3.1.90 \(\int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [90]

3.1.90.1 Optimal result
3.1.90.2 Mathematica [A] (verified)
3.1.90.3 Rubi [A] (verified)
3.1.90.4 Maple [A] (verified)
3.1.90.5 Fricas [A] (verification not implemented)
3.1.90.6 Sympy [F(-1)]
3.1.90.7 Maxima [A] (verification not implemented)
3.1.90.8 Giac [A] (verification not implemented)
3.1.90.9 Mupad [B] (verification not implemented)

3.1.90.1 Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{a^2 d}-\frac {9 \cot ^7(c+d x)}{7 a^2 d}-\frac {7 \cot ^9(c+d x)}{9 a^2 d}-\frac {2 \cot ^{11}(c+d x)}{11 a^2 d}-\frac {2 \csc ^9(c+d x)}{9 a^2 d}+\frac {2 \csc ^{11}(c+d x)}{11 a^2 d} \]

output
-1/3*cot(d*x+c)^3/a^2/d-cot(d*x+c)^5/a^2/d-9/7*cot(d*x+c)^7/a^2/d-7/9*cot( 
d*x+c)^9/a^2/d-2/11*cot(d*x+c)^11/a^2/d-2/9*csc(d*x+c)^9/a^2/d+2/11*csc(d* 
x+c)^11/a^2/d
 
3.1.90.2 Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.86 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\csc (c) \csc ^7(c+d x) \sec ^2(c+d x) (630784 \sin (c)-1103872 \sin (d x)-218834 \sin (c+d x)-79576 \sin (2 (c+d x))+119364 \sin (3 (c+d x))+79576 \sin (4 (c+d x))-28420 \sin (5 (c+d x))-34104 \sin (6 (c+d x))-1421 \sin (7 (c+d x))+5684 \sin (8 (c+d x))+1421 \sin (9 (c+d x))+1419264 \sin (2 c+d x)+114688 \sin (c+2 d x)-172032 \sin (2 c+3 d x)-114688 \sin (3 c+4 d x)+40960 \sin (4 c+5 d x)+49152 \sin (5 c+6 d x)+2048 \sin (6 c+7 d x)-8192 \sin (7 c+8 d x)-2048 \sin (8 c+9 d x))}{22708224 a^2 d (1+\sec (c+d x))^2} \]

input
Integrate[Csc[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]
 
output
-1/22708224*(Csc[c]*Csc[c + d*x]^7*Sec[c + d*x]^2*(630784*Sin[c] - 1103872 
*Sin[d*x] - 218834*Sin[c + d*x] - 79576*Sin[2*(c + d*x)] + 119364*Sin[3*(c 
 + d*x)] + 79576*Sin[4*(c + d*x)] - 28420*Sin[5*(c + d*x)] - 34104*Sin[6*( 
c + d*x)] - 1421*Sin[7*(c + d*x)] + 5684*Sin[8*(c + d*x)] + 1421*Sin[9*(c 
+ d*x)] + 1419264*Sin[2*c + d*x] + 114688*Sin[c + 2*d*x] - 172032*Sin[2*c 
+ 3*d*x] - 114688*Sin[3*c + 4*d*x] + 40960*Sin[4*c + 5*d*x] + 49152*Sin[5* 
c + 6*d*x] + 2048*Sin[6*c + 7*d*x] - 8192*Sin[7*c + 8*d*x] - 2048*Sin[8*c 
+ 9*d*x]))/(a^2*d*(1 + Sec[c + d*x])^2)
 
3.1.90.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4360, 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.

\(\displaystyle \int \frac {\csc ^8(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^8 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^8 \left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}dx\)

\(\Big \downarrow \) 3354

\(\displaystyle \frac {\int (a-a \cos (c+d x))^2 \cot ^2(c+d x) \csc ^{10}(c+d x)dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{12}}dx}{a^4}\)

\(\Big \downarrow \) 3352

\(\displaystyle \frac {\int \left (a^2 \cot ^2(c+d x) \csc ^{10}(c+d x)-2 a^2 \cot ^3(c+d x) \csc ^9(c+d x)+a^2 \cot ^4(c+d x) \csc ^8(c+d x)\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 a^2 \cot ^{11}(c+d x)}{11 d}-\frac {7 a^2 \cot ^9(c+d x)}{9 d}-\frac {9 a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^5(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {2 a^2 \csc ^{11}(c+d x)}{11 d}-\frac {2 a^2 \csc ^9(c+d x)}{9 d}}{a^4}\)

input
Int[Csc[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]
 
output
(-1/3*(a^2*Cot[c + d*x]^3)/d - (a^2*Cot[c + d*x]^5)/d - (9*a^2*Cot[c + d*x 
]^7)/(7*d) - (7*a^2*Cot[c + d*x]^9)/(9*d) - (2*a^2*Cot[c + d*x]^11)/(11*d) 
 - (2*a^2*Csc[c + d*x]^9)/(9*d) + (2*a^2*Csc[c + d*x]^11)/(11*d))/a^4
 

3.1.90.3.1 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]
 

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]
 
3.1.90.4 Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90

method result size
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{11}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {8}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{512 d \,a^{2}}\) \(112\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{11}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {8}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{512 d \,a^{2}}\) \(112\)
parallelrisch \(\frac {63 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}+385 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+792 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-99 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-693 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-3234 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-1848 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-9702 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{354816 a^{2} d}\) \(112\)
risch \(\frac {32 i \left (693 \,{\mathrm e}^{10 i \left (d x +c \right )}+308 \,{\mathrm e}^{9 i \left (d x +c \right )}+539 \,{\mathrm e}^{8 i \left (d x +c \right )}-56 \,{\mathrm e}^{7 i \left (d x +c \right )}+84 \,{\mathrm e}^{6 i \left (d x +c \right )}+56 \,{\mathrm e}^{5 i \left (d x +c \right )}-20 \,{\mathrm e}^{4 i \left (d x +c \right )}-24 \,{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{693 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{11} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{7}}\) \(148\)

input
int(csc(d*x+c)^8/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/512/d/a^2*(1/11*tan(1/2*d*x+1/2*c)^11+5/9*tan(1/2*d*x+1/2*c)^9+8/7*tan(1 
/2*d*x+1/2*c)^7-14/3*tan(1/2*d*x+1/2*c)^3-14*tan(1/2*d*x+1/2*c)-1/7/tan(1/ 
2*d*x+1/2*c)^7-1/tan(1/2*d*x+1/2*c)^5-8/3/tan(1/2*d*x+1/2*c)^3)
 
3.1.90.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.63 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {16 \, \cos \left (d x + c\right )^{9} + 32 \, \cos \left (d x + c\right )^{8} - 40 \, \cos \left (d x + c\right )^{7} - 112 \, \cos \left (d x + c\right )^{6} + 14 \, \cos \left (d x + c\right )^{5} + 140 \, \cos \left (d x + c\right )^{4} + 35 \, \cos \left (d x + c\right )^{3} - 70 \, \cos \left (d x + c\right )^{2} + 56 \, \cos \left (d x + c\right ) + 28}{693 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} + 2 \, a^{2} d \cos \left (d x + c\right )^{7} - 2 \, a^{2} d \cos \left (d x + c\right )^{6} - 6 \, a^{2} d \cos \left (d x + c\right )^{5} + 6 \, a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )} \]

input
integrate(csc(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 
output
1/693*(16*cos(d*x + c)^9 + 32*cos(d*x + c)^8 - 40*cos(d*x + c)^7 - 112*cos 
(d*x + c)^6 + 14*cos(d*x + c)^5 + 140*cos(d*x + c)^4 + 35*cos(d*x + c)^3 - 
 70*cos(d*x + c)^2 + 56*cos(d*x + c) + 28)/((a^2*d*cos(d*x + c)^8 + 2*a^2* 
d*cos(d*x + c)^7 - 2*a^2*d*cos(d*x + c)^6 - 6*a^2*d*cos(d*x + c)^5 + 6*a^2 
*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 - 2*a^2*d*cos(d*x + c) - a^2*d) 
*sin(d*x + c))
 
3.1.90.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

input
integrate(csc(d*x+c)**8/(a+a*sec(d*x+c))**2,x)
 
output
Timed out
 
3.1.90.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.39 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {9702 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3234 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {792 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {385 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {63 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a^{2}} + \frac {33 \, {\left (\frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {56 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{7}}{a^{2} \sin \left (d x + c\right )^{7}}}{354816 \, d} \]

input
integrate(csc(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 
output
-1/354816*((9702*sin(d*x + c)/(cos(d*x + c) + 1) + 3234*sin(d*x + c)^3/(co 
s(d*x + c) + 1)^3 - 792*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 385*sin(d*x 
+ c)^9/(cos(d*x + c) + 1)^9 - 63*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)/a^ 
2 + 33*(21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 56*sin(d*x + c)^4/(cos(d* 
x + c) + 1)^4 + 3)*(cos(d*x + c) + 1)^7/(a^2*sin(d*x + c)^7))/d
 
3.1.90.8 Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {33 \, {\left (56 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}} - \frac {63 \, a^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 385 \, a^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 792 \, a^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3234 \, a^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9702 \, a^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{22}}}{354816 \, d} \]

input
integrate(csc(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 
output
-1/354816*(33*(56*tan(1/2*d*x + 1/2*c)^4 + 21*tan(1/2*d*x + 1/2*c)^2 + 3)/ 
(a^2*tan(1/2*d*x + 1/2*c)^7) - (63*a^20*tan(1/2*d*x + 1/2*c)^11 + 385*a^20 
*tan(1/2*d*x + 1/2*c)^9 + 792*a^20*tan(1/2*d*x + 1/2*c)^7 - 3234*a^20*tan( 
1/2*d*x + 1/2*c)^3 - 9702*a^20*tan(1/2*d*x + 1/2*c))/a^22)/d
 
3.1.90.9 Mupad [B] (verification not implemented)

Time = 14.01 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {99\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+693\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1848\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+9702\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3234\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-792\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-385\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-63\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}}{354816\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

input
int(1/(sin(c + d*x)^8*(a + a/cos(c + d*x))^2),x)
 
output
-(99*cos(c/2 + (d*x)/2)^18 - 63*sin(c/2 + (d*x)/2)^18 - 385*cos(c/2 + (d*x 
)/2)^2*sin(c/2 + (d*x)/2)^16 - 792*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2) 
^14 + 3234*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^10 + 9702*cos(c/2 + (d* 
x)/2)^10*sin(c/2 + (d*x)/2)^8 + 1848*cos(c/2 + (d*x)/2)^14*sin(c/2 + (d*x) 
/2)^4 + 693*cos(c/2 + (d*x)/2)^16*sin(c/2 + (d*x)/2)^2)/(354816*a^2*d*cos( 
c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2)^7)