\(\int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [747]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {7 \text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {4 \cot ^3(c+d x)}{3 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^3 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d} \] Output:

-7/8*arctanh(cos(d*x+c))/a^3/d-4/3*cot(d*x+c)^3/a^3/d-1/5*cot(d*x+c)^5/a^3 
/d+1/8*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 = 3.53 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^5(c+d x) \left (560 \cos (c+d x)-40 \cos (3 (c+d x))-136 \cos (5 (c+d x))+1050 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-1050 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-780 \sin (2 (c+d x))-525 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+525 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+30 \sin (4 (c+d x))+105 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-105 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{1920 a^3 d} \] Input:

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^3,x]
 

Output:

-1/1920*(Csc[c + d*x]^5*(560*Cos[c + d*x] - 40*Cos[3*(c + d*x)] - 136*Cos[ 
5*(c + d*x)] + 1050*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 1050*Log[Sin[(c + 
 d*x)/2]]*Sin[c + d*x] - 780*Sin[2*(c + d*x)] - 525*Log[Cos[(c + d*x)/2]]* 
Sin[3*(c + d*x)] + 525*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 30*Sin[4*( 
c + d*x)] + 105*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 105*Log[Sin[(c + 
d*x)/2]]*Sin[5*(c + d*x)]))/(a^3*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 104, 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, 3352, 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.

\(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^6 (a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3354

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\cos (c+d x)^2 (a-a \sin (c+d x))^3}{\sin (c+d x)^6}dx}{a^6}\)

\(\Big \downarrow \) 3352

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {7 a^3 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^3 \cot ^5(c+d x)}{5 d}-\frac {4 a^3 \cot ^3(c+d x)}{3 d}+\frac {3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a^3 \cot (c+d x) \csc (c+d x)}{8 d}}{a^6}\)

Input:

Int[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^3,x]
 

Output:

((-7*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (4*a^3*Cot[c + d*x]^3)/(3*d) - (a^ 
3*Cot[c + d*x]^5)/(5*d) + (a^3*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (3*a^3*C 
ot[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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 = 5.32 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {13}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+28 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{3}}\)	\(150\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {13}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+28 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{3}}\)	\(150\)
risch	\(-\frac {-360 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+960 i {\mathrm e}^{6 i \left (d x +c \right )}-390 \,{\mathrm e}^{7 i \left (d x +c \right )}-400 i {\mathrm e}^{4 i \left (d x +c \right )}+320 i {\mathrm e}^{2 i \left (d x +c \right )}+390 \,{\mathrm e}^{3 i \left (d x +c \right )}-136 i-15 \,{\mathrm e}^{i \left (d x +c \right )}}{60 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}\)	\(158\)

Input:

int(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/32/d/a^3*(1/5*tan(1/2*d*x+1/2*c)^5-3/2*tan(1/2*d*x+1/2*c)^4+13/3*tan(1/2 
*d*x+1/2*c)^3-4*tan(1/2*d*x+1/2*c)^2-14*tan(1/2*d*x+1/2*c)-1/5/tan(1/2*d*x 
+1/2*c)^5+14/tan(1/2*d*x+1/2*c)+4/tan(1/2*d*x+1/2*c)^2-13/3/tan(1/2*d*x+1/ 
2*c)^3+3/2/tan(1/2*d*x+1/2*c)^4+28*ln(tan(1/2*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {272 \, \cos \left (d x + c\right )^{5} - 320 \, \cos \left (d x + c\right )^{3} - 105 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 105 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (\cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

1/240*(272*cos(d*x + c)^5 - 320*cos(d*x + c)^3 - 105*(cos(d*x + c)^4 - 2*c 
os(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 105*(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*(cos(d*x + c)^3 - 7*cos(d*x + c))*sin(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]

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

integrate(cos(d*x+c)**2*cot(d*x+c)**6/(a+a*sin(d*x+c))**3,x)
 

Output:

Integral(cos(c + d*x)**2*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. 235 vs. \(2 (90) = 180\).

Time = 0.04 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.35 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {420 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {130 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {840 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {130 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {420 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{3} \sin \left (d x + c\right )^{5}}}{960 \, d} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

-1/960*((420*sin(d*x + c)/(cos(d*x + c) + 1) + 120*sin(d*x + c)^2/(cos(d*x 
 + c) + 1)^2 - 130*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 - 840*l 
og(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (45*sin(d*x + c)/(cos(d*x + c) + 
 1) - 130*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 120*sin(d*x + c)^3/(cos(d* 
x + c) + 1)^3 + 420*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (90) = 180\).

Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {1918 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 130 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {6 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 130 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 420 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{960 \, d} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

1/960*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (1918*tan(1/2*d*x + 1/2*c) 
^5 - 420*tan(1/2*d*x + 1/2*c)^4 - 120*tan(1/2*d*x + 1/2*c)^3 + 130*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 + 130* 
a^12*tan(1/2*d*x + 1/2*c)^3 - 120*a^12*tan(1/2*d*x + 1/2*c)^2 - 420*a^12*t 
an(1/2*d*x + 1/2*c))/a^15)/d
 

Mupad [B] (verification not implemented)

Time = 31.96 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.91 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-45\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+45\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+130\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-130\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+840\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:

int((cos(c + d*x)^2*cot(c + d*x)^6)/(a + a*sin(c + d*x))^3,x)
 

Output:

(6*sin(c/2 + (d*x)/2)^10 - 6*cos(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) + 130*c 
os(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 120*cos(c/2 + (d*x)/2)^3*sin(c/ 
2 + (d*x)/2)^7 - 420*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 420*cos(c 
/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + 
(d*x)/2)^3 - 130*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 840*log(sin(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 = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {136 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-112 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+90 \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \cos \left (d x +c \right )+105 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}}{120 \sin \left (d x +c \right )^{5} a^{3} d} \] Input:

int(cos(d*x+c)^2*cot(d*x+c)^6/(a+a*sin(d*x+c))^3,x)
 

Output:

(136*cos(c + d*x)*sin(c + d*x)**4 + 15*cos(c + d*x)*sin(c + d*x)**3 - 112* 
cos(c + d*x)*sin(c + d*x)**2 + 90*cos(c + d*x)*sin(c + d*x) - 24*cos(c + d 
*x) + 105*log(tan((c + d*x)/2))*sin(c + d*x)**5)/(120*sin(c + d*x)**5*a**3 
*d)