a^2 x}{16}-\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d} \] Output:
-25/16*a^2*x-2*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+2/3*a^2*cos(d* x+c)^3/d+2/5*a^2*cos(d*x+c)^5/d-a^2*cot(d*x+c)/d-7/16*a^2*cos(d*x+c)*sin(d *x+c)/d-7/24*a^2*cos(d*x+c)*sin(d*x+c)^3/d+1/6*a^2*cos(d*x+c)*sin(d*x+c)^5 /d
Time = 0.78 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-1500 c-1500 d x+2640 \cos (c+d x)+280 \cos (3 (c+d x))+24 \cos (5 (c+d x))-960 \cot (c+d x)-1920 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1920 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-255 \sin (2 (c+d x))+15 \sin (4 (c+d x))+5 \sin (6 (c+d x))\right )}{960 d} \] Input:
Integrate[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(-1500*c - 1500*d*x + 2640*Cos[c + d*x] + 280*Cos[3*(c + d*x)] + 24*C os[5*(c + d*x)] - 960*Cot[c + d*x] - 1920*Log[Cos[(c + d*x)/2]] + 1920*Log [Sin[(c + d*x)/2]] - 255*Sin[2*(c + d*x)] + 15*Sin[4*(c + d*x)] + 5*Sin[6* (c + d*x)]))/(960*d)
Time = 0.50 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {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.
\(\displaystyle \int \cos ^4(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^2}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3351 |
\(\displaystyle \frac {\int \left (-\sin ^6(c+d x) a^8-2 \sin ^5(c+d x) a^8+2 \sin ^4(c+d x) a^8+6 \sin ^3(c+d x) a^8+\csc ^2(c+d x) a^8+2 \csc (c+d x) a^8-6 \sin (c+d x) a^8-2 a^8\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a^8 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^8 \cos ^5(c+d x)}{5 d}+\frac {2 a^8 \cos ^3(c+d x)}{3 d}+\frac {2 a^8 \cos (c+d x)}{d}-\frac {a^8 \cot (c+d x)}{d}+\frac {a^8 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^8 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^8 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {25 a^8 x}{16}}{a^6}\) |
Input:
Int[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
((-25*a^8*x)/16 - (2*a^8*ArcTanh[Cos[c + d*x]])/d + (2*a^8*Cos[c + d*x])/d + (2*a^8*Cos[c + d*x]^3)/(3*d) + (2*a^8*Cos[c + d*x]^5)/(5*d) - (a^8*Cot[ c + d*x])/d - (7*a^8*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (7*a^8*Cos[c + d* x]*Sin[c + d*x]^3)/(24*d) + (a^8*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d))/a^6
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]))
Time = 9.46 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(166\) |
default | \(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(166\) |
risch | \(-\frac {25 a^{2} x}{16}+\frac {17 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}+\frac {11 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {11 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {17 i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} \cos \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a^{2} \cos \left (3 d x +3 c \right )}{24 d}\) | \(208\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(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 /16*d*x+5/16*c)+2*a^2*(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)))+a^2*(-1/sin(d*x+c)*cos(d*x+c)^7-(cos(d*x+c)^5+5/4*cos (d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)-15/8*d*x-15/8*c))
Time = 0.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {40 \, a^{2} \cos \left (d x + c\right )^{7} - 50 \, a^{2} \cos \left (d x + c\right )^{5} - 125 \, a^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 375 \, a^{2} \cos \left (d x + c\right ) - {\left (96 \, a^{2} \cos \left (d x + c\right )^{5} + 160 \, a^{2} \cos \left (d x + c\right )^{3} - 375 \, a^{2} d x + 480 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/240*(40*a^2*cos(d*x + c)^7 - 50*a^2*cos(d*x + c)^5 - 125*a^2*cos(d*x + c)^3 + 240*a^2*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 240*a^2*log(-1/2 *cos(d*x + c) + 1/2)*sin(d*x + c) + 375*a^2*cos(d*x + c) - (96*a^2*cos(d*x + c)^5 + 160*a^2*cos(d*x + c)^3 - 375*a^2*d*x + 480*a^2*cos(d*x + c))*sin (d*x + c))/(d*sin(d*x + c))
\[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cos(c + d*x)**4*cot(c + d*x)**2, x) + Integral(2*sin(c + d* x)*cos(c + d*x)**4*cot(c + d*x)**2, x) + Integral(sin(c + d*x)**2*cos(c + d*x)**4*cot(c + d*x)**2, x))
Time = 0.13 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {64 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{960 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/960*(64*(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^2 - 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^2 - 120*( 15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^ 5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a^2)/d
Time = 0.21 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {120 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6720 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2976 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 736 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/240*(375*(d*x + c)*a^2 - 480*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 120*a ^2*tan(1/2*d*x + 1/2*c) + 120*(4*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d *x + 1/2*c) - 2*(105*a^2*tan(1/2*d*x + 1/2*c)^11 + 1440*a^2*tan(1/2*d*x + 1/2*c)^10 + 595*a^2*tan(1/2*d*x + 1/2*c)^9 + 4320*a^2*tan(1/2*d*x + 1/2*c) ^8 - 150*a^2*tan(1/2*d*x + 1/2*c)^7 + 7360*a^2*tan(1/2*d*x + 1/2*c)^6 + 15 0*a^2*tan(1/2*d*x + 1/2*c)^5 + 6720*a^2*tan(1/2*d*x + 1/2*c)^4 - 595*a^2*t an(1/2*d*x + 1/2*c)^3 + 2976*a^2*tan(1/2*d*x + 1/2*c)^2 - 105*a^2*tan(1/2* d*x + 1/2*c) + 736*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
Time = 32.73 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.54 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {25\,a^2\,\mathrm {atan}\left (\frac {625\,a^4}{64\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{4}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {47\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{12}+72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {368\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+112\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {299\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {184\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \] Input:
int(cos(c + d*x)^4*cot(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
Output:
(2*a^2*log(tan(c/2 + (d*x)/2)))/d + (25*a^2*atan((625*a^4)/(64*((25*a^4)/2 + (625*a^4*tan(c/2 + (d*x)/2))/64)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25 *a^4)/2 + (625*a^4*tan(c/2 + (d*x)/2))/64))))/(8*d) + ((248*a^2*tan(c/2 + (d*x)/2)^3)/5 - (31*a^2*tan(c/2 + (d*x)/2)^2)/4 - (299*a^2*tan(c/2 + (d*x) /2)^4)/12 + 112*a^2*tan(c/2 + (d*x)/2)^5 - (35*a^2*tan(c/2 + (d*x)/2)^6)/2 + (368*a^2*tan(c/2 + (d*x)/2)^7)/3 - (35*a^2*tan(c/2 + (d*x)/2)^8)/2 + 72 *a^2*tan(c/2 + (d*x)/2)^9 + (47*a^2*tan(c/2 + (d*x)/2)^10)/12 + 24*a^2*tan (c/2 + (d*x)/2)^11 + (3*a^2*tan(c/2 + (d*x)/2)^12)/4 - a^2 + (184*a^2*tan( c/2 + (d*x)/2))/15)/(d*(2*tan(c/2 + (d*x)/2) + 12*tan(c/2 + (d*x)/2)^3 + 3 0*tan(c/2 + (d*x)/2)^5 + 40*tan(c/2 + (d*x)/2)^7 + 30*tan(c/2 + (d*x)/2)^9 + 12*tan(c/2 + (d*x)/2)^11 + 2*tan(c/2 + (d*x)/2)^13)) + (a^2*tan(c/2 + ( d*x)/2))/(2*d)
Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.98 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-352 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+736 \cos \left (d x +c \right ) \sin \left (d x +c \right )-240 \cos \left (d x +c \right )+480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-375 \sin \left (d x +c \right ) d x -736 \sin \left (d x +c \right )\right )}{240 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(40*cos(c + d*x)*sin(c + d*x)**6 + 96*cos(c + d*x)*sin(c + d*x)**5 - 70*cos(c + d*x)*sin(c + d*x)**4 - 352*cos(c + d*x)*sin(c + d*x)**3 - 105* cos(c + d*x)*sin(c + d*x)**2 + 736*cos(c + d*x)*sin(c + d*x) - 240*cos(c + d*x) + 480*log(tan((c + d*x)/2))*sin(c + d*x) - 375*sin(c + d*x)*d*x - 73 6*sin(c + d*x)))/(240*sin(c + d*x)*d)