3.2.66 \(\int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx\) [166]

3.2.66.1 Optimal result
3.2.66.2 Mathematica [A] (verified)
3.2.66.3 Rubi [A] (warning: unable to verify)
3.2.66.4 Maple [A] (verified)
3.2.66.5 Fricas [B] (verification not implemented)
3.2.66.6 Sympy [F]
3.2.66.7 Maxima [A] (verification not implemented)
3.2.66.8 Giac [B] (verification not implemented)
3.2.66.9 Mupad [B] (verification not implemented)

3.2.66.1 Optimal result

Integrand size = 19, antiderivative size = 100 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {3 a \text {arctanh}(\cos (c+d x))}{8 d}-\frac {b \cot ^2(c+d x)}{d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {b \log (\tan (c+d x))}{d} \]

output
-3/8*a*arctanh(cos(d*x+c))/d-b*cot(d*x+c)^2/d-1/4*b*cot(d*x+c)^4/d-3/8*a*c 
ot(d*x+c)*csc(d*x+c)/d-1/4*a*cot(d*x+c)*csc(d*x+c)^3/d+b*ln(tan(d*x+c))/d
 
3.2.66.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.76 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {3 a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {b \csc ^2(c+d x)}{2 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {b \log (\cos (c+d x))}{d}+\frac {3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {b \log (\sin (c+d x))}{d}+\frac {3 a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \]

input
Integrate[Csc[c + d*x]^5*(a + b*Sec[c + d*x]),x]
 
output
(-3*a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) - (b*Csc[ 
c + d*x]^2)/(2*d) - (b*Csc[c + d*x]^4)/(4*d) - (3*a*Log[Cos[(c + d*x)/2]]) 
/(8*d) - (b*Log[Cos[c + d*x]])/d + (3*a*Log[Sin[(c + d*x)/2]])/(8*d) + (b* 
Log[Sin[c + d*x]])/d + (3*a*Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/ 
2]^4)/(64*d)
 
3.2.66.3 Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.947, Rules used = {3042, 4360, 25, 25, 3042, 25, 3313, 25, 3042, 3100, 243, 49, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int -\left (\csc ^5(c+d x) \sec (c+d x) (-a \cos (c+d x)-b)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\left ((b+a \cos (c+d x)) \csc ^5(c+d x) \sec (c+d x)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \csc ^5(c+d x) \sec (c+d x) (a \cos (c+d x)+b)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3313

\(\displaystyle a \int \csc ^5(c+d x)dx-b \int -\csc ^5(c+d x) \sec (c+d x)dx\)

\(\Big \downarrow \) 25

\(\displaystyle a \int \csc ^5(c+d x)dx+b \int \csc ^5(c+d x) \sec (c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \csc (c+d x)^5dx+b \int \csc (c+d x)^5 \sec (c+d x)dx\)

\(\Big \downarrow \) 3100

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input
Int[Csc[c + d*x]^5*(a + b*Sec[c + d*x]),x]
 
output
a*(-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) + (b*(-2*Cot[c + d*x] - Cot[c + 
d*x]^2/2 + Log[Tan[c + d*x]^2]))/(2*d)
 

3.2.66.3.1 Defintions of rubi rules used

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

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 3100
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] 
:> Simp[1/f   Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] 
, x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
 

rule 3313
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[Cos[e + f*x]^ 
p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[Cos[e + f*x]^p*(d*Sin[e + f*x 
])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 
] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | 
| LtQ[p + 1, -n, 2*p + 1])
 

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]
 
3.2.66.4 Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83

method result size
derivativedivides \(\frac {a \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )+b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) \(83\)
default \(\frac {a \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )+b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) \(83\)
parallelrisch \(\frac {-64 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-64 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (24 a +64 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a -b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-8 a -12 b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )+8 a -12 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d}\) \(129\)
norman \(\frac {-\frac {a +b}{64 d}+\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{64 d}+\frac {\left (2 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16 d}-\frac {\left (2 a +3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {\left (3 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\) \(148\)
risch \(\frac {3 a \,{\mathrm e}^{7 i \left (d x +c \right )}+8 b \,{\mathrm e}^{6 i \left (d x +c \right )}-11 a \,{\mathrm e}^{5 i \left (d x +c \right )}-32 b \,{\mathrm e}^{4 i \left (d x +c \right )}-11 a \,{\mathrm e}^{3 i \left (d x +c \right )}+8 b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} a}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) \(193\)

input
int(csc(d*x+c)^5*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d*(a*((-1/4*csc(d*x+c)^3-3/8*csc(d*x+c))*cot(d*x+c)+3/8*ln(-cot(d*x+c)+c 
sc(d*x+c)))+b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c))))
 
3.2.66.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (92) = 184\).

Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.01 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {6 \, a \cos \left (d x + c\right )^{3} + 8 \, b \cos \left (d x + c\right )^{2} - 10 \, a \cos \left (d x + c\right ) - 16 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left ({\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{2} + 3 \, a - 8 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{2} + 3 \, a + 8 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, b}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

input
integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="fricas")
 
output
1/16*(6*a*cos(d*x + c)^3 + 8*b*cos(d*x + c)^2 - 10*a*cos(d*x + c) - 16*(b* 
cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-cos(d*x + c)) - ((3*a - 8*b) 
*cos(d*x + c)^4 - 2*(3*a - 8*b)*cos(d*x + c)^2 + 3*a - 8*b)*log(1/2*cos(d* 
x + c) + 1/2) + ((3*a + 8*b)*cos(d*x + c)^4 - 2*(3*a + 8*b)*cos(d*x + c)^2 
 + 3*a + 8*b)*log(-1/2*cos(d*x + c) + 1/2) - 12*b)/(d*cos(d*x + c)^4 - 2*d 
*cos(d*x + c)^2 + d)
 
3.2.66.6 Sympy [F]

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

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

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (3 \, a - 8 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (3 \, a \cos \left (d x + c\right )^{3} + 4 \, b \cos \left (d x + c\right )^{2} - 5 \, a \cos \left (d x + c\right ) - 6 \, b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{16 \, d} \]

input
integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="maxima")
 
output
-1/16*((3*a - 8*b)*log(cos(d*x + c) + 1) - (3*a + 8*b)*log(cos(d*x + c) - 
1) + 16*b*log(cos(d*x + c)) - 2*(3*a*cos(d*x + c)^3 + 4*b*cos(d*x + c)^2 - 
 5*a*cos(d*x + c) - 6*b)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/d
 
3.2.66.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (92) = 184\).

Time = 0.32 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.66 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {4 \, {\left (3 \, a + 8 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 64 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a + b - \frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {12 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{64 \, d} \]

input
integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="giac")
 
output
1/64*(4*(3*a + 8*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 64 
*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) - (a + b - 8*a*(co 
s(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*b*(cos(d*x + c) - 1)/(cos(d*x + c) 
 + 1) + 18*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b*(cos(d*x + c 
) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 
 8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*b*(cos(d*x + c) - 1)/(cos( 
d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + 
 c) - 1)^2/(cos(d*x + c) + 1)^2)/d
 
3.2.66.9 Mupad [B] (verification not implemented)

Time = 13.91 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.17 \[ \int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a}{16}+\frac {b}{2}\right )}{d}-\frac {-\frac {3\,a\,{\cos \left (c+d\,x\right )}^3}{8}-\frac {b\,{\cos \left (c+d\,x\right )}^2}{2}+\frac {5\,a\,\cos \left (c+d\,x\right )}{8}+\frac {3\,b}{4}}{d\,\left ({\cos \left (c+d\,x\right )}^4-2\,{\cos \left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a}{16}-\frac {b}{2}\right )}{d}-\frac {b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

input
int((a + b/cos(c + d*x))/sin(c + d*x)^5,x)
 
output
(log(cos(c + d*x) - 1)*((3*a)/16 + b/2))/d - ((3*b)/4 + (5*a*cos(c + d*x)) 
/8 - (3*a*cos(c + d*x)^3)/8 - (b*cos(c + d*x)^2)/2)/(d*(cos(c + d*x)^4 - 2 
*cos(c + d*x)^2 + 1)) - (log(cos(c + d*x) + 1)*((3*a)/16 - b/2))/d - (b*lo 
g(cos(c + d*x)))/d