Integrand size = 29, antiderivative size = 173 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {15 a^3 x}{16}-\frac {3 a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {11 a^3 \cos (c+d x) \sin ^3(c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{2 d} \] Output:
-15/16*a^3*x-3*a^3*arctanh(cos(d*x+c))/d+3*a^3*cos(d*x+c)/d+a^3*cos(d*x+c) ^3/d+3/5*a^3*cos(d*x+c)^5/d-1/7*a^3*cos(d*x+c)^7/d-a^3*cot(d*x+c)/d+15/16* a^3*cos(d*x+c)*sin(d*x+c)/d-11/8*a^3*cos(d*x+c)*sin(d*x+c)^3/d+1/2*a^3*cos (d*x+c)*sin(d*x+c)^5/d
Time = 8.10 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {(a+a \sin (c+d x))^3 \left (-2100 (c+d x)+9065 \cos (c+d x)+875 \cos (3 (c+d x))+49 \cos (5 (c+d x))-5 \cos (7 (c+d x))-1120 \cot \left (\frac {1}{2} (c+d x)\right )-6720 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6720 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+455 \sin (2 (c+d x))+245 \sin (4 (c+d x))+35 \sin (6 (c+d x))+1120 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2240 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \] Input:
Integrate[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
((a + a*Sin[c + d*x])^3*(-2100*(c + d*x) + 9065*Cos[c + d*x] + 875*Cos[3*( c + d*x)] + 49*Cos[5*(c + d*x)] - 5*Cos[7*(c + d*x)] - 1120*Cot[(c + d*x)/ 2] - 6720*Log[Cos[(c + d*x)/2]] + 6720*Log[Sin[(c + d*x)/2]] + 455*Sin[2*( c + d*x)] + 245*Sin[4*(c + d*x)] + 35*Sin[6*(c + d*x)] + 1120*Tan[(c + d*x )/2]))/(2240*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)
Time = 0.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.02, 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.
| \(\displaystyle \int \cos ^4(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^3}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (-\sin ^7(c+d x) a^9-3 \sin ^6(c+d x) a^9+8 \sin ^4(c+d x) a^9+6 \sin ^3(c+d x) a^9+\csc ^2(c+d x) a^9-6 \sin ^2(c+d x) a^9+3 \csc (c+d x) a^9-8 \sin (c+d x) a^9\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3 a^9 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^9 \cos ^7(c+d x)}{7 d}+\frac {3 a^9 \cos ^5(c+d x)}{5 d}+\frac {a^9 \cos ^3(c+d x)}{d}+\frac {3 a^9 \cos (c+d x)}{d}-\frac {a^9 \cot (c+d x)}{d}+\frac {a^9 \sin ^5(c+d x) \cos (c+d x)}{2 d}-\frac {11 a^9 \sin ^3(c+d x) \cos (c+d x)}{8 d}+\frac {15 a^9 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {15 a^9 x}{16}}{a^6}\) |
Input:
Int[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
Output:
((-15*a^9*x)/16 - (3*a^9*ArcTanh[Cos[c + d*x]])/d + (3*a^9*Cos[c + d*x])/d + (a^9*Cos[c + d*x]^3)/d + (3*a^9*Cos[c + d*x]^5)/(5*d) - (a^9*Cos[c + d* x]^7)/(7*d) - (a^9*Cot[c + d*x])/d + (15*a^9*Cos[c + d*x]*Sin[c + d*x])/(1 6*d) - (11*a^9*Cos[c + d*x]*Sin[c + d*x]^3)/(8*d) + (a^9*Cos[c + d*x]*Sin[ c + d*x]^5)/(2*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 = 25.81 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{7}}{7}+3 a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+3 a^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(180\) |
| default | \(\frac {-\frac {a^{3} \cos \left (d x +c \right )^{7}}{7}+3 a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+3 a^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(180\) |
| risch | \(-\frac {15 a^{3} x}{16}+\frac {13 i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}-\frac {13 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}+\frac {259 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{128 d}+\frac {259 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{128 d}-\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{3} \cos \left (7 d x +7 c \right )}{448 d}+\frac {a^{3} \sin \left (6 d x +6 c \right )}{64 d}+\frac {7 a^{3} \cos \left (5 d x +5 c \right )}{320 d}+\frac {7 a^{3} \sin \left (4 d x +4 c \right )}{64 d}+\frac {25 a^{3} \cos \left (3 d x +3 c \right )}{64 d}\) | \(225\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/7*a^3*cos(d*x+c)^7+3*a^3*(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)+3*a^3*(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^3*(-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 = 173, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {280 \, a^{3} \cos \left (d x + c\right )^{7} - 70 \, a^{3} \cos \left (d x + c\right )^{5} - 175 \, a^{3} \cos \left (d x + c\right )^{3} + 840 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 840 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 525 \, a^{3} \cos \left (d x + c\right ) + {\left (80 \, a^{3} \cos \left (d x + c\right )^{7} - 336 \, a^{3} \cos \left (d x + c\right )^{5} - 560 \, a^{3} \cos \left (d x + c\right )^{3} + 525 \, a^{3} d x - 1680 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{560 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/560*(280*a^3*cos(d*x + c)^7 - 70*a^3*cos(d*x + c)^5 - 175*a^3*cos(d*x + c)^3 + 840*a^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 840*a^3*log(-1/ 2*cos(d*x + c) + 1/2)*sin(d*x + c) + 525*a^3*cos(d*x + c) + (80*a^3*cos(d* x + c)^7 - 336*a^3*cos(d*x + c)^5 - 560*a^3*cos(d*x + c)^3 + 525*a^3*d*x - 1680*a^3*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))
\[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)**2*(a+a*sin(d*x+c))**3,x)
Output:
a**3*(Integral(cos(c + d*x)**4*cot(c + d*x)**2, x) + Integral(3*sin(c + d* x)*cos(c + d*x)**4*cot(c + d*x)**2, x) + Integral(3*sin(c + d*x)**2*cos(c + d*x)**4*cot(c + d*x)**2, x) + Integral(sin(c + d*x)**3*cos(c + d*x)**4*c ot(c + d*x)**2, x))
Time = 0.11 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.08 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {320 \, a^{3} \cos \left (d x + c\right )^{7} - 224 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 280 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3}}{2240 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
-1/2240*(320*a^3*cos(d*x + c)^7 - 224*(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^3 + 35*(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^3 + 280*(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^3)/d
Time = 0.24 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.68 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {525 \, {\left (d x + c\right )} a^{3} - 1680 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 280 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {280 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (525 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 4480 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 980 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 20160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 945 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 38080 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 49280 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 945 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 32256 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 980 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12992 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 525 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2496 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{560 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-1/560*(525*(d*x + c)*a^3 - 1680*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 280* a^3*tan(1/2*d*x + 1/2*c) + 280*(6*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2* d*x + 1/2*c) + 2*(525*a^3*tan(1/2*d*x + 1/2*c)^13 - 4480*a^3*tan(1/2*d*x + 1/2*c)^12 - 980*a^3*tan(1/2*d*x + 1/2*c)^11 - 20160*a^3*tan(1/2*d*x + 1/2 *c)^10 + 945*a^3*tan(1/2*d*x + 1/2*c)^9 - 38080*a^3*tan(1/2*d*x + 1/2*c)^8 - 49280*a^3*tan(1/2*d*x + 1/2*c)^6 - 945*a^3*tan(1/2*d*x + 1/2*c)^5 - 322 56*a^3*tan(1/2*d*x + 1/2*c)^4 + 980*a^3*tan(1/2*d*x + 1/2*c)^3 - 12992*a^3 *tan(1/2*d*x + 1/2*c)^2 - 525*a^3*tan(1/2*d*x + 1/2*c) - 2496*a^3)/(tan(1/ 2*d*x + 1/2*c)^2 + 1)^7)/d
Time = 33.92 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.48 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {15\,a^3\,\mathrm {atan}\left (\frac {225\,a^6}{64\,\left (\frac {45\,a^6}{4}+\frac {225\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {45\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {45\,a^6}{4}+\frac {225\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{4}-32\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-144\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {111\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{4}-272\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-352\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {113\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}-\frac {1152\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {464\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {624\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+14\,{\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 )} \] Input:
int(cos(c + d*x)^4*cot(c + d*x)^2*(a + a*sin(c + d*x))^3,x)
Output:
(3*a^3*log(tan(c/2 + (d*x)/2)))/d + (15*a^3*atan((225*a^6)/(64*((45*a^6)/4 + (225*a^6*tan(c/2 + (d*x)/2))/64)) - (45*a^6*tan(c/2 + (d*x)/2))/(4*((45 *a^6)/4 + (225*a^6*tan(c/2 + (d*x)/2))/64))))/(8*d) + (a^3*tan(c/2 + (d*x) /2))/(2*d) - ((13*a^3*tan(c/2 + (d*x)/2)^2)/4 - (464*a^3*tan(c/2 + (d*x)/2 )^3)/5 + 28*a^3*tan(c/2 + (d*x)/2)^4 - (1152*a^3*tan(c/2 + (d*x)/2)^5)/5 + (113*a^3*tan(c/2 + (d*x)/2)^6)/4 - 352*a^3*tan(c/2 + (d*x)/2)^7 + 35*a^3* tan(c/2 + (d*x)/2)^8 - 272*a^3*tan(c/2 + (d*x)/2)^9 + (111*a^3*tan(c/2 + ( d*x)/2)^10)/4 - 144*a^3*tan(c/2 + (d*x)/2)^11 - 32*a^3*tan(c/2 + (d*x)/2)^ 13 + (19*a^3*tan(c/2 + (d*x)/2)^14)/4 + a^3 - (624*a^3*tan(c/2 + (d*x)/2)) /35)/(d*(2*tan(c/2 + (d*x)/2) + 14*tan(c/2 + (d*x)/2)^3 + 42*tan(c/2 + (d* x)/2)^5 + 70*tan(c/2 + (d*x)/2)^7 + 70*tan(c/2 + (d*x)/2)^9 + 42*tan(c/2 + (d*x)/2)^11 + 14*tan(c/2 + (d*x)/2)^13 + 2*tan(c/2 + (d*x)/2)^15))
Time = 5.54 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (80 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-770 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-992 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+525 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+2496 \cos \left (d x +c \right ) \sin \left (d x +c \right )-560 \cos \left (d x +c \right )+1680 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-525 \sin \left (d x +c \right ) d x -2496 \sin \left (d x +c \right )\right )}{560 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^4*cot(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*(80*cos(c + d*x)*sin(c + d*x)**7 + 280*cos(c + d*x)*sin(c + d*x)**6 + 96*cos(c + d*x)*sin(c + d*x)**5 - 770*cos(c + d*x)*sin(c + d*x)**4 - 992 *cos(c + d*x)*sin(c + d*x)**3 + 525*cos(c + d*x)*sin(c + d*x)**2 + 2496*co s(c + d*x)*sin(c + d*x) - 560*cos(c + d*x) + 1680*log(tan((c + d*x)/2))*si n(c + d*x) - 525*sin(c + d*x)*d*x - 2496*sin(c + d*x)))/(560*sin(c + d*x)* d)