Integrand size = 21, antiderivative size = 226 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35}{256 d (a+a \sin (c+d x))}-\frac {a^9}{160 d \left (a^2+a^2 \sin (c+d x)\right )^5}+\frac {21 a^9}{512 d \left (a^5-a^5 \sin (c+d x)\right )^2}-\frac {35 a^9}{512 d \left (a^5+a^5 \sin (c+d x)\right )^2} \] Output:
63/256*arctanh(sin(d*x+c))/a/d+1/256*a^3/d/(a-a*sin(d*x+c))^4+1/64*a^2/d/( a-a*sin(d*x+c))^3+7/64/d/(a-a*sin(d*x+c))-5/256*a^3/d/(a+a*sin(d*x+c))^4-5 /128*a^2/d/(a+a*sin(d*x+c))^3-35/256/d/(a+a*sin(d*x+c))-1/160*a^9/d/(a^2+a ^2*sin(d*x+c))^5+21/512*a^9/d/(a^5-a^5*sin(d*x+c))^2-35/512*a^9/d/(a^5+a^5 *sin(d*x+c))^2
Time = 1.54 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^8(c+d x) \left (-128+315 \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^8 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{10}+837 \sin (c+d x)+837 \sin ^2(c+d x)-1533 \sin ^3(c+d x)-1533 \sin ^4(c+d x)+1155 \sin ^5(c+d x)+1155 \sin ^6(c+d x)-315 \sin ^7(c+d x)-315 \sin ^8(c+d x)\right )}{1280 a d (1+\sin (c+d x))} \] Input:
Integrate[Sec[c + d*x]^9/(a + a*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^8*(-128 + 315*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2])^8*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^10 + 837*Sin[c + d*x ] + 837*Sin[c + d*x]^2 - 1533*Sin[c + d*x]^3 - 1533*Sin[c + d*x]^4 + 1155* Sin[c + d*x]^5 + 1155*Sin[c + d*x]^6 - 315*Sin[c + d*x]^7 - 315*Sin[c + d* x]^8))/(1280*a*d*(1 + Sin[c + d*x]))
Time = 0.39 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 54, 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.
| \(\displaystyle \int \frac {\sec ^9(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^9 \int \frac {1}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^9 \int \left (\frac {7}{64 a^9 (a-a \sin (c+d x))^2}+\frac {35}{256 a^9 (\sin (c+d x) a+a)^2}+\frac {21}{256 a^8 (a-a \sin (c+d x))^3}+\frac {35}{256 a^8 (\sin (c+d x) a+a)^3}+\frac {3}{64 a^7 (a-a \sin (c+d x))^4}+\frac {15}{128 a^7 (\sin (c+d x) a+a)^4}+\frac {1}{64 a^6 (a-a \sin (c+d x))^5}+\frac {5}{64 a^6 (\sin (c+d x) a+a)^5}+\frac {1}{32 a^5 (\sin (c+d x) a+a)^6}+\frac {63}{256 a^9 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^9 \left (\frac {63 \text {arctanh}(\sin (c+d x))}{256 a^{10}}+\frac {7}{64 a^9 (a-a \sin (c+d x))}-\frac {35}{256 a^9 (a \sin (c+d x)+a)}+\frac {21}{512 a^8 (a-a \sin (c+d x))^2}-\frac {35}{512 a^8 (a \sin (c+d x)+a)^2}+\frac {1}{64 a^7 (a-a \sin (c+d x))^3}-\frac {5}{128 a^7 (a \sin (c+d x)+a)^3}+\frac {1}{256 a^6 (a-a \sin (c+d x))^4}-\frac {5}{256 a^6 (a \sin (c+d x)+a)^4}-\frac {1}{160 a^5 (a \sin (c+d x)+a)^5}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^9/(a + a*Sin[c + d*x]),x]
Output:
(a^9*((63*ArcTanh[Sin[c + d*x]])/(256*a^10) + 1/(256*a^6*(a - a*Sin[c + d* x])^4) + 1/(64*a^7*(a - a*Sin[c + d*x])^3) + 21/(512*a^8*(a - a*Sin[c + d* x])^2) + 7/(64*a^9*(a - a*Sin[c + d*x])) - 1/(160*a^5*(a + a*Sin[c + d*x]) ^5) - 5/(256*a^6*(a + a*Sin[c + d*x])^4) - 5/(128*a^7*(a + a*Sin[c + d*x]) ^3) - 35/(512*a^8*(a + a*Sin[c + d*x])^2) - 35/(256*a^9*(a + a*Sin[c + d*x ]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 3.66 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.62
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| default | \(\frac {-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}}{d a}\) | \(139\) |
| risch | \(-\frac {i \left (-30318 i {\mathrm e}^{8 i \left (d x +c \right )}+315 \,{\mathrm e}^{17 i \left (d x +c \right )}+4830 i {\mathrm e}^{14 i \left (d x +c \right )}+2100 \,{\mathrm e}^{15 i \left (d x +c \right )}-16086 i {\mathrm e}^{6 i \left (d x +c \right )}+5628 \,{\mathrm e}^{13 i \left (d x +c \right )}+16086 i {\mathrm e}^{12 i \left (d x +c \right )}+7116 \,{\mathrm e}^{11 i \left (d x +c \right )}+630 i {\mathrm e}^{16 i \left (d x +c \right )}+2450 \,{\mathrm e}^{9 i \left (d x +c \right )}-630 i {\mathrm e}^{2 i \left (d x +c \right )}+7116 \,{\mathrm e}^{7 i \left (d x +c \right )}+30318 i {\mathrm e}^{10 i \left (d x +c \right )}+5628 \,{\mathrm e}^{5 i \left (d x +c \right )}-4830 i {\mathrm e}^{4 i \left (d x +c \right )}+2100 \,{\mathrm e}^{3 i \left (d x +c \right )}+315 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}+\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}\) | \(277\) |
| parallelrisch | \(\frac {\left (-8820 \sin \left (3 d x +3 c \right )-6300 \sin \left (5 d x +5 c \right )-2205 \sin \left (7 d x +7 c \right )-315 \sin \left (9 d x +9 c \right )-35280 \cos \left (2 d x +2 c \right )-17640 \cos \left (4 d x +4 c \right )-5040 \cos \left (6 d x +6 c \right )-630 \cos \left (8 d x +8 c \right )-4410 \sin \left (d x +c \right )-22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (8820 \sin \left (3 d x +3 c \right )+6300 \sin \left (5 d x +5 c \right )+2205 \sin \left (7 d x +7 c \right )+315 \sin \left (9 d x +9 c \right )+35280 \cos \left (2 d x +2 c \right )+17640 \cos \left (4 d x +4 c \right )+5040 \cos \left (6 d x +6 c \right )+630 \cos \left (8 d x +8 c \right )+4410 \sin \left (d x +c \right )+22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+35756 \sin \left (3 d x +3 c \right )+12220 \sin \left (5 d x +5 c \right )+2156 \sin \left (7 d x +7 c \right )+128 \sin \left (9 d x +9 c \right )+104 \cos \left (2 d x +2 c \right )-4088 \cos \left (4 d x +4 c \right )-2152 \cos \left (6 d x +6 c \right )-374 \cos \left (8 d x +8 c \right )+62428 \sin \left (d x +c \right )+6510}{1280 a d \left (\sin \left (9 d x +9 c \right )+7 \sin \left (7 d x +7 c \right )+20 \sin \left (5 d x +5 c \right )+28 \sin \left (3 d x +3 c \right )+14 \sin \left (d x +c \right )+2 \cos \left (8 d x +8 c \right )+16 \cos \left (6 d x +6 c \right )+56 \cos \left (4 d x +4 c \right )+112 \cos \left (2 d x +2 c \right )+70\right )}\) | \(427\) |
Input:
int(sec(d*x+c)^9/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/160/(1+sin(d*x+c))^5-5/256/(1+sin(d*x+c))^4-5/128/(1+sin(d*x+c)) ^3-35/512/(1+sin(d*x+c))^2-35/256/(1+sin(d*x+c))+63/512*ln(1+sin(d*x+c))+1 /256/(sin(d*x+c)-1)^4-1/64/(sin(d*x+c)-1)^3+21/512/(sin(d*x+c)-1)^2-7/64/( sin(d*x+c)-1)-63/512*ln(sin(d*x+c)-1))
Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {630 \, \cos \left (d x + c\right )^{8} - 210 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{4} - 48 \, \cos \left (d x + c\right )^{2} - 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 70 \, \cos \left (d x + c\right )^{4} + 56 \, \cos \left (d x + c\right )^{2} + 48\right )} \sin \left (d x + c\right ) - 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sec(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/2560*(630*cos(d*x + c)^8 - 210*cos(d*x + c)^6 - 84*cos(d*x + c)^4 - 48* cos(d*x + c)^2 - 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(si n(d*x + c) + 1) + 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(- sin(d*x + c) + 1) - 6*(105*cos(d*x + c)^6 + 70*cos(d*x + c)^4 + 56*cos(d*x + c)^2 + 48)*sin(d*x + c) - 32)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*co s(d*x + c)^8)
Timed out. \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**9/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} + 315 \, \sin \left (d x + c\right )^{7} - 1155 \, \sin \left (d x + c\right )^{6} - 1155 \, \sin \left (d x + c\right )^{5} + 1533 \, \sin \left (d x + c\right )^{4} + 1533 \, \sin \left (d x + c\right )^{3} - 837 \, \sin \left (d x + c\right )^{2} - 837 \, \sin \left (d x + c\right ) + 128\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \] Input:
integrate(sec(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2560*(2*(315*sin(d*x + c)^8 + 315*sin(d*x + c)^7 - 1155*sin(d*x + c)^6 - 1155*sin(d*x + c)^5 + 1533*sin(d*x + c)^4 + 1533*sin(d*x + c)^3 - 837*si n(d*x + c)^2 - 837*sin(d*x + c) + 128)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^ 8 - 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin (d*x + c)^4 - 4*a*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 + a*sin(d*x + c) + a ) - 315*log(sin(d*x + c) + 1)/a + 315*log(sin(d*x + c) - 1)/a)/d
Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {63 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {315 \, \sin \left (d x + c\right )^{8} + 315 \, \sin \left (d x + c\right )^{7} - 1155 \, \sin \left (d x + c\right )^{6} - 1155 \, \sin \left (d x + c\right )^{5} + 1533 \, \sin \left (d x + c\right )^{4} + 1533 \, \sin \left (d x + c\right )^{3} - 837 \, \sin \left (d x + c\right )^{2} - 837 \, \sin \left (d x + c\right ) + 128}{1280 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
63/512*log(abs(sin(d*x + c) + 1))/(a*d) - 63/512*log(abs(sin(d*x + c) - 1) )/(a*d) - 1/1280*(315*sin(d*x + c)^8 + 315*sin(d*x + c)^7 - 1155*sin(d*x + c)^6 - 1155*sin(d*x + c)^5 + 1533*sin(d*x + c)^4 + 1533*sin(d*x + c)^3 - 837*sin(d*x + c)^2 - 837*sin(d*x + c) + 128)/(a*d*(sin(d*x + c) + 1)^5*(si n(d*x + c) - 1)^4)
Time = 32.96 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{256\,a\,d}-\frac {\frac {63\,{\sin \left (c+d\,x\right )}^8}{256}+\frac {63\,{\sin \left (c+d\,x\right )}^7}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^6}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^5}{256}+\frac {1533\,{\sin \left (c+d\,x\right )}^4}{1280}+\frac {1533\,{\sin \left (c+d\,x\right )}^3}{1280}-\frac {837\,{\sin \left (c+d\,x\right )}^2}{1280}-\frac {837\,\sin \left (c+d\,x\right )}{1280}+\frac {1}{10}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^9+a\,{\sin \left (c+d\,x\right )}^8-4\,a\,{\sin \left (c+d\,x\right )}^7-4\,a\,{\sin \left (c+d\,x\right )}^6+6\,a\,{\sin \left (c+d\,x\right )}^5+6\,a\,{\sin \left (c+d\,x\right )}^4-4\,a\,{\sin \left (c+d\,x\right )}^3-4\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \] Input:
int(1/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)
Output:
(63*atanh(sin(c + d*x)))/(256*a*d) - ((1533*sin(c + d*x)^3)/1280 - (837*si n(c + d*x)^2)/1280 - (837*sin(c + d*x))/1280 + (1533*sin(c + d*x)^4)/1280 - (231*sin(c + d*x)^5)/256 - (231*sin(c + d*x)^6)/256 + (63*sin(c + d*x)^7 )/256 + (63*sin(c + d*x)^8)/256 + 1/10)/(d*(a + a*sin(c + d*x) - 4*a*sin(c + d*x)^2 - 4*a*sin(c + d*x)^3 + 6*a*sin(c + d*x)^4 + 6*a*sin(c + d*x)^5 - 4*a*sin(c + d*x)^6 - 4*a*sin(c + d*x)^7 + a*sin(c + d*x)^8 + a*sin(c + d* x)^9))
Time = 0.19 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.64 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^9/(a+a*sin(d*x+c)),x)
Output:
( - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**9 - 315*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**8 + 1260*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 + 1260*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 1890*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**5 - 1890*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 1260*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 1260*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 315*l og(tan((c + d*x)/2) - 1) + 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**9 + 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 - 1260*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7 - 1260*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 1 890*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 + 1890*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 1260*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 12 60*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 315*log(tan((c + d*x)/2) + 1) - 837*sin(c + d*x)**9 - 11 52*sin(c + d*x)**8 + 3033*sin(c + d*x)**7 + 4503*sin(c + d*x)**6 - 3867*si n(c + d*x)**5 - 6555*sin(c + d*x)**4 + 1815*sin(c + d*x)**3 + 4185*sin(c + d*x)**2 - 965)/(1280*a*d*(sin(c + d*x)**9 + sin(c + d*x)**8 - 4*sin(c + d *x)**7 - 4*sin(c + d*x)**6 + 6*sin(c + d*x)**5 + 6*sin(c + d*x)**4 - 4*sin (c + d*x)**3 - 4*sin(c + d*x)**2 + sin(c + d*x) + 1))