3.3.8 \(\int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx\) [208]

3.3.8.1 Optimal result
3.3.8.2 Mathematica [A] (verified)
3.3.8.3 Rubi [A] (verified)
3.3.8.4 Maple [A] (verified)
3.3.8.5 Fricas [B] (verification not implemented)
3.3.8.6 Sympy [F]
3.3.8.7 Maxima [F(-2)]
3.3.8.8 Giac [B] (verification not implemented)
3.3.8.9 Mupad [B] (verification not implemented)

3.3.8.1 Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 a^5 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {\left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right ) \csc (c+d x)}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^3(c+d x)}{15 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^5(c+d x)}{5 \left (a^2-b^2\right ) d} \]

output
-2*a^5*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/( 
a+b)^(7/2)/d+1/15*(15*a^4*b-a*(8*a^4+9*a^2*b^2-2*b^4)*cos(d*x+c))*csc(d*x+ 
c)/(a^2-b^2)^3/d+1/15*(5*a^2*b-a*(4*a^2+b^2)*cos(d*x+c))*csc(d*x+c)^3/(a^2 
-b^2)^2/d+1/5*(b-a*cos(d*x+c))*csc(d*x+c)^5/(a^2-b^2)/d
 
3.3.8.2 Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.38 \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (\frac {960 a^5 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}-\frac {2 \left (64 a^2+43 a b+9 b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3}+\frac {8 (19 a-9 b) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}+\frac {96 \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )}{a-b}-\frac {(19 a+9 b) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{2 (a+b)^2}-\frac {3 \csc ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{2 (a+b)}+\frac {2 \left (64 a^2-43 a b+9 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3}\right )}{480 d (a+b \sec (c+d x))} \]

input
Integrate[Csc[c + d*x]^6/(a + b*Sec[c + d*x]),x]
 
output
((b + a*Cos[c + d*x])*Sec[c + d*x]*((960*a^5*b*ArcTanh[((-a + b)*Tan[(c + 
d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) - (2*(64*a^2 + 43*a*b + 9*b^2 
)*Cot[(c + d*x)/2])/(a + b)^3 + (8*(19*a - 9*b)*Csc[c + d*x]^3*Sin[(c + d* 
x)/2]^4)/(a - b)^2 + (96*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6)/(a - b) - ((19 
*a + 9*b)*Csc[(c + d*x)/2]^4*Sin[c + d*x])/(2*(a + b)^2) - (3*Csc[(c + d*x 
)/2]^6*Sin[c + d*x])/(2*(a + b)) + (2*(64*a^2 - 43*a*b + 9*b^2)*Tan[(c + d 
*x)/2])/(a - b)^3))/(480*d*(a + b*Sec[c + d*x]))
 
3.3.8.3 Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.20, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4360, 25, 25, 3042, 25, 3345, 3042, 3345, 3042, 3345, 27, 3042, 3138, 221}

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.

\(\displaystyle \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^6 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {\cot (c+d x) \csc ^5(c+d x)}{a \cos (c+d x)+b}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^6 \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\)

\(\Big \downarrow \) 3345

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\int \frac {4 \sin \left (c+d x-\frac {\pi }{2}\right ) a^2+b a}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3345

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\int \frac {\left (a b \left (7 a^2-2 b^2\right )-2 a^2 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{b+a \cos (c+d x)}dx}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\int \frac {2 \left (4 a^2+b^2\right ) \sin \left (c+d x-\frac {\pi }{2}\right ) a^2+b \left (7 a^2-2 b^2\right ) a}{\cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3345

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\frac {\int \frac {15 a^5 b}{b+a \cos (c+d x)}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\frac {15 a^5 b \int \frac {1}{b+a \cos (c+d x)}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\frac {15 a^5 b \int \frac {1}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\frac {30 a^5 b \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\csc ^5(c+d x) (b-a \cos (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {\frac {\frac {30 a^5 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\csc (c+d x) \left (15 a^4 b-a \left (8 a^4+9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}-\frac {\csc ^3(c+d x) \left (5 a^2 b-a \left (4 a^2+b^2\right ) \cos (c+d x)\right )}{3 d \left (a^2-b^2\right )}}{5 \left (a^2-b^2\right )}\)

input
Int[Csc[c + d*x]^6/(a + b*Sec[c + d*x]),x]
 
output
((b - a*Cos[c + d*x])*Csc[c + d*x]^5)/(5*(a^2 - b^2)*d) - (-1/3*((5*a^2*b 
- a*(4*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^3)/((a^2 - b^2)*d) + ((30*a^5 
*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[ 
a + b]*(a^2 - b^2)*d) - ((15*a^4*b - a*(8*a^4 + 9*a^2*b^2 - 2*b^4)*Cos[c + 
 d*x])*Csc[c + d*x])/((a^2 - b^2)*d))/(3*(a^2 - b^2)))/(5*(a^2 - b^2))
 

3.3.8.3.1 Defintions of rubi rules used

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

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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

rule 3138
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3345
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co 
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 
Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 
 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 
 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], 
x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[p, -1] && IntegerQ[2*m]
 

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]
 
3.3.8.4 Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.40

method result size
derivativedivides \(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 \left (a -b \right )^{3}}-\frac {2 a^{5} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{160 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5 a +3 b}{96 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {10 a^{2}+8 a b +2 b^{2}}{32 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) \(282\)
default \(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 \left (a -b \right )^{3}}-\frac {2 a^{5} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{160 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5 a +3 b}{96 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {10 a^{2}+8 a b +2 b^{2}}{32 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) \(282\)
risch \(-\frac {2 i \left (15 a^{4} b \,{\mathrm e}^{9 i \left (d x +c \right )}-15 a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-80 a^{4} b \,{\mathrm e}^{7 i \left (d x +c \right )}+20 a^{2} b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+90 a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-30 a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+178 a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}-136 a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+48 b^{5} {\mathrm e}^{5 i \left (d x +c \right )}-80 a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-10 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-80 a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}+20 a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+40 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+30 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-10 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}-8 a^{5}-9 a^{3} b^{2}+2 a \,b^{4}\right )}{15 \left (-a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}+b^{6}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} d}-\frac {b \,a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {b \,a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) \(499\)

input
int(csc(d*x+c)^6/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d*(1/32/(a-b)^3*(1/5*a^2*tan(1/2*d*x+1/2*c)^5-2/5*a*b*tan(1/2*d*x+1/2*c) 
^5+1/5*b^2*tan(1/2*d*x+1/2*c)^5+5/3*a^2*tan(1/2*d*x+1/2*c)^3-8/3*a*b*tan(1 
/2*d*x+1/2*c)^3+b^2*tan(1/2*d*x+1/2*c)^3+10*a^2*tan(1/2*d*x+1/2*c)-8*a*b*t 
an(1/2*d*x+1/2*c)+2*b^2*tan(1/2*d*x+1/2*c))-2/(a-b)^3/(a+b)^3*a^5*b/((a-b) 
*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))-1/160/ 
(a+b)/tan(1/2*d*x+1/2*c)^5-1/96*(5*a+3*b)/(a+b)^2/tan(1/2*d*x+1/2*c)^3-1/3 
2/(a+b)^3*(10*a^2+8*a*b+2*b^2)/tan(1/2*d*x+1/2*c))
 
3.3.8.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (187) = 374\).

Time = 0.32 (sec) , antiderivative size = 861, normalized size of antiderivative = 4.28 \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {46 \, a^{6} b - 68 \, a^{4} b^{3} + 28 \, a^{2} b^{5} - 6 \, b^{7} - 2 \, {\left (8 \, a^{7} + a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 30 \, {\left (a^{6} b - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{7} - a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (a^{5} b \cos \left (d x + c\right )^{4} - 2 \, a^{5} b \cos \left (d x + c\right )^{2} + a^{5} b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 10 \, {\left (7 \, a^{6} b - 8 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 30 \, {\left (a^{7} - a^{5} b^{2}\right )} \cos \left (d x + c\right )}{30 \, {\left ({\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d\right )} \sin \left (d x + c\right )}, \frac {23 \, a^{6} b - 34 \, a^{4} b^{3} + 14 \, a^{2} b^{5} - 3 \, b^{7} - {\left (8 \, a^{7} + a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left (a^{6} b - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (4 \, a^{7} - a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (a^{5} b \cos \left (d x + c\right )^{4} - 2 \, a^{5} b \cos \left (d x + c\right )^{2} + a^{5} b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 5 \, {\left (7 \, a^{6} b - 8 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (a^{7} - a^{5} b^{2}\right )} \cos \left (d x + c\right )}{15 \, {\left ({\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

input
integrate(csc(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="fricas")
 
output
[1/30*(46*a^6*b - 68*a^4*b^3 + 28*a^2*b^5 - 6*b^7 - 2*(8*a^7 + a^5*b^2 - 1 
1*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^5 + 30*(a^6*b - a^4*b^3)*cos(d*x + c)^4 
+ 10*(4*a^7 - a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 15*(a^5*b*cos( 
d*x + c)^4 - 2*a^5*b*cos(d*x + c)^2 + a^5*b)*sqrt(a^2 - b^2)*log((2*a*b*co 
s(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + 
 c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + 
 c) + b^2))*sin(d*x + c) - 10*(7*a^6*b - 8*a^4*b^3 + a^2*b^5)*cos(d*x + c) 
^2 - 30*(a^7 - a^5*b^2)*cos(d*x + c))/(((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a 
^2*b^6 + b^8)*d*cos(d*x + c)^4 - 2*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^ 
6 + b^8)*d*cos(d*x + c)^2 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8 
)*d)*sin(d*x + c)), 1/15*(23*a^6*b - 34*a^4*b^3 + 14*a^2*b^5 - 3*b^7 - (8* 
a^7 + a^5*b^2 - 11*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^5 + 15*(a^6*b - a^4*b^3 
)*cos(d*x + c)^4 + 5*(4*a^7 - a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 
- 15*(a^5*b*cos(d*x + c)^4 - 2*a^5*b*cos(d*x + c)^2 + a^5*b)*sqrt(-a^2 + b 
^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c 
)))*sin(d*x + c) - 5*(7*a^6*b - 8*a^4*b^3 + a^2*b^5)*cos(d*x + c)^2 - 15*( 
a^7 - a^5*b^2)*cos(d*x + c))/(((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + 
b^8)*d*cos(d*x + c)^4 - 2*(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)* 
d*cos(d*x + c)^2 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sin( 
d*x + c))]
 
3.3.8.6 Sympy [F]

\[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

input
integrate(csc(d*x+c)**6/(a+b*sec(d*x+c)),x)
 
output
Integral(csc(c + d*x)**6/(a + b*sec(c + d*x)), x)
 
3.3.8.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]

input
integrate(csc(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 
3.3.8.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (187) = 374\).

Time = 0.34 (sec) , antiderivative size = 541, normalized size of antiderivative = 2.69 \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {960 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{5} b}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 150 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 420 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 420 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 180 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} + \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

input
integrate(csc(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="giac")
 
output
-1/480*(960*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a 
*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*a^5*b/( 
(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - (3*a^4*tan(1/2*d*x 
 + 1/2*c)^5 - 12*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*b^2*tan(1/2*d*x + 1 
/2*c)^5 - 12*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*b^4*tan(1/2*d*x + 1/2*c)^5 + 
 25*a^4*tan(1/2*d*x + 1/2*c)^3 - 90*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 120*a^2 
*b^2*tan(1/2*d*x + 1/2*c)^3 - 70*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*b^4*tan 
(1/2*d*x + 1/2*c)^3 + 150*a^4*tan(1/2*d*x + 1/2*c) - 420*a^3*b*tan(1/2*d*x 
 + 1/2*c) + 420*a^2*b^2*tan(1/2*d*x + 1/2*c) - 180*a*b^3*tan(1/2*d*x + 1/2 
*c) + 30*b^4*tan(1/2*d*x + 1/2*c))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^ 
3 + 5*a*b^4 - b^5) + (150*a^2*tan(1/2*d*x + 1/2*c)^4 + 120*a*b*tan(1/2*d*x 
 + 1/2*c)^4 + 30*b^2*tan(1/2*d*x + 1/2*c)^4 + 25*a^2*tan(1/2*d*x + 1/2*c)^ 
2 + 40*a*b*tan(1/2*d*x + 1/2*c)^2 + 15*b^2*tan(1/2*d*x + 1/2*c)^2 + 3*a^2 
+ 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(1/2*d*x + 1/2*c)^5)) 
/d
 
3.3.8.9 Mupad [B] (verification not implemented)

Time = 14.35 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.93 \[ \int \frac {\csc ^6(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,d\,\left (32\,a-32\,b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {4}{3\,\left (32\,a-32\,b\right )}+\frac {32\,a+32\,b}{3\,{\left (32\,a-32\,b\right )}^2}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5}{32\,a-32\,b}+\frac {\left (\frac {4}{32\,a-32\,b}+\frac {32\,a+32\,b}{{\left (32\,a-32\,b\right )}^2}\right )\,\left (32\,a+32\,b\right )}{32\,a-32\,b}\right )}{d}-\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{5\,\left (a+b\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-5\,a^5+11\,a^4\,b-4\,a^3\,b^2-4\,a^2\,b^3+a\,b^4+b^5\right )}{{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^4-12\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3-3\,b^4\right )}{3\,{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (32\,a^3-96\,a^2\,b+96\,a\,b^2-32\,b^3\right )}-\frac {2\,a^5\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{5/2}}\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

input
int(1/(sin(c + d*x)^6*(a + b/cos(c + d*x))),x)
 
output
tan(c/2 + (d*x)/2)^5/(5*d*(32*a - 32*b)) + (tan(c/2 + (d*x)/2)^3*(4/(3*(32 
*a - 32*b)) + (32*a + 32*b)/(3*(32*a - 32*b)^2)))/d + (tan(c/2 + (d*x)/2)* 
(5/(32*a - 32*b) + ((4/(32*a - 32*b) + (32*a + 32*b)/(32*a - 32*b)^2)*(32* 
a + 32*b))/(32*a - 32*b)))/d - ((3*a*b^2 - 3*a^2*b + a^3 - b^3)/(5*(a + b) 
) - (2*tan(c/2 + (d*x)/2)^4*(a*b^4 + 11*a^4*b - 5*a^5 + b^5 - 4*a^2*b^3 - 
4*a^3*b^2))/(a + b)^3 + (tan(c/2 + (d*x)/2)^2*(4*a*b^3 - 12*a^3*b + 5*a^4 
- 3*b^4 + 6*a^2*b^2))/(3*(a + b)^2))/(d*tan(c/2 + (d*x)/2)^5*(96*a*b^2 - 9 
6*a^2*b + 32*a^3 - 32*b^3)) - (2*a^5*b*atanh((tan(c/2 + (d*x)/2)*(a^6 - b^ 
6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(5/2))))/(d*(a + b)^(7/ 
2)*(a - b)^(7/2))