Integrand size = 21, antiderivative size = 109 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cot ^3(c+d x)}{3 a d}+\frac {4 \cot ^5(c+d x)}{5 a d}+\frac {6 \cot ^7(c+d x)}{7 a d}+\frac {4 \cot ^9(c+d x)}{9 a d}+\frac {\cot ^{11}(c+d x)}{11 a d}-\frac {\csc ^{11}(c+d x)}{11 a d} \]
1/3*cot(d*x+c)^3/a/d+4/5*cot(d*x+c)^5/a/d+6/7*cot(d*x+c)^7/a/d+4/9*cot(d*x +c)^9/a/d+1/11*cot(d*x+c)^11/a/d-1/11*csc(d*x+c)^11/a/d
Leaf count is larger than twice the leaf count of optimal. \(242\) vs. \(2(109)=218\).
Time = 3.18 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.22 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\csc (c) \csc ^9(c+d x) \sec (c+d x) (-45416448 \sin (c)+8257536 \sin (d x)+5000940 \sin (c+d x)+833490 \sin (2 (c+d x))-3333960 \sin (3 (c+d x))-952560 \sin (4 (c+d x))+1428840 \sin (5 (c+d x))+535815 \sin (6 (c+d x))-357210 \sin (7 (c+d x))-158760 \sin (8 (c+d x))+39690 \sin (9 (c+d x))+19845 \sin (10 (c+d x))+1376256 \sin (c+2 d x)-5505024 \sin (2 c+3 d x)-1572864 \sin (3 c+4 d x)+2359296 \sin (4 c+5 d x)+884736 \sin (5 c+6 d x)-589824 \sin (6 c+7 d x)-262144 \sin (7 c+8 d x)+65536 \sin (8 c+9 d x)+32768 \sin (9 c+10 d x))}{454164480 a d (1+\sec (c+d x))} \]
(Csc[c]*Csc[c + d*x]^9*Sec[c + d*x]*(-45416448*Sin[c] + 8257536*Sin[d*x] + 5000940*Sin[c + d*x] + 833490*Sin[2*(c + d*x)] - 3333960*Sin[3*(c + d*x)] - 952560*Sin[4*(c + d*x)] + 1428840*Sin[5*(c + d*x)] + 535815*Sin[6*(c + d*x)] - 357210*Sin[7*(c + d*x)] - 158760*Sin[8*(c + d*x)] + 39690*Sin[9*(c + d*x)] + 19845*Sin[10*(c + d*x)] + 1376256*Sin[c + 2*d*x] - 5505024*Sin[ 2*c + 3*d*x] - 1572864*Sin[3*c + 4*d*x] + 2359296*Sin[4*c + 5*d*x] + 88473 6*Sin[5*c + 6*d*x] - 589824*Sin[6*c + 7*d*x] - 262144*Sin[7*c + 8*d*x] + 6 5536*Sin[8*c + 9*d*x] + 32768*Sin[9*c + 10*d*x]))/(454164480*a*d*(1 + Sec[ c + d*x]))
Time = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81, 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, 3318, 25, 3042, 25, 3086, 15, 3087, 244, 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.
| \(\displaystyle \int \frac {\csc ^{10}(c+d x)}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^{10} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cot (c+d x) \csc ^9(c+d x)}{a (-\cos (c+d x))-a}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cot (c+d x) \csc ^9(c+d x)}{\cos (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cot (c+d x) \csc ^9(c+d x)}{a \cos (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^{10} \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^{10} \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle -\frac {\int \cot ^2(c+d x) \csc ^{10}(c+d x)dx}{a}-\frac {\int -\cot (c+d x) \csc ^{11}(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \cot (c+d x) \csc ^{11}(c+d x)dx}{a}-\frac {\int \cot ^2(c+d x) \csc ^{10}(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^{11} \tan \left (c+d x-\frac {\pi }{2}\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^{10} \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^{11} \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^{10} \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle -\frac {\int \csc ^{10}(c+d x)d\csc (c+d x)}{a d}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^{10} \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle -\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^{10} \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}-\frac {\csc ^{11}(c+d x)}{11 a d}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle -\frac {\int \cot ^2(c+d x) \left (\cot ^2(c+d x)+1\right )^4d(-\cot (c+d x))}{a d}-\frac {\csc ^{11}(c+d x)}{11 a d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle -\frac {\int \left (\cot ^{10}(c+d x)+4 \cot ^8(c+d x)+6 \cot ^6(c+d x)+4 \cot ^4(c+d x)+\cot ^2(c+d x)\right )d(-\cot (c+d x))}{a d}-\frac {\csc ^{11}(c+d x)}{11 a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {1}{11} \cot ^{11}(c+d x)-\frac {4}{9} \cot ^9(c+d x)-\frac {6}{7} \cot ^7(c+d x)-\frac {4}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)}{a d}-\frac {\csc ^{11}(c+d x)}{11 a d}\) |
-((-1/3*Cot[c + d*x]^3 - (4*Cot[c + d*x]^5)/5 - (6*Cot[c + d*x]^7)/7 - (4* Cot[c + d*x]^9)/9 - Cot[c + d*x]^11/11)/(a*d)) - Csc[c + d*x]^11/(11*a*d)
3.1.73.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
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]
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]
Time = 0.77 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.27
| method | result | size |
| parallelrisch | \(\frac {-315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-385 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-3080 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-3960 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-13365 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-18711 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-33264 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-55440 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-48510 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-145530 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{3548160 d a}\) | \(138\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{11}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {1}{9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {8}{7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}-\frac {42}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {27}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}}{1024 d a}\) | \(140\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{11}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {1}{9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {8}{7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}-\frac {42}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {27}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}}{1024 d a}\) | \(140\) |
| risch | \(-\frac {256 i \left (1386 \,{\mathrm e}^{10 i \left (d x +c \right )}+252 \,{\mathrm e}^{9 i \left (d x +c \right )}+42 \,{\mathrm e}^{8 i \left (d x +c \right )}-168 \,{\mathrm e}^{7 i \left (d x +c \right )}-48 \,{\mathrm e}^{6 i \left (d x +c \right )}+72 \,{\mathrm e}^{5 i \left (d x +c \right )}+27 \,{\mathrm e}^{4 i \left (d x +c \right )}-18 \,{\mathrm e}^{3 i \left (d x +c \right )}-8 \,{\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{3465 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{11} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{9}}\) | \(148\) |
1/3548160*(-315*tan(1/2*d*x+1/2*c)^11-385*cot(1/2*d*x+1/2*c)^9-3080*tan(1/ 2*d*x+1/2*c)^9-3960*cot(1/2*d*x+1/2*c)^7-13365*tan(1/2*d*x+1/2*c)^7-18711* cot(1/2*d*x+1/2*c)^5-33264*tan(1/2*d*x+1/2*c)^5-55440*cot(1/2*d*x+1/2*c)^3 -48510*tan(1/2*d*x+1/2*c)^3-145530*cot(1/2*d*x+1/2*c))/d/a
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (97) = 194\).
Time = 0.27 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.01 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {128 \, \cos \left (d x + c\right )^{10} + 128 \, \cos \left (d x + c\right )^{9} - 576 \, \cos \left (d x + c\right )^{8} - 576 \, \cos \left (d x + c\right )^{7} + 1008 \, \cos \left (d x + c\right )^{6} + 1008 \, \cos \left (d x + c\right )^{5} - 840 \, \cos \left (d x + c\right )^{4} - 840 \, \cos \left (d x + c\right )^{3} + 315 \, \cos \left (d x + c\right )^{2} + 315 \, \cos \left (d x + c\right ) + 315}{3465 \, {\left (a d \cos \left (d x + c\right )^{9} + a d \cos \left (d x + c\right )^{8} - 4 \, a d \cos \left (d x + c\right )^{7} - 4 \, a d \cos \left (d x + c\right )^{6} + 6 \, a d \cos \left (d x + c\right )^{5} + 6 \, a d \cos \left (d x + c\right )^{4} - 4 \, a d \cos \left (d x + c\right )^{3} - 4 \, a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )} \]
-1/3465*(128*cos(d*x + c)^10 + 128*cos(d*x + c)^9 - 576*cos(d*x + c)^8 - 5 76*cos(d*x + c)^7 + 1008*cos(d*x + c)^6 + 1008*cos(d*x + c)^5 - 840*cos(d* x + c)^4 - 840*cos(d*x + c)^3 + 315*cos(d*x + c)^2 + 315*cos(d*x + c) + 31 5)/((a*d*cos(d*x + c)^9 + a*d*cos(d*x + c)^8 - 4*a*d*cos(d*x + c)^7 - 4*a* d*cos(d*x + c)^6 + 6*a*d*cos(d*x + c)^5 + 6*a*d*cos(d*x + c)^4 - 4*a*d*cos (d*x + c)^3 - 4*a*d*cos(d*x + c)^2 + a*d*cos(d*x + c) + a*d)*sin(d*x + c))
Timed out. \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (97) = 194\).
Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.98 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {\frac {48510 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {33264 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {13365 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {3080 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {315 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a} + \frac {11 \, {\left (\frac {360 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1701 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {5040 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {13230 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 35\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{9}}{a \sin \left (d x + c\right )^{9}}}{3548160 \, d} \]
-1/3548160*((48510*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 33264*sin(d*x + c )^5/(cos(d*x + c) + 1)^5 + 13365*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 308 0*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 315*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)/a + 11*(360*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1701*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5040*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 132 30*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 35)*(cos(d*x + c) + 1)^9/(a*sin(d *x + c)^9))/d
Time = 0.34 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.48 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {11 \, {\left (13230 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 5040 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1701 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}} + \frac {315 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3080 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 13365 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33264 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48510 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{a^{11}}}{3548160 \, d} \]
-1/3548160*(11*(13230*tan(1/2*d*x + 1/2*c)^8 + 5040*tan(1/2*d*x + 1/2*c)^6 + 1701*tan(1/2*d*x + 1/2*c)^4 + 360*tan(1/2*d*x + 1/2*c)^2 + 35)/(a*tan(1 /2*d*x + 1/2*c)^9) + (315*a^10*tan(1/2*d*x + 1/2*c)^11 + 3080*a^10*tan(1/2 *d*x + 1/2*c)^9 + 13365*a^10*tan(1/2*d*x + 1/2*c)^7 + 33264*a^10*tan(1/2*d *x + 1/2*c)^5 + 48510*a^10*tan(1/2*d*x + 1/2*c)^3)/a^11)/d
Time = 15.91 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int \frac {\csc ^{10}(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {63\,\cos \left (c+d\,x\right )+\frac {21\,\cos \left (2\,c+2\,d\,x\right )}{2}-42\,\cos \left (3\,c+3\,d\,x\right )-12\,\cos \left (4\,c+4\,d\,x\right )+18\,\cos \left (5\,c+5\,d\,x\right )+\frac {27\,\cos \left (6\,c+6\,d\,x\right )}{4}-\frac {9\,\cos \left (7\,c+7\,d\,x\right )}{2}-2\,\cos \left (8\,c+8\,d\,x\right )+\frac {\cos \left (9\,c+9\,d\,x\right )}{2}+\frac {\cos \left (10\,c+10\,d\,x\right )}{4}+\frac {693}{2}}{3548160\,a\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]
-(63*cos(c + d*x) + (21*cos(2*c + 2*d*x))/2 - 42*cos(3*c + 3*d*x) - 12*cos (4*c + 4*d*x) + 18*cos(5*c + 5*d*x) + (27*cos(6*c + 6*d*x))/4 - (9*cos(7*c + 7*d*x))/2 - 2*cos(8*c + 8*d*x) + cos(9*c + 9*d*x)/2 + cos(10*c + 10*d*x )/4 + 693/2)/(3548160*a*d*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2)^9)