\(\int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [711]

Optimal result

Integrand size = 29, antiderivative size = 133 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 x}{8 a}+\frac {\text {arctanh}(\cos (c+d x))}{a d}-\frac {\cos (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cot (c+d x)}{a d}-\frac {9 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a d} \] Output:

-15/8*x/a+arctanh(cos(d*x+c))/a/d-cos(d*x+c)/a/d-1/3*cos(d*x+c)^3/a/d-1/5* 
cos(d*x+c)^5/a/d-cot(d*x+c)/a/d-9/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+ 
c)*sin(d*x+c)^3/a/d
 

Mathematica [A] (verified)

Time = 2.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (1200 \cos (c+d x)-225 \cos (3 (c+d x))-15 \cos (5 (c+d x))+1800 c \sin (c+d x)+1800 d x \sin (c+d x)-960 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+960 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+590 \sin (2 (c+d x))+64 \sin (4 (c+d x))+6 \sin (6 (c+d x))\right )}{1920 a d} \] Input:

Integrate[(Cos[c + d*x]^6*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

-1/1920*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(1200*Cos[c + d*x] - 225*Cos[3* 
(c + d*x)] - 15*Cos[5*(c + d*x)] + 1800*c*Sin[c + d*x] + 1800*d*x*Sin[c + 
d*x] - 960*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 960*Log[Sin[(c + d*x)/2]]* 
Sin[c + d*x] + 590*Sin[2*(c + d*x)] + 64*Sin[4*(c + d*x)] + 6*Sin[6*(c + d 
*x)]))/(a*d)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3318, 3042, 25, 3071, 252, 252, 262, 216, 3072, 254, 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.

\(\displaystyle \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^2 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3318

\(\displaystyle \frac {\int \cos ^4(c+d x) \cot ^2(c+d x)dx}{a}-\frac {\int \cos ^5(c+d x) \cot (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \tan \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}-\frac {\int -\sin \left (c+d x+\frac {\pi }{2}\right )^5 \tan \left (c+d x+\frac {\pi }{2}\right )dx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \tan \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}+\frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}\)

\(\Big \downarrow \) 3071

\(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}-\frac {\int \frac {\cot ^6(c+d x)}{\left (\cot ^2(c+d x)+1\right )^3}d\cot (c+d x)}{a d}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}-\frac {\frac {5}{4} \int \frac {\cot ^4(c+d x)}{\left (\cot ^2(c+d x)+1\right )^2}d\cot (c+d x)-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}-\frac {\frac {5}{4} \left (\frac {3}{2} \int \frac {\cot ^2(c+d x)}{\cot ^2(c+d x)+1}d\cot (c+d x)-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}-\frac {\frac {5}{4} \left (\frac {3}{2} \left (\cot (c+d x)-\int \frac {1}{\cot ^2(c+d x)+1}d\cot (c+d x)\right )-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx}{a}-\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 3072

\(\displaystyle \frac {\int \frac {\cos ^6(c+d x)}{1-\cos ^2(c+d x)}d\cos (c+d x)}{a d}-\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 254

\(\displaystyle \frac {\int \left (-\cos ^4(c+d x)-\cos ^2(c+d x)+\frac {1}{1-\cos ^2(c+d x)}-1\right )d\cos (c+d x)}{a d}-\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\text {arctanh}(\cos (c+d x))-\frac {1}{5} \cos ^5(c+d x)-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)}{a d}-\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}}{a d}\)

Input:

Int[(Cos[c + d*x]^6*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

(ArcTanh[Cos[c + d*x]] - Cos[c + d*x] - Cos[c + d*x]^3/3 - Cos[c + d*x]^5/ 
5)/(a*d) - (-1/4*Cot[c + d*x]^5/(1 + Cot[c + d*x]^2)^2 + (5*((3*(-ArcTan[C 
ot[c + d*x]] + Cot[c + d*x]))/2 - Cot[c + d*x]^3/(2*(1 + Cot[c + d*x]^2))) 
)/4)/(a*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 254

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]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3071

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f)   Subst[I 
nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], 
x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
 

rule 3072

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]
 

rule 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 7.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {23}{15}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\)	\(177\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {23}{15}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\)	\(177\)
risch	\(-\frac {15 x}{8 a}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 d a}-\frac {11 \,{\mathrm e}^{i \left (d x +c \right )}}{16 a d}-\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 a d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}-\frac {2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}-\frac {\cos \left (5 d x +5 c \right )}{80 a d}-\frac {\sin \left (4 d x +4 c \right )}{32 d a}-\frac {7 \cos \left (3 d x +3 c \right )}{48 a d}\)	\(190\)

Input:

int(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/2/d/a*(tan(1/2*d*x+1/2*c)-4*(-9/8*tan(1/2*d*x+1/2*c)^9+3*tan(1/2*d*x+1/2 
*c)^8-5/4*tan(1/2*d*x+1/2*c)^7+6*tan(1/2*d*x+1/2*c)^6+28/3*tan(1/2*d*x+1/2 
*c)^4+5/4*tan(1/2*d*x+1/2*c)^3+14/3*tan(1/2*d*x+1/2*c)^2+9/8*tan(1/2*d*x+1 
/2*c)+23/15)/(1+tan(1/2*d*x+1/2*c)^2)^5-15/2*arctan(tan(1/2*d*x+1/2*c))-1/ 
tan(1/2*d*x+1/2*c)-2*ln(tan(1/2*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{5} + 75 \, \cos \left (d x + c\right )^{3} - {\left (24 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{3} + 225 \, d x + 120 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 60 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 225 \, \cos \left (d x + c\right )}{120 \, a d \sin \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

1/120*(30*cos(d*x + c)^5 + 75*cos(d*x + c)^3 - (24*cos(d*x + c)^5 + 40*cos 
(d*x + c)^3 + 225*d*x + 120*cos(d*x + c))*sin(d*x + c) + 60*log(1/2*cos(d* 
x + c) + 1/2)*sin(d*x + c) - 60*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) 
- 225*cos(d*x + c))/(a*d*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{6}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate(cos(d*x+c)**6*cot(d*x+c)**2/(a+a*sin(d*x+c)),x)
 

Output:

Integral(cos(c + d*x)**6*cot(c + d*x)**2/(sin(c + d*x) + 1), x)/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (123) = 246\).

Time = 0.14 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.85 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {184 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {285 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {560 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {450 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1120 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {300 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {720 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {360 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {105 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + 30}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {10 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {10 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {5 \, a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {a \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}} + \frac {225 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {30 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{60 \, d} \] Input:

integrate(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

-1/60*((184*sin(d*x + c)/(cos(d*x + c) + 1) + 285*sin(d*x + c)^2/(cos(d*x 
+ c) + 1)^2 + 560*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 450*sin(d*x + c)^4 
/(cos(d*x + c) + 1)^4 + 1120*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 300*sin 
(d*x + c)^6/(cos(d*x + c) + 1)^6 + 720*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 
 + 360*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 105*sin(d*x + c)^10/(cos(d*x 
+ c) + 1)^10 + 30)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a*sin(d*x + c)^3 
/(cos(d*x + c) + 1)^3 + 10*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 10*a*si 
n(d*x + c)^7/(cos(d*x + c) + 1)^7 + 5*a*sin(d*x + c)^9/(cos(d*x + c) + 1)^ 
9 + a*sin(d*x + c)^11/(cos(d*x + c) + 1)^11) + 225*arctan(sin(d*x + c)/(co 
s(d*x + c) + 1))/a + 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 30*sin(d* 
x + c)/(a*(cos(d*x + c) + 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {225 \, {\left (d x + c\right )}}{a} + \frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {60 \, {\left (2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 150 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 720 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 150 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 560 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 184\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a}}{120 \, d} \] Input:

integrate(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

-1/120*(225*(d*x + c)/a + 120*log(abs(tan(1/2*d*x + 1/2*c)))/a - 60*tan(1/ 
2*d*x + 1/2*c)/a - 60*(2*tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c) 
) - 2*(135*tan(1/2*d*x + 1/2*c)^9 - 360*tan(1/2*d*x + 1/2*c)^8 + 150*tan(1 
/2*d*x + 1/2*c)^7 - 720*tan(1/2*d*x + 1/2*c)^6 - 1120*tan(1/2*d*x + 1/2*c) 
^4 - 150*tan(1/2*d*x + 1/2*c)^3 - 560*tan(1/2*d*x + 1/2*c)^2 - 135*tan(1/2 
*d*x + 1/2*c) - 184)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a))/d
 

Mupad [B] (verification not implemented)

Time = 31.94 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.23 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15\,\mathrm {atan}\left (\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-\frac {15}{2}\right )}+\frac {225}{16\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-\frac {15}{2}\right )}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {112\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {92\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \] Input:

int((cos(c + d*x)^6*cot(c + d*x)^2)/(a + a*sin(c + d*x)),x)
 

Output:

(15*atan((15*tan(c/2 + (d*x)/2))/(2*((225*tan(c/2 + (d*x)/2))/16 - 15/2)) 
+ 225/(16*((225*tan(c/2 + (d*x)/2))/16 - 15/2))))/(4*a*d) - log(tan(c/2 + 
(d*x)/2))/(a*d) - ((92*tan(c/2 + (d*x)/2))/15 + (19*tan(c/2 + (d*x)/2)^2)/ 
2 + (56*tan(c/2 + (d*x)/2)^3)/3 + 15*tan(c/2 + (d*x)/2)^4 + (112*tan(c/2 + 
 (d*x)/2)^5)/3 + 10*tan(c/2 + (d*x)/2)^6 + 24*tan(c/2 + (d*x)/2)^7 + 12*ta 
n(c/2 + (d*x)/2)^9 - (7*tan(c/2 + (d*x)/2)^10)/2 + 1)/(d*(2*a*tan(c/2 + (d 
*x)/2) + 10*a*tan(c/2 + (d*x)/2)^3 + 20*a*tan(c/2 + (d*x)/2)^5 + 20*a*tan( 
c/2 + (d*x)/2)^7 + 10*a*tan(c/2 + (d*x)/2)^9 + 2*a*tan(c/2 + (d*x)/2)^11)) 
 + tan(c/2 + (d*x)/2)/(2*a*d)
 

Reduce [F]

\[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{6} \cot \left (d x +c \right )^{2}}{\sin \left (d x +c \right ) a +a}d x \] Input:

int(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x)
 

Output:

int(cos(d*x+c)^6*cot(d*x+c)^2/(a+a*sin(d*x+c)),x)