\(\int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [712]

Optimal result

Integrand size = 29, antiderivative size = 139 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 x}{8 a}+\frac {5 \text {arctanh}(\cos (c+d x))}{2 a d}-\frac {2 \cos (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {9 \cos (c+d x) \sin (c+d x)}{8 a d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a d} \] Output:

15/8*x/a+5/2*arctanh(cos(d*x+c))/a/d-2*cos(d*x+c)/a/d-1/3*cos(d*x+c)^3/a/d 
+cot(d*x+c)/a/d-1/2*cot(d*x+c)*csc(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 = 1.81 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-360 c-360 d x+400 \cos (c+d x)-200 \cos (3 (c+d x))-8 \cos (5 (c+d x))-480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 \cos (2 (c+d x)) \left (3 c+3 d x+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-285 \sin (2 (c+d x))+42 \sin (4 (c+d x))+3 \sin (6 (c+d x))\right )}{1536 a d (1+\sin (c+d x))} \] Input:

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

-1/1536*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(-360*c - 360*d*x + 400*C 
os[c + d*x] - 200*Cos[3*(c + d*x)] - 8*Cos[5*(c + d*x)] - 480*Log[Cos[(c + 
 d*x)/2]] + 120*Cos[2*(c + d*x)]*(3*c + 3*d*x + 4*Log[Cos[(c + d*x)/2]] - 
4*Log[Sin[(c + d*x)/2]]) + 480*Log[Sin[(c + d*x)/2]] - 285*Sin[2*(c + d*x) 
] + 42*Sin[4*(c + d*x)] + 3*Sin[6*(c + d*x)]))/(a*d*(1 + Sin[c + d*x]))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3318, 3042, 25, 3071, 252, 252, 262, 216, 3072, 252, 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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^3 (a \sin (c+d x)+a)}dx\)

\(\Big \downarrow \) 3318

\(\displaystyle \frac {\int \cos ^3(c+d x) \cot ^3(c+d x)dx}{a}-\frac {\int \cos ^4(c+d x) \cot ^2(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3071

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3072

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 254

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

-((Cos[c + d*x]^5/(2*(1 - Cos[c + d*x]^2)) - (5*(ArcTanh[Cos[c + d*x]] - C 
os[c + d*x] - Cos[c + d*x]^3/3))/2)/(a*d)) + (-1/4*Cot[c + d*x]^5/(1 + Cot 
[c + d*x]^2)^2 + (5*((3*(-ArcTan[Cot[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 = 5.18 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-56 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {152 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {56}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}\)	\(192\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-56 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {152 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {56}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}\)	\(192\)
risch	\(\frac {15 x}{8 a}-\frac {{\mathrm e}^{3 i \left (d x +c \right )}}{24 a d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 d a}-\frac {9 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {9 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}-\frac {{\mathrm e}^{-3 i \left (d x +c \right )}}{24 a d}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+2 i {\mathrm e}^{2 i \left (d x +c \right )}-2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}+\frac {\sin \left (4 d x +4 c \right )}{32 d a}\)	\(222\)

Input:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4/d/a*(1/2*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)+8*(-9/8*tan(1/2*d*x 
+1/2*c)^7-3*tan(1/2*d*x+1/2*c)^6-1/8*tan(1/2*d*x+1/2*c)^5-7*tan(1/2*d*x+1/ 
2*c)^4+1/8*tan(1/2*d*x+1/2*c)^3-19/3*tan(1/2*d*x+1/2*c)^2+9/8*tan(1/2*d*x+ 
1/2*c)-7/3)/(1+tan(1/2*d*x+1/2*c)^2)^4+15*arctan(tan(1/2*d*x+1/2*c))-1/2/t 
an(1/2*d*x+1/2*c)^2+2/tan(1/2*d*x+1/2*c)-10*ln(tan(1/2*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {8 \, \cos \left (d x + c\right )^{5} - 45 \, d x \cos \left (d x + c\right )^{2} + 40 \, \cos \left (d x + c\right )^{3} + 45 \, d x - 30 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 30 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{3} - 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 60 \, \cos \left (d x + c\right )}{24 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

-1/24*(8*cos(d*x + c)^5 - 45*d*x*cos(d*x + c)^2 + 40*cos(d*x + c)^3 + 45*d 
*x - 30*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) + 30*(cos(d*x + c 
)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) - 3*(2*cos(d*x + c)^5 + 5*cos(d*x + 
c)^3 - 15*cos(d*x + c))*sin(d*x + c) - 60*cos(d*x + c))/(a*d*cos(d*x + c)^ 
2 - a*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*cot(d*x+c)**3/(a+a*sin(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 383 vs. \(2 (127) = 254\).

Time = 0.11 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.76 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {124 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {102 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {322 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {78 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {348 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {42 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {147 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {42 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 3}{\frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {3 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a} + \frac {90 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{24 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

1/24*((12*sin(d*x + c)/(cos(d*x + c) + 1) - 124*sin(d*x + c)^2/(cos(d*x + 
c) + 1)^2 + 102*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 322*sin(d*x + c)^4/( 
cos(d*x + c) + 1)^4 + 78*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 348*sin(d*x 
 + c)^6/(cos(d*x + c) + 1)^6 + 42*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 14 
7*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 42*sin(d*x + c)^9/(cos(d*x + c) + 
1)^9 - 3)/(a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a*sin(d*x + c)^4/(cos 
(d*x + c) + 1)^4 + 6*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a*sin(d*x + 
 c)^8/(cos(d*x + c) + 1)^8 + a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) - 3* 
(4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/ 
a + 90*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 60*log(sin(d*x + c)/(co 
s(d*x + c) + 1))/a)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.55 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {45 \, {\left (d x + c\right )}}{a} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{2}} + \frac {3 \, {\left (30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {2 \, {\left (27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 168 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 152 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 56\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/24*(45*(d*x + c)/a - 60*log(abs(tan(1/2*d*x + 1/2*c)))/a + 3*(a*tan(1/2* 
d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c))/a^2 + 3*(30*tan(1/2*d*x + 1/2*c 
)^2 + 4*tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c)^2) - 2*(27*tan(1 
/2*d*x + 1/2*c)^7 + 72*tan(1/2*d*x + 1/2*c)^6 + 3*tan(1/2*d*x + 1/2*c)^5 + 
 168*tan(1/2*d*x + 1/2*c)^4 - 3*tan(1/2*d*x + 1/2*c)^3 + 152*tan(1/2*d*x + 
 1/2*c)^2 - 27*tan(1/2*d*x + 1/2*c) + 56)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4* 
a))/d
 

Mupad [B] (verification not implemented)

Time = 31.80 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.18 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {15\,\mathrm {atan}\left (\frac {225}{16\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {75}{4}\right )}-\frac {75\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {75}{4}\right )}\right )}{4\,a\,d}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {49\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \] Input:

int((cos(c + d*x)^5*cot(c + d*x)^3)/(a + a*sin(c + d*x)),x)
 

Output:

tan(c/2 + (d*x)/2)^2/(8*a*d) - (15*atan(225/(16*((225*tan(c/2 + (d*x)/2))/ 
16 + 75/4)) - (75*tan(c/2 + (d*x)/2))/(4*((225*tan(c/2 + (d*x)/2))/16 + 75 
/4))))/(4*a*d) - (5*log(tan(c/2 + (d*x)/2)))/(2*a*d) - ((62*tan(c/2 + (d*x 
)/2)^2)/3 - 2*tan(c/2 + (d*x)/2) - 17*tan(c/2 + (d*x)/2)^3 + (161*tan(c/2 
+ (d*x)/2)^4)/3 - 13*tan(c/2 + (d*x)/2)^5 + 58*tan(c/2 + (d*x)/2)^6 - 7*ta 
n(c/2 + (d*x)/2)^7 + (49*tan(c/2 + (d*x)/2)^8)/2 + 7*tan(c/2 + (d*x)/2)^9 
+ 1/2)/(d*(4*a*tan(c/2 + (d*x)/2)^2 + 16*a*tan(c/2 + (d*x)/2)^4 + 24*a*tan 
(c/2 + (d*x)/2)^6 + 16*a*tan(c/2 + (d*x)/2)^8 + 4*a*tan(c/2 + (d*x)/2)^10) 
) - tan(c/2 + (d*x)/2)/(2*a*d)
 

Reduce [F]

\[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{5} \cot \left (d x +c \right )^{3}}{\sin \left (d x +c \right ) a +a}d x \] Input:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)
 

Output:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)