\(\int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [729]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 29, antiderivative size = 97 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 x}{a^2}+\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{a^2 d} \] Output:

3*x/a^2+1/2*arctanh(cos(d*x+c))/a^2/d+1/3*cos(d*x+c)^3/a^2/d+2*cot(d*x+c)/ 
a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a^2/d+cos(d*x+c)*sin(d*x+c)/a^2/d
 

Mathematica [A] (verified)

Time = 3.60 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (6 \cos (c+d x)+2 \cos (3 (c+d x))+3 \left (24 c+24 d x+8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+4 \sin (2 (c+d x))-8 \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{24 a^2 d (1+\sin (c+d x))^2} \] Input:

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
 

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(6*Cos[c + d*x] + 2*Cos[3*(c + d* 
x)] + 3*(24*c + 24*d*x + 8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 + 4*Log[C 
os[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + 4*Sin[2* 
(c + d*x)] - 8*Tan[(c + d*x)/2])))/(24*a^2*d*(1 + Sin[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^3 (a \sin (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3354

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3351

\(\displaystyle \frac {\int \left (\csc ^3(c+d x) a^6+\sin ^3(c+d x) a^6-2 \csc ^2(c+d x) a^6-2 \sin ^2(c+d x) a^6-\csc (c+d x) a^6-\sin (c+d x) a^6+4 a^6\right )dx}{a^8}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^6 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^6 \cos ^3(c+d x)}{3 d}+\frac {2 a^6 \cot (c+d x)}{d}+\frac {a^6 \sin (c+d x) \cos (c+d x)}{d}-\frac {a^6 \cot (c+d x) \csc (c+d x)}{2 d}+3 a^6 x}{a^8}\)

Input:

Int[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
 

Output:

(3*a^6*x + (a^6*ArcTanh[Cos[c + d*x]])/(2*d) + (a^6*Cos[c + d*x]^3)/(3*d) 
+ (2*a^6*Cot[c + d*x])/d - (a^6*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (a^6*Co 
s[c + d*x]*Sin[c + d*x])/d)/a^8
 

Defintions of rubi rules used

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 3351
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]))
 

rule 3354
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* 
m)   Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] 
)^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && 
ILtQ[m, 0]
 
Maple [A] (verified)

Time = 3.74 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.44

method result size
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+24 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) \(140\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+24 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) \(140\)
risch \(\frac {3 x}{a^{2}}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}}{24 d \,a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 d \,a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d \,a^{2}}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d \,a^{2}}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+4 i {\mathrm e}^{2 i \left (d x +c \right )}-4 i}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}\) \(205\)

Input:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/4/d/a^2*(1/2*tan(1/2*d*x+1/2*c)^2-4*tan(1/2*d*x+1/2*c)+16*(-1/2*tan(1/2* 
d*x+1/2*c)^5+1/2*tan(1/2*d*x+1/2*c)^4+1/2*tan(1/2*d*x+1/2*c)+1/6)/(1+tan(1 
/2*d*x+1/2*c)^2)^3+24*arctan(tan(1/2*d*x+1/2*c))-1/2/tan(1/2*d*x+1/2*c)^2+ 
4/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 = 140, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \, \cos \left (d x + c\right )^{5} + 36 \, d x \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right )^{3} - 36 \, d x + 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

1/12*(4*cos(d*x + c)^5 + 36*d*x*cos(d*x + c)^2 - 4*cos(d*x + c)^3 - 36*d*x 
 + 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - 3*(cos(d*x + c)^2 
- 1)*log(-1/2*cos(d*x + c) + 1/2) + 12*(cos(d*x + c)^3 - 3*cos(d*x + c))*s 
in(d*x + c) + 6*cos(d*x + c))/(a^2*d*cos(d*x + c)^2 - a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*cot(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (91) = 182\).

Time = 0.11 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.40 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {24 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {72 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {45 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {24 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 3}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {3 \, {\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{2}} + \frac {144 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{24 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

1/24*((24*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^2/(cos(d*x + c) 
 + 1)^2 + 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 9*sin(d*x + c)^4/(cos( 
d*x + c) + 1)^4 + 72*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 45*sin(d*x + c) 
^6/(cos(d*x + c) + 1)^6 - 24*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3)/(a^2 
*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) 
+ 1)^4 + 3*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8/(c 
os(d*x + c) + 1)^8) - 3*(8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^ 
2/(cos(d*x + c) + 1)^2)/a^2 + 144*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/ 
a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.73 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {72 \, {\left (d x + c\right )}}{a^{2}} - \frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {3 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{4}} + \frac {3 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {16 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}}}{24 \, d} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

1/24*(72*(d*x + c)/a^2 - 12*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*(a^2*ta 
n(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + 3*(6*tan(1/2*d*x 
+ 1/2*c)^2 + 8*tan(1/2*d*x + 1/2*c) - 1)/(a^2*tan(1/2*d*x + 1/2*c)^2) - 16 
*(3*tan(1/2*d*x + 1/2*c)^5 - 3*tan(1/2*d*x + 1/2*c)^4 - 3*tan(1/2*d*x + 1/ 
2*c) - 1)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 31.28 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.78 \[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {6\,\mathrm {atan}\left (\frac {36}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^2\,d}+\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6}+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \] Input:

int((cos(c + d*x)^5*cot(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)
 

Output:

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (6*atan(36/(36*tan(c/2 + (d*x)/2) + 6) - 
(6*tan(c/2 + (d*x)/2))/(36*tan(c/2 + (d*x)/2) + 6)))/(a^2*d) + (4*tan(c/2 
+ (d*x)/2) + (7*tan(c/2 + (d*x)/2)^2)/6 + 20*tan(c/2 + (d*x)/2)^3 - (3*tan 
(c/2 + (d*x)/2)^4)/2 + 12*tan(c/2 + (d*x)/2)^5 + (15*tan(c/2 + (d*x)/2)^6) 
/2 - 4*tan(c/2 + (d*x)/2)^7 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 12*a^2 
*tan(c/2 + (d*x)/2)^4 + 12*a^2*tan(c/2 + (d*x)/2)^6 + 4*a^2*tan(c/2 + (d*x 
)/2)^8)) - log(tan(c/2 + (d*x)/2))/(2*a^2*d) - tan(c/2 + (d*x)/2)/(a^2*d)
 

Reduce [F]

\[ \int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int \frac {\cos \left (d x +c \right )^{5} \cot \left (d x +c \right )^{3}}{\left (\sin \left (d x +c \right ) a +a \right )^{2}}d x \] Input:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
 

Output:

int(cos(d*x+c)^5*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x)