Integrand size = 27, antiderivative size = 153 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=5 a^2 x+\frac {5 a^2 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {4 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{d} \] Output:
5*a^2*x+5/8*a^2*arctanh(cos(d*x+c))/d-a^2*cos(d*x+c)/d-1/3*a^2*cos(d*x+c)^ 3/d+4*a^2*cot(d*x+c)/d-2/3*a^2*cot(d*x+c)^3/d+5/8*a^2*cot(d*x+c)*csc(d*x+c )/d-1/4*a^2*cot(d*x+c)*csc(d*x+c)^3/d+a^2*cos(d*x+c)*sin(d*x+c)/d
Time = 7.22 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.48 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (1+\sin (c+d x))^2 \left (960 (c+d x)-240 \cos (c+d x)-16 \cos (3 (c+d x))+448 \cot \left (\frac {1}{2} (c+d x)\right )+30 \csc ^2\left (\frac {1}{2} (c+d x)\right )-3 \csc ^4\left (\frac {1}{2} (c+d x)\right )+120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-30 \sec ^2\left (\frac {1}{2} (c+d x)\right )+3 \sec ^4\left (\frac {1}{2} (c+d x)\right )+128 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-8 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+96 \sin (2 (c+d x))-448 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{192 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \] Input:
Integrate[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(1 + Sin[c + d*x])^2*(960*(c + d*x) - 240*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 448*Cot[(c + d*x)/2] + 30*Csc[(c + d*x)/2]^2 - 3*Csc[(c + d*x)/2 ]^4 + 120*Log[Cos[(c + d*x)/2]] - 120*Log[Sin[(c + d*x)/2]] - 30*Sec[(c + d*x)/2]^2 + 3*Sec[(c + d*x)/2]^4 + 128*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 8*Csc[(c + d*x)/2]^4*Sin[c + d*x] + 96*Sin[2*(c + d*x)] - 448*Tan[(c + d* x)/2]))/(192*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)
Time = 0.46 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
| \(\displaystyle \int \cos (c+d x) \cot ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^2}{\sin (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\csc ^5(c+d x) a^8+2 \csc ^4(c+d x) a^8-2 \csc ^3(c+d x) a^8-\sin ^3(c+d x) a^8-6 \csc ^2(c+d x) a^8-2 \sin ^2(c+d x) a^8+2 \sin (c+d x) a^8+6 a^8\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {5 a^8 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^8 \cos ^3(c+d x)}{3 d}-\frac {a^8 \cos (c+d x)}{d}-\frac {2 a^8 \cot ^3(c+d x)}{3 d}+\frac {4 a^8 \cot (c+d x)}{d}+\frac {a^8 \sin (c+d x) \cos (c+d x)}{d}-\frac {a^8 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {5 a^8 \cot (c+d x) \csc (c+d x)}{8 d}+5 a^8 x}{a^6}\) |
Input:
Int[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]
Output:
(5*a^8*x + (5*a^8*ArcTanh[Cos[c + d*x]])/(8*d) - (a^8*Cos[c + d*x])/d - (a ^8*Cos[c + d*x]^3)/(3*d) + (4*a^8*Cot[c + d*x])/d - (2*a^8*Cot[c + d*x]^3) /(3*d) + (5*a^8*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^8*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d) + (a^8*Cos[c + d*x]*Sin[c + d*x])/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 = 3.65 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \cos \left (d x +c \right )^{7}}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \cos \left (d x +c \right )^{5}}{8}+\frac {5 \cos \left (d x +c \right )^{3}}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(246\) |
| default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{4 \sin \left (d x +c \right )^{4}}+\frac {3 \cos \left (d x +c \right )^{7}}{8 \sin \left (d x +c \right )^{2}}+\frac {3 \cos \left (d x +c \right )^{5}}{8}+\frac {5 \cos \left (d x +c \right )^{3}}{8}+\frac {15 \cos \left (d x +c \right )}{8}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(246\) |
| risch | \(5 a^{2} x -\frac {a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {5 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {5 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {a^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {a^{2} \left (15 \,{\mathrm e}^{7 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-144 i {\mathrm e}^{6 i \left (d x +c \right )}+9 \,{\mathrm e}^{3 i \left (d x +c \right )}+336 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}-304 i {\mathrm e}^{2 i \left (d x +c \right )}+112 i\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\) | \(256\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^5*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^7-1/2*cos(d*x+c)^5-5/6*cos(d*x+c)^3 -5/2*cos(d*x+c)-5/2*ln(csc(d*x+c)-cot(d*x+c)))+2*a^2*(-1/3/sin(d*x+c)^3*co s(d*x+c)^7+4/3/sin(d*x+c)*cos(d*x+c)^7+4/3*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+ 15/8*cos(d*x+c))*sin(d*x+c)+5/2*d*x+5/2*c)+a^2*(-1/4/sin(d*x+c)^4*cos(d*x+ c)^7+3/8/sin(d*x+c)^2*cos(d*x+c)^7+3/8*cos(d*x+c)^5+5/8*cos(d*x+c)^3+15/8* cos(d*x+c)+15/8*ln(csc(d*x+c)-cot(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.60 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {16 \, a^{2} \cos \left (d x + c\right )^{7} - 240 \, a^{2} d x \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{5} + 480 \, a^{2} d x \cos \left (d x + c\right )^{2} - 50 \, a^{2} \cos \left (d x + c\right )^{3} - 240 \, a^{2} d x + 30 \, a^{2} \cos \left (d x + c\right ) - 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/48*(16*a^2*cos(d*x + c)^7 - 240*a^2*d*x*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^5 + 480*a^2*d*x*cos(d*x + c)^2 - 50*a^2*cos(d*x + c)^3 - 240*a^2*d*x + 30*a^2*cos(d*x + c) - 15*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a ^2)*log(1/2*cos(d*x + c) + 1/2) + 15*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x + c) + 1/2) - 16*(3*a^2*cos(d*x + c)^5 - 20* a^2*cos(d*x + c)^3 + 15*a^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)
\[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**5*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cos(c + d*x)*cot(c + d*x)**5, x) + Integral(2*sin(c + d*x)* cos(c + d*x)*cot(c + d*x)**5, x) + Integral(sin(c + d*x)**2*cos(c + d*x)*c ot(c + d*x)**5, x))
Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.35 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 16 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2} + 3 \, a^{2} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/48*(4*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos( d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 - 16*( 15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^ 5 + tan(d*x + c)^3))*a^2 + 3*a^2*(2*(9*cos(d*x + c)^3 - 7*cos(d*x + c))/(c os(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x + c) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)))/d
Time = 0.20 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.69 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 960 \, {\left (d x + c\right )} a^{2} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 432 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {128 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac {250 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 432 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*a ^2*tan(1/2*d*x + 1/2*c)^2 + 960*(d*x + c)*a^2 - 120*a^2*log(abs(tan(1/2*d* x + 1/2*c))) - 432*a^2*tan(1/2*d*x + 1/2*c) - 128*(3*a^2*tan(1/2*d*x + 1/2 *c)^5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^ 2*tan(1/2*d*x + 1/2*c) + 4*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 + (250*a^2* tan(1/2*d*x + 1/2*c)^4 + 432*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*a^2*tan(1/2*d *x + 1/2*c)^2 - 16*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^ 4)/d
Time = 33.22 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.44 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-62\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {233\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+136\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {449\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {10\,a^2\,\mathrm {atan}\left (\frac {100\,a^4}{\frac {25\,a^4}{2}+100\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+100\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d}-\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \] Input:
int(cos(c + d*x)*cot(c + d*x)^5*(a + a*sin(c + d*x))^2,x)
Output:
(a^2*tan(c/2 + (d*x)/2)^3)/(12*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^ 2*tan(c/2 + (d*x)/2)^4)/(64*d) + ((5*a^2*tan(c/2 + (d*x)/2)^2)/4 + 32*a^2* tan(c/2 + (d*x)/2)^3 - (449*a^2*tan(c/2 + (d*x)/2)^4)/12 + 136*a^2*tan(c/2 + (d*x)/2)^5 - (233*a^2*tan(c/2 + (d*x)/2)^6)/4 + (320*a^2*tan(c/2 + (d*x )/2)^7)/3 - 62*a^2*tan(c/2 + (d*x)/2)^8 + 4*a^2*tan(c/2 + (d*x)/2)^9 - a^2 /4 - (4*a^2*tan(c/2 + (d*x)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 48*tan(c/ 2 + (d*x)/2)^6 + 48*tan(c/2 + (d*x)/2)^8 + 16*tan(c/2 + (d*x)/2)^10)) - (5 *a^2*log(tan(c/2 + (d*x)/2)))/(8*d) - (10*a^2*atan((100*a^4)/((25*a^4)/2 + 100*a^4*tan(c/2 + (d*x)/2)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25*a^4)/2 + 100*a^4*tan(c/2 + (d*x)/2)))))/d - (9*a^2*tan(c/2 + (d*x)/2))/(4*d)
Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+896 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )-48 \cos \left (d x +c \right )-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+960 \sin \left (d x +c \right )^{4} d x +193 \sin \left (d x +c \right )^{4}\right )}{192 \sin \left (d x +c \right )^{4} d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^5*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(64*cos(c + d*x)*sin(c + d*x)**6 + 192*cos(c + d*x)*sin(c + d*x)**5 - 256*cos(c + d*x)*sin(c + d*x)**4 + 896*cos(c + d*x)*sin(c + d*x)**3 + 12 0*cos(c + d*x)*sin(c + d*x)**2 - 128*cos(c + d*x)*sin(c + d*x) - 48*cos(c + d*x) - 120*log(tan((c + d*x)/2))*sin(c + d*x)**4 + 960*sin(c + d*x)**4*d *x + 193*sin(c + d*x)**4))/(192*sin(c + d*x)**4*d)