\(\int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [652]

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

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {13 \text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d}-\frac {5 \cot ^3(c+d x)}{3 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^3 d}+\frac {13 \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d} \] Output:

13/8*arctanh(cos(d*x+c))/a^3/d-4*cot(d*x+c)/a^3/d-5/3*cot(d*x+c)^3/a^3/d-1 
/5*cot(d*x+c)^5/a^3/d+13/8*cot(d*x+c)*csc(d*x+c)/a^3/d+3/4*cot(d*x+c)*csc( 
d*x+c)^3/a^3/d
 

Mathematica [A] (verified)

Time = 2.85 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.66 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\csc ^5(c+d x) \left (-1600 \cos (c+d x)+1520 \cos (3 (c+d x))-304 \cos (5 (c+d x))+1950 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-1950 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+1500 \sin (2 (c+d x))-975 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+975 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-390 \sin (4 (c+d x))+195 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-195 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{1920 a^3 d} \] Input:

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]
 

Output:

(Csc[c + d*x]^5*(-1600*Cos[c + d*x] + 1520*Cos[3*(c + d*x)] - 304*Cos[5*(c 
 + d*x)] + 1950*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 1950*Log[Sin[(c + d*x 
)/2]]*Sin[c + d*x] + 1500*Sin[2*(c + d*x)] - 975*Log[Cos[(c + d*x)/2]]*Sin 
[3*(c + d*x)] + 975*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 390*Sin[4*(c 
+ d*x)] + 195*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 195*Log[Sin[(c + d* 
x)/2]]*Sin[5*(c + d*x)]))/(1920*a^3*d)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3187, 3042, 3236, 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.

\(\displaystyle \int \frac {\cot ^6(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^6 (a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3187

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3}{\sin (c+d x)^6}dx}{a^6}\)

\(\Big \downarrow \) 3236

\(\displaystyle \frac {\int \left (a^3 \csc ^6(c+d x)-3 a^3 \csc ^5(c+d x)+3 a^3 \csc ^4(c+d x)-a^3 \csc ^3(c+d x)\right )dx}{a^6}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {13 a^3 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^3 \cot ^5(c+d x)}{5 d}-\frac {5 a^3 \cot ^3(c+d x)}{3 d}-\frac {4 a^3 \cot (c+d x)}{d}+\frac {3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {13 a^3 \cot (c+d x) \csc (c+d x)}{8 d}}{a^6}\)

Input:

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]
 

Output:

((13*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (4*a^3*Cot[c + d*x])/d - (5*a^3*Co 
t[c + d*x]^3)/(3*d) - (a^3*Cot[c + d*x]^5)/(5*d) + (13*a^3*Cot[c + d*x]*Cs 
c[c + d*x])/(8*d) + (3*a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d))/a^6
 

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 3187
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_), x_Symbol] :> Simp[a^p   Int[Sin[e + f*x]^p/(a - b*Sin[e + f*x])^m, x], 
x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ 
[p, 2*m]
 

rule 3236
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + 
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt 
Q[m, 0] && RationalQ[n]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 8.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.28

method result size
risch \(-\frac {195 \,{\mathrm e}^{9 i \left (d x +c \right )}-720 i {\mathrm e}^{6 i \left (d x +c \right )}-750 \,{\mathrm e}^{7 i \left (d x +c \right )}+2320 i {\mathrm e}^{4 i \left (d x +c \right )}-1520 i {\mathrm e}^{2 i \left (d x +c \right )}+750 \,{\mathrm e}^{3 i \left (d x +c \right )}+304 i-195 \,{\mathrm e}^{i \left (d x +c \right )}}{60 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}\) \(146\)
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-52 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {17}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {46}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{3}}\) \(150\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+46 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-52 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {17}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {46}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{3}}\) \(150\)

Input:

int(cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/60*(195*exp(9*I*(d*x+c))-720*I*exp(6*I*(d*x+c))-750*exp(7*I*(d*x+c))+23 
20*I*exp(4*I*(d*x+c))-1520*I*exp(2*I*(d*x+c))+750*exp(3*I*(d*x+c))+304*I-1 
95*exp(I*(d*x+c)))/a^3/d/(exp(2*I*(d*x+c))-1)^5-13/8/d/a^3*ln(exp(I*(d*x+c 
))-1)+13/8/d/a^3*ln(exp(I*(d*x+c))+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.57 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {608 \, \cos \left (d x + c\right )^{5} - 1520 \, \cos \left (d x + c\right )^{3} - 195 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 195 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (13 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 960 \, \cos \left (d x + c\right )}{240 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/240*(608*cos(d*x + c)^5 - 1520*cos(d*x + c)^3 - 195*(cos(d*x + c)^4 - 2 
*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 195*(cos(d 
*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c 
) + 30*(13*cos(d*x + c)^3 - 19*cos(d*x + c))*sin(d*x + c) + 960*cos(d*x + 
c))/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cot ^{6}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:

integrate(cot(d*x+c)**6/(a+a*sin(d*x+c))**3,x)
 

Output:

Integral(cot(c + d*x)**6/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + 
d*x) + 1), x)/a**3
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (104) = 208\).

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.05 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {1380 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {480 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {170 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {1560 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {{\left (\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {170 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {480 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1380 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{3} \sin \left (d x + c\right )^{5}}}{960 \, d} \] Input:

integrate(cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/960*((1380*sin(d*x + c)/(cos(d*x + c) + 1) - 480*sin(d*x + c)^2/(cos(d*x 
 + c) + 1)^2 + 170*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 45*sin(d*x + c)^4 
/(cos(d*x + c) + 1)^4 + 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 1560* 
log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + (45*sin(d*x + c)/(cos(d*x + c) 
+ 1) - 170*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 480*sin(d*x + c)^3/(cos(d 
*x + c) + 1)^3 - 1380*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6)*(cos(d*x + 
c) + 1)^5/(a^3*sin(d*x + c)^5))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.64 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {1560 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {3562 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1380 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 170 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {6 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 170 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 480 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1380 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{960 \, d} \] Input:

integrate(cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

-1/960*(1560*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (3562*tan(1/2*d*x + 1/2* 
c)^5 - 1380*tan(1/2*d*x + 1/2*c)^4 + 480*tan(1/2*d*x + 1/2*c)^3 - 170*tan( 
1/2*d*x + 1/2*c)^2 + 45*tan(1/2*d*x + 1/2*c) - 6)/(a^3*tan(1/2*d*x + 1/2*c 
)^5) - (6*a^12*tan(1/2*d*x + 1/2*c)^5 - 45*a^12*tan(1/2*d*x + 1/2*c)^4 + 1 
70*a^12*tan(1/2*d*x + 1/2*c)^3 - 480*a^12*tan(1/2*d*x + 1/2*c)^2 + 1380*a^ 
12*tan(1/2*d*x + 1/2*c))/a^15)/d
 

Mupad [B] (verification not implemented)

Time = 34.62 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.55 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+45\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-45\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-170\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+480\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-1380\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+1380\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-480\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+170\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1560\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:

int(cot(c + d*x)^6/(a + a*sin(c + d*x))^3,x)
 

Output:

-(6*cos(c/2 + (d*x)/2)^10 - 6*sin(c/2 + (d*x)/2)^10 + 45*cos(c/2 + (d*x)/2 
)*sin(c/2 + (d*x)/2)^9 - 45*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 170* 
cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 480*cos(c/2 + (d*x)/2)^3*sin(c 
/2 + (d*x)/2)^7 - 1380*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 1380*co 
s(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 - 480*cos(c/2 + (d*x)/2)^7*sin(c/2 
 + (d*x)/2)^3 + 170*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 1560*log(s 
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x) 
/2)^5)/(960*a^3*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-304 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+195 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-152 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+90 \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \cos \left (d x +c \right )-195 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}}{120 \sin \left (d x +c \right )^{5} a^{3} d} \] Input:

int(cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x)
 

Output:

( - 304*cos(c + d*x)*sin(c + d*x)**4 + 195*cos(c + d*x)*sin(c + d*x)**3 - 
152*cos(c + d*x)*sin(c + d*x)**2 + 90*cos(c + d*x)*sin(c + d*x) - 24*cos(c 
 + d*x) - 195*log(tan((c + d*x)/2))*sin(c + d*x)**5)/(120*sin(c + d*x)**5* 
a**3*d)