Integrand size = 27, antiderivative size = 249 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {193 \log (1-\sin (c+d x))}{512 a d}-\frac {319 \log (1+\sin (c+d x))}{512 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {7 a^2}{192 d (a-a \sin (c+d x))^3}-\frac {61}{128 d (a-a \sin (c+d x))}+\frac {15 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {95 a^2}{384 d (a+a \sin (c+d x))^3}-\frac {315}{256 d (a+a \sin (c+d x))}-\frac {a^9}{160 d \left (a^2+a^2 \sin (c+d x)\right )^5}+\frac {81 a^9}{512 d \left (a^5-a^5 \sin (c+d x)\right )^2}+\frac {325 a^9}{512 d \left (a^5+a^5 \sin (c+d x)\right )^2} \] Output:
-193/512*ln(1-sin(d*x+c))/a/d-319/512*ln(1+sin(d*x+c))/a/d+1/256*a^3/d/(a- a*sin(d*x+c))^4-7/192*a^2/d/(a-a*sin(d*x+c))^3-61/128/d/(a-a*sin(d*x+c))+1 5/256*a^3/d/(a+a*sin(d*x+c))^4-95/384*a^2/d/(a+a*sin(d*x+c))^3-315/256/d/( a+a*sin(d*x+c))-1/160*a^9/d/(a^2+a^2*sin(d*x+c))^5+81/512*a^9/d/(a^5-a^5*s in(d*x+c))^2+325/512*a^9/d/(a^5+a^5*sin(d*x+c))^2
Time = 5.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.55 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2895 \log (1-\sin (c+d x))+4785 \log (1+\sin (c+d x))+\frac {2 \left (4384+3439 \sin (c+d x)-16561 \sin ^2(c+d x)-12151 \sin ^3(c+d x)+23049 \sin ^4(c+d x)+14985 \sin ^5(c+d x)-13815 \sin ^6(c+d x)-6705 \sin ^7(c+d x)+2895 \sin ^8(c+d x)\right )}{(-1+\sin (c+d x))^4 (1+\sin (c+d x))^5}}{7680 a d} \] Input:
Integrate[(Sin[c + d*x]*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
Output:
-1/7680*(2895*Log[1 - Sin[c + d*x]] + 4785*Log[1 + Sin[c + d*x]] + (2*(438 4 + 3439*Sin[c + d*x] - 16561*Sin[c + d*x]^2 - 12151*Sin[c + d*x]^3 + 2304 9*Sin[c + d*x]^4 + 14985*Sin[c + d*x]^5 - 13815*Sin[c + d*x]^6 - 6705*Sin[ c + d*x]^7 + 2895*Sin[c + d*x]^8))/((-1 + Sin[c + d*x])^4*(1 + Sin[c + d*x ])^5))/(a*d)
Time = 0.45 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.
| \(\displaystyle \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^{10}}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^9 \int \frac {\sin ^{10}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^{10} \sin ^{10}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {a^5}{32 (\sin (c+d x) a+a)^6}+\frac {a^4}{64 (a-a \sin (c+d x))^5}-\frac {15 a^4}{64 (\sin (c+d x) a+a)^5}-\frac {7 a^3}{64 (a-a \sin (c+d x))^4}+\frac {95 a^3}{128 (\sin (c+d x) a+a)^4}+\frac {81 a^2}{256 (a-a \sin (c+d x))^3}-\frac {325 a^2}{256 (\sin (c+d x) a+a)^3}-\frac {61 a}{128 (a-a \sin (c+d x))^2}+\frac {315 a}{256 (\sin (c+d x) a+a)^2}+\frac {193}{512 (a-a \sin (c+d x))}-\frac {319}{512 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^5}{160 (a \sin (c+d x)+a)^5}+\frac {a^4}{256 (a-a \sin (c+d x))^4}+\frac {15 a^4}{256 (a \sin (c+d x)+a)^4}-\frac {7 a^3}{192 (a-a \sin (c+d x))^3}-\frac {95 a^3}{384 (a \sin (c+d x)+a)^3}+\frac {81 a^2}{512 (a-a \sin (c+d x))^2}+\frac {325 a^2}{512 (a \sin (c+d x)+a)^2}-\frac {61 a}{128 (a-a \sin (c+d x))}-\frac {315 a}{256 (a \sin (c+d x)+a)}-\frac {193}{512} \log (a-a \sin (c+d x))-\frac {319}{512} \log (a \sin (c+d x)+a)}{a d}\) |
Input:
Int[(Sin[c + d*x]*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
Output:
((-193*Log[a - a*Sin[c + d*x]])/512 - (319*Log[a + a*Sin[c + d*x]])/512 + a^4/(256*(a - a*Sin[c + d*x])^4) - (7*a^3)/(192*(a - a*Sin[c + d*x])^3) + (81*a^2)/(512*(a - a*Sin[c + d*x])^2) - (61*a)/(128*(a - a*Sin[c + d*x])) - a^5/(160*(a + a*Sin[c + d*x])^5) + (15*a^4)/(256*(a + a*Sin[c + d*x])^4) - (95*a^3)/(384*(a + a*Sin[c + d*x])^3) + (325*a^2)/(512*(a + a*Sin[c + d *x])^2) - (315*a)/(256*(a + a*Sin[c + d*x])))/(a*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.56
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {7}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {81}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {61}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {193 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {15}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {95}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {325}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {315}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {319 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(139\) |
| default | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {7}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {81}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {61}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {193 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {15}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {95}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {325}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {315}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {319 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(139\) |
| risch | \(\frac {i x}{a}+\frac {2 i c}{a d}-\frac {i \left (3146 i {\mathrm e}^{8 i \left (d x +c \right )}+2895 \,{\mathrm e}^{17 i \left (d x +c \right )}-26010 i {\mathrm e}^{14 i \left (d x +c \right )}+32100 \,{\mathrm e}^{15 i \left (d x +c \right )}+71042 i {\mathrm e}^{6 i \left (d x +c \right )}+118284 \,{\mathrm e}^{13 i \left (d x +c \right )}-71042 i {\mathrm e}^{12 i \left (d x +c \right )}+251548 \,{\mathrm e}^{11 i \left (d x +c \right )}-13410 i {\mathrm e}^{16 i \left (d x +c \right )}+312650 \,{\mathrm e}^{9 i \left (d x +c \right )}+13410 i {\mathrm e}^{2 i \left (d x +c \right )}+251548 \,{\mathrm e}^{7 i \left (d x +c \right )}-3146 i {\mathrm e}^{10 i \left (d x +c \right )}+118284 \,{\mathrm e}^{5 i \left (d x +c \right )}+26010 i {\mathrm e}^{4 i \left (d x +c \right )}+32100 \,{\mathrm e}^{3 i \left (d x +c \right )}+2895 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{1920 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} d a}-\frac {319 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {193 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}\) | \(294\) |
Input:
int(sin(d*x+c)*tan(d*x+c)^9/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/256/(sin(d*x+c)-1)^4+7/192/(sin(d*x+c)-1)^3+81/512/(sin(d*x+c)-1) ^2+61/128/(sin(d*x+c)-1)-193/512*ln(sin(d*x+c)-1)-1/160/(1+sin(d*x+c))^5+1 5/256/(1+sin(d*x+c))^4-95/384/(1+sin(d*x+c))^3+325/512/(1+sin(d*x+c))^2-31 5/256/(1+sin(d*x+c))-319/512*ln(1+sin(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.75 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5790 \, \cos \left (d x + c\right )^{8} + 4470 \, \cos \left (d x + c\right )^{6} - 2052 \, \cos \left (d x + c\right )^{4} + 656 \, \cos \left (d x + c\right )^{2} + 4785 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2895 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6705 \, \cos \left (d x + c\right )^{6} - 5130 \, \cos \left (d x + c\right )^{4} + 2296 \, \cos \left (d x + c\right )^{2} - 432\right )} \sin \left (d x + c\right ) - 96}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/7680*(5790*cos(d*x + c)^8 + 4470*cos(d*x + c)^6 - 2052*cos(d*x + c)^4 + 656*cos(d*x + c)^2 + 4785*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)* log(sin(d*x + c) + 1) + 2895*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8 )*log(-sin(d*x + c) + 1) + 2*(6705*cos(d*x + c)^6 - 5130*cos(d*x + c)^4 + 2296*cos(d*x + c)^2 - 432)*sin(d*x + c) - 96)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)
\[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \tan ^{9}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)**9/(a+a*sin(d*x+c)),x)
Output:
Integral(sin(c + d*x)*tan(c + d*x)**9/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.86 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (2895 \, \sin \left (d x + c\right )^{8} - 6705 \, \sin \left (d x + c\right )^{7} - 13815 \, \sin \left (d x + c\right )^{6} + 14985 \, \sin \left (d x + c\right )^{5} + 23049 \, \sin \left (d x + c\right )^{4} - 12151 \, \sin \left (d x + c\right )^{3} - 16561 \, \sin \left (d x + c\right )^{2} + 3439 \, \sin \left (d x + c\right ) + 4384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {4785 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {2895 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/7680*(2*(2895*sin(d*x + c)^8 - 6705*sin(d*x + c)^7 - 13815*sin(d*x + c) ^6 + 14985*sin(d*x + c)^5 + 23049*sin(d*x + c)^4 - 12151*sin(d*x + c)^3 - 16561*sin(d*x + c)^2 + 3439*sin(d*x + c) + 4384)/(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) + 4785*log(sin(d*x + c) + 1)/a + 2895*log(sin(d*x + c) - 1)/a) /d
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.58 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {319 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {193 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {2895 \, \sin \left (d x + c\right )^{8} - 6705 \, \sin \left (d x + c\right )^{7} - 13815 \, \sin \left (d x + c\right )^{6} + 14985 \, \sin \left (d x + c\right )^{5} + 23049 \, \sin \left (d x + c\right )^{4} - 12151 \, \sin \left (d x + c\right )^{3} - 16561 \, \sin \left (d x + c\right )^{2} + 3439 \, \sin \left (d x + c\right ) + 4384}{3840 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-319/512*log(abs(sin(d*x + c) + 1))/(a*d) - 193/512*log(abs(sin(d*x + c) - 1))/(a*d) - 1/3840*(2895*sin(d*x + c)^8 - 6705*sin(d*x + c)^7 - 13815*sin (d*x + c)^6 + 14985*sin(d*x + c)^5 + 23049*sin(d*x + c)^4 - 12151*sin(d*x + c)^3 - 16561*sin(d*x + c)^2 + 3439*sin(d*x + c) + 4384)/(a*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^4)
Time = 32.69 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.16 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int((sin(c + d*x)*tan(c + d*x)^9)/(a + a*sin(c + d*x)),x)
Output:
((63*tan(c/2 + (d*x)/2))/128 - (65*tan(c/2 + (d*x)/2)^2)/64 - (233*tan(c/2 + (d*x)/2)^3)/32 + (413*tan(c/2 + (d*x)/2)^4)/64 + (6527*tan(c/2 + (d*x)/ 2)^5)/160 - (14911*tan(c/2 + (d*x)/2)^6)/960 - (59737*tan(c/2 + (d*x)/2)^7 )/480 + (12763*tan(c/2 + (d*x)/2)^8)/960 + (45791*tan(c/2 + (d*x)/2)^9)/19 2 + (12763*tan(c/2 + (d*x)/2)^10)/960 - (59737*tan(c/2 + (d*x)/2)^11)/480 - (14911*tan(c/2 + (d*x)/2)^12)/960 + (6527*tan(c/2 + (d*x)/2)^13)/160 + ( 413*tan(c/2 + (d*x)/2)^14)/64 - (233*tan(c/2 + (d*x)/2)^15)/32 - (65*tan(c /2 + (d*x)/2)^16)/64 + (63*tan(c/2 + (d*x)/2)^17)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 7*a*tan(c/2 + (d*x)/2)^2 - 16*a*tan(c/2 + (d*x)/2)^3 + 20*a* tan(c/2 + (d*x)/2)^4 + 56*a*tan(c/2 + (d*x)/2)^5 - 28*a*tan(c/2 + (d*x)/2) ^6 - 112*a*tan(c/2 + (d*x)/2)^7 + 14*a*tan(c/2 + (d*x)/2)^8 + 140*a*tan(c/ 2 + (d*x)/2)^9 + 14*a*tan(c/2 + (d*x)/2)^10 - 112*a*tan(c/2 + (d*x)/2)^11 - 28*a*tan(c/2 + (d*x)/2)^12 + 56*a*tan(c/2 + (d*x)/2)^13 + 20*a*tan(c/2 + (d*x)/2)^14 - 16*a*tan(c/2 + (d*x)/2)^15 - 7*a*tan(c/2 + (d*x)/2)^16 + 2* a*tan(c/2 + (d*x)/2)^17 + a*tan(c/2 + (d*x)/2)^18)) - (193*log(tan(c/2 + ( d*x)/2) - 1))/(256*a*d) - (319*log(tan(c/2 + (d*x)/2) + 1))/(256*a*d) + lo g(tan(c/2 + (d*x)/2)^2 + 1)/(a*d)
Time = 0.18 (sec) , antiderivative size = 1476, normalized size of antiderivative = 5.93 \[ \int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)*tan(d*x+c)^9/(a+a*sin(d*x+c)),x)
Output:
(1920*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**9 + 1920*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**8 - 7680*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**7 - 768 0*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**6 + 11520*log(tan(c + d*x)**2 + 1 )*sin(c + d*x)**5 + 11520*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4 - 7680* log(tan(c + d*x)**2 + 1)*sin(c + d*x)**3 - 7680*log(tan(c + d*x)**2 + 1)*s in(c + d*x)**2 + 1920*log(tan(c + d*x)**2 + 1)*sin(c + d*x) + 1920*log(tan (c + d*x)**2 + 1) + 945*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**9 + 945*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 - 3780*log(tan((c + d*x)/2) - 1)*s in(c + d*x)**7 - 3780*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 + 5670*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**5 + 5670*log(tan((c + d*x)/2) - 1)*si n(c + d*x)**4 - 3780*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 - 3780*log( tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 945*log(tan((c + d*x)/2) - 1)*sin( c + d*x) + 945*log(tan((c + d*x)/2) - 1) - 945*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**9 - 945*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 + 3780*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**7 + 3780*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**6 - 5670*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 5670*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**4 + 3780*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**3 + 3780*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 945*log(tan ((c + d*x)/2) + 1)*sin(c + d*x) - 945*log(tan((c + d*x)/2) + 1) + 480*sin( c + d*x)**9*tan(c + d*x)**8 - 640*sin(c + d*x)**9*tan(c + d*x)**6 + 960...