\(\int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [306]

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 = 29, antiderivative size = 114 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x)}{a d}-\frac {2 \cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \] Output:

3/8*arctanh(cos(d*x+c))/a/d-cot(d*x+c)/a/d-2/3*cot(d*x+c)^3/a/d-1/5*cot(d* 
x+c)^5/a/d+3/8*cot(d*x+c)*csc(d*x+c)/a/d+1/4*cot(d*x+c)*csc(d*x+c)^3/a/d
 

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.66 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^5(c+d x) \left (-640 \cos (c+d x)+320 \cos (3 (c+d x))-64 \cos (5 (c+d x))+450 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-450 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+420 \sin (2 (c+d x))-225 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+225 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-90 \sin (4 (c+d x))+45 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-45 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{1920 a d} \] Input:

Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
 

Output:

(Csc[c + d*x]^5*(-640*Cos[c + d*x] + 320*Cos[3*(c + d*x)] - 64*Cos[5*(c + 
d*x)] + 450*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 450*Log[Sin[(c + d*x)/2]] 
*Sin[c + d*x] + 420*Sin[2*(c + d*x)] - 225*Log[Cos[(c + d*x)/2]]*Sin[3*(c 
+ d*x)] + 225*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 90*Sin[4*(c + d*x)] 
 + 45*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 45*Log[Sin[(c + d*x)/2]]*Si 
n[5*(c + d*x)]))/(1920*a*d)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3318, 3042, 4254, 2009, 4255, 3042, 4255, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\int \csc (c+d x)^5dx}{a}-\frac {\frac {1}{5} \cot ^5(c+d x)+\frac {2}{3} \cot ^3(c+d x)+\cot (c+d x)}{a d}\)

\(\Big \downarrow \) 4255

\(\displaystyle -\frac {\frac {3}{4} \int \csc ^3(c+d x)dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\frac {1}{5} \cot ^5(c+d x)+\frac {2}{3} \cot ^3(c+d x)+\cot (c+d x)}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{4} \int \csc (c+d x)^3dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\frac {1}{5} \cot ^5(c+d x)+\frac {2}{3} \cot ^3(c+d x)+\cot (c+d x)}{a d}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[(Cot[c + d*x]^2*Csc[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
 

Output:

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

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 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]
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18

method result size
risch \(-\frac {45 \,{\mathrm e}^{9 i \left (d x +c \right )}-210 \,{\mathrm e}^{7 i \left (d x +c \right )}+640 i {\mathrm e}^{4 i \left (d x +c \right )}-320 i {\mathrm e}^{2 i \left (d x +c \right )}+210 \,{\mathrm e}^{3 i \left (d x +c \right )}+64 i-45 \,{\mathrm e}^{i \left (d x +c \right )}}{60 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\) \(134\)
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {5}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{32 d a}\) \(150\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {5}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{32 d a}\) \(150\)

Input:

int(cot(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-1/60*(45*exp(9*I*(d*x+c))-210*exp(7*I*(d*x+c))+640*I*exp(4*I*(d*x+c))-320 
*I*exp(2*I*(d*x+c))+210*exp(3*I*(d*x+c))+64*I-45*exp(I*(d*x+c)))/d/a/(exp( 
2*I*(d*x+c))-1)^5-3/8/d/a*ln(exp(I*(d*x+c))-1)+3/8/d/a*ln(exp(I*(d*x+c))+1 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.52 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {128 \, \cos \left (d x + c\right )^{5} - 320 \, \cos \left (d x + c\right )^{3} - 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 240 \, \cos \left (d x + c\right )}{240 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

-1/240*(128*cos(d*x + c)^5 - 320*cos(d*x + c)^3 - 45*(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) + 45*(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*(3*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 240*cos(d*x + c))/( 
(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{2}{\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate(cot(d*x+c)**2*csc(d*x+c)**4/(a+a*sin(d*x+c)),x)
 

Output:

Integral(cot(c + d*x)**2*csc(c + d*x)**4/(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. 234 vs. \(2 (104) = 208\).

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.05 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {300 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {50 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac {360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {50 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {300 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{960 \, d} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
                                                                                    
                                                                                    
 

Output:

1/960*((300*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x 
+ c) + 1)^2 + 50*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/( 
cos(d*x + c) + 1)^4 + 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 360*log(s 
in(d*x + c)/(cos(d*x + c) + 1))/a + (15*sin(d*x + c)/(cos(d*x + c) + 1) - 
50*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 120*sin(d*x + c)^3/(cos(d*x + c) 
+ 1)^3 - 300*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6)*(cos(d*x + c) + 1)^5 
/(a*sin(d*x + c)^5))/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.64 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 50 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 300 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 300 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

-1/960*(360*log(abs(tan(1/2*d*x + 1/2*c)))/a - (6*a^4*tan(1/2*d*x + 1/2*c) 
^5 - 15*a^4*tan(1/2*d*x + 1/2*c)^4 + 50*a^4*tan(1/2*d*x + 1/2*c)^3 - 120*a 
^4*tan(1/2*d*x + 1/2*c)^2 + 300*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (822*tan(1 
/2*d*x + 1/2*c)^5 - 300*tan(1/2*d*x + 1/2*c)^4 + 120*tan(1/2*d*x + 1/2*c)^ 
3 - 50*tan(1/2*d*x + 1/2*c)^2 + 15*tan(1/2*d*x + 1/2*c) - 6)/(a*tan(1/2*d* 
x + 1/2*c)^5))/d
 

Mupad [B] (verification not implemented)

Time = 18.03 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.55 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-50\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-300\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+300\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+50\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960\,a\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:

int(cot(c + d*x)^2/(sin(c + d*x)^4*(a + a*sin(c + d*x))),x)
 

Output:

-(6*cos(c/2 + (d*x)/2)^10 - 6*sin(c/2 + (d*x)/2)^10 + 15*cos(c/2 + (d*x)/2 
)*sin(c/2 + (d*x)/2)^9 - 15*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 50*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 - 300*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 300*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 + 50*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 360*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*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x) \csc ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \cos \left (d x +c \right )-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}}{120 \sin \left (d x +c \right )^{5} a d} \] Input:

int(cot(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x)
 

Output:

( - 64*cos(c + d*x)*sin(c + d*x)**4 + 45*cos(c + d*x)*sin(c + d*x)**3 - 32 
*cos(c + d*x)*sin(c + d*x)**2 + 30*cos(c + d*x)*sin(c + d*x) - 24*cos(c + 
d*x) - 45*log(tan((c + d*x)/2))*sin(c + d*x)**5)/(120*sin(c + d*x)**5*a*d)