3.1.64 \(\int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx\) [64]

3.1.64.1 Optimal result
3.1.64.2 Mathematica [A] (verified)
3.1.64.3 Rubi [A] (verified)
3.1.64.4 Maple [A] (verified)
3.1.64.5 Fricas [B] (verification not implemented)
3.1.64.6 Sympy [F]
3.1.64.7 Maxima [A] (verification not implemented)
3.1.64.8 Giac [A] (verification not implemented)
3.1.64.9 Mupad [B] (verification not implemented)

3.1.64.1 Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{16 a d}-\frac {\cot (c+d x) \csc (c+d x)}{16 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\csc ^6(c+d x)}{6 a d} \]

output
-1/16*arctanh(cos(d*x+c))/a/d-1/16*cot(d*x+c)*csc(d*x+c)/a/d-1/24*cot(d*x+ 
c)*csc(d*x+c)^3/a/d+1/6*cot(d*x+c)*csc(d*x+c)^5/a/d-1/6*csc(d*x+c)^6/a/d
 
3.1.64.2 Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (12 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \csc ^4\left (\frac {1}{2} (c+d x)\right )+24 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \sec ^4\left (\frac {1}{2} (c+d x)\right )+2 \sec ^6\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{192 a d (1+\sec (c+d x))} \]

input
Integrate[Csc[c + d*x]^5/(a + a*Sec[c + d*x]),x]
 
output
-1/192*(Cos[(c + d*x)/2]^2*(12*Csc[(c + d*x)/2]^2 + 3*Csc[(c + d*x)/2]^4 + 
 24*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) + 3*Sec[(c + d*x)/2]^4 
 + 2*Sec[(c + d*x)/2]^6)*Sec[c + d*x])/(a*d*(1 + Sec[c + d*x]))
 
3.1.64.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.905, Rules used = {3042, 4360, 25, 25, 3042, 25, 3314, 25, 3042, 25, 3086, 15, 3091, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^5 \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3314

\(\displaystyle -\frac {\int \cot ^2(c+d x) \csc ^5(c+d x)dx}{a}-\frac {\int -\cot (c+d x) \csc ^6(c+d x)dx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \cot (c+d x) \csc ^6(c+d x)dx}{a}-\frac {\int \cot ^2(c+d x) \csc ^5(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^6 \tan \left (c+d x-\frac {\pi }{2}\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^5 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^6 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^5 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\)

\(\Big \downarrow \) 3086

\(\displaystyle -\frac {\int \csc ^5(c+d x)d\csc (c+d x)}{a d}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^5 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^5 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}-\frac {\csc ^6(c+d x)}{6 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle -\frac {-\frac {1}{6} \int \csc ^5(c+d x)dx-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 d}}{a}-\frac {\csc ^6(c+d x)}{6 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {1}{6} \int \csc (c+d x)^5dx-\frac {\cot (c+d x) \csc ^5(c+d x)}{6 d}}{a}-\frac {\csc ^6(c+d x)}{6 a d}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input
Int[Csc[c + d*x]^5/(a + a*Sec[c + d*x]),x]
 
output
-1/6*Csc[c + d*x]^6/(a*d) - (-1/6*(Cot[c + d*x]*Csc[c + d*x]^5)/d + ((Cot[ 
c + d*x]*Csc[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Cos[c + d*x]]/d - (Cot[c 
 + d*x]*Csc[c + d*x])/(2*d)))/4)/6)/a
 

3.1.64.3.1 Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3086
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[a/f   Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 
), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 
] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])
 

rule 3091
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3314
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( 
a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[Cos[e + f 
*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d)   Int[Cos[e + f*x]^(p 
 - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & 
& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p 
+ 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
 

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]
 

rule 4360
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 
3.1.64.4 Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75

method result size
derivativedivides \(\frac {-\frac {1}{32 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {1}{16 \cos \left (d x +c \right )-16}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{32}}{d a}\) \(79\)
default \(\frac {-\frac {1}{32 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {1}{16 \cos \left (d x +c \right )-16}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{32}}{d a}\) \(79\)
parallelrisch \(\frac {-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-3 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-18 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d a}\) \(87\)
norman \(\frac {-\frac {1}{128 a d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{32 d a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{128 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{192 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a d}\) \(117\)
risch \(\frac {3 \,{\mathrm e}^{9 i \left (d x +c \right )}+6 \,{\mathrm e}^{8 i \left (d x +c \right )}-8 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{6 i \left (d x +c \right )}-150 \,{\mathrm e}^{5 i \left (d x +c \right )}-22 \,{\mathrm e}^{4 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{24 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{16 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{16 d a}\) \(176\)

input
int(csc(d*x+c)^5/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(-1/32/(cos(d*x+c)-1)^2+1/16/(cos(d*x+c)-1)+1/32*ln(cos(d*x+c)-1)-1/ 
24/(cos(d*x+c)+1)^3-1/32/(cos(d*x+c)+1)^2-1/32*ln(cos(d*x+c)+1))
 
3.1.64.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (96) = 192\).

Time = 0.26 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.05 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {6 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} - 10 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 10 \, \cos \left (d x + c\right ) - 16}{96 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{3} - 2 \, a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right ) + a d\right )}} \]

input
integrate(csc(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="fricas")
 
output
1/96*(6*cos(d*x + c)^4 + 6*cos(d*x + c)^3 - 10*cos(d*x + c)^2 - 3*(cos(d*x 
 + c)^5 + cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + 
 c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^5 + cos(d*x + c)^4 
- 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c) + 1)*log(-1/2*cos(d*x 
 + c) + 1/2) - 10*cos(d*x + c) - 16)/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c 
)^4 - 2*a*d*cos(d*x + c)^3 - 2*a*d*cos(d*x + c)^2 + a*d*cos(d*x + c) + a*d 
)
 
3.1.64.6 Sympy [F]

\[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

input
integrate(csc(d*x+c)**5/(a+a*sec(d*x+c)),x)
 
output
Integral(csc(c + d*x)**5/(sec(c + d*x) + 1), x)/a
 
3.1.64.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.23 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) - 8\right )}}{a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) + a} - \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \]

input
integrate(csc(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="maxima")
 
output
1/96*(2*(3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - 5*cos(d* 
x + c) - 8)/(a*cos(d*x + c)^5 + a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^3 - 2* 
a*cos(d*x + c)^2 + a*cos(d*x + c) + a) - 3*log(cos(d*x + c) + 1)/a + 3*log 
(cos(d*x + c) - 1)/a)/d
 
3.1.64.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.72 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {3 \, {\left (\frac {6 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac {12 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac {\frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {9 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}}}{384 \, d} \]

input
integrate(csc(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="giac")
 
output
1/384*(3*(6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*(cos(d*x + c) - 1)^2 
/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^2/(a*(cos(d*x + c) - 1)^2) + 
 12*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (12*a^2*(cos(d*x 
 + c) - 1)/(cos(d*x + c) + 1) - 9*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 
 1)^2 + 2*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^3)/d
 
3.1.64.9 Mupad [B] (verification not implemented)

Time = 13.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {-\frac {{\cos \left (c+d\,x\right )}^4}{16}-\frac {{\cos \left (c+d\,x\right )}^3}{16}+\frac {5\,{\cos \left (c+d\,x\right )}^2}{48}+\frac {5\,\cos \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\cos \left (c+d\,x\right )}^5+a\,{\cos \left (c+d\,x\right )}^4-2\,a\,{\cos \left (c+d\,x\right )}^3-2\,a\,{\cos \left (c+d\,x\right )}^2+a\,\cos \left (c+d\,x\right )+a\right )} \]

input
int(1/(sin(c + d*x)^5*(a + a/cos(c + d*x))),x)
 
output
- atanh(cos(c + d*x))/(16*a*d) - ((5*cos(c + d*x))/48 + (5*cos(c + d*x)^2) 
/48 - cos(c + d*x)^3/16 - cos(c + d*x)^4/16 + 1/6)/(d*(a + a*cos(c + d*x) 
- 2*a*cos(c + d*x)^2 - 2*a*cos(c + d*x)^3 + a*cos(c + d*x)^4 + a*cos(c + d 
*x)^5))