Integrand size = 29, antiderivative size = 263 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {437 \log (1-\sin (c+d x))}{512 a d}+\frac {949 \log (1+\sin (c+d x))}{512 a d}-\frac {\sin (c+d x)}{a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {a^2}{24 d (a-a \sin (c+d x))^3}-\frac {203}{256 d (a-a \sin (c+d x))}-\frac {17 a^3}{256 d (a+a \sin (c+d x))^4}+\frac {125 a^2}{384 d (a+a \sin (c+d x))^3}+\frac {5}{2 d (a+a \sin (c+d x))}+\frac {a^9}{160 d \left (a^2+a^2 \sin (c+d x)\right )^5}+\frac {109 a^9}{512 d \left (a^5-a^5 \sin (c+d x)\right )^2}-\frac {515 a^9}{512 d \left (a^5+a^5 \sin (c+d x)\right )^2} \] Output:
-437/512*ln(1-sin(d*x+c))/a/d+949/512*ln(1+sin(d*x+c))/a/d-sin(d*x+c)/a/d+ 1/256*a^3/d/(a-a*sin(d*x+c))^4-1/24*a^2/d/(a-a*sin(d*x+c))^3-203/256/d/(a- a*sin(d*x+c))-17/256*a^3/d/(a+a*sin(d*x+c))^4+125/384*a^2/d/(a+a*sin(d*x+c ))^3+5/2/d/(a+a*sin(d*x+c))+1/160*a^9/d/(a^2+a^2*sin(d*x+c))^5+109/512*a^9 /d/(a^5-a^5*sin(d*x+c))^2-515/512*a^9/d/(a^5+a^5*sin(d*x+c))^2
Time = 6.18 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6555 \log (1-\sin (c+d x))-14235 \log (1+\sin (c+d x))-\frac {30}{(1-\sin (c+d x))^4}+\frac {320}{(1-\sin (c+d x))^3}-\frac {1635}{(1-\sin (c+d x))^2}+\frac {6090}{1-\sin (c+d x)}+7680 \sin (c+d x)-\frac {48}{(1+\sin (c+d x))^5}+\frac {510}{(1+\sin (c+d x))^4}-\frac {2500}{(1+\sin (c+d x))^3}+\frac {7725}{(1+\sin (c+d x))^2}-\frac {19200}{1+\sin (c+d x)}}{7680 a d} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
Output:
-1/7680*(6555*Log[1 - Sin[c + d*x]] - 14235*Log[1 + Sin[c + d*x]] - 30/(1 - Sin[c + d*x])^4 + 320/(1 - Sin[c + d*x])^3 - 1635/(1 - Sin[c + d*x])^2 + 6090/(1 - Sin[c + d*x]) + 7680*Sin[c + d*x] - 48/(1 + Sin[c + d*x])^5 + 5 10/(1 + Sin[c + d*x])^4 - 2500/(1 + Sin[c + d*x])^3 + 7725/(1 + Sin[c + d* x])^2 - 19200/(1 + Sin[c + d*x]))/(a*d)
Time = 0.49 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^{11}}{\cos (c+d x)^9 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^9 \int \frac {\sin ^{11}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^{11} \sin ^{11}(c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a^6}{32 (\sin (c+d x) a+a)^6}+\frac {a^5}{64 (a-a \sin (c+d x))^5}+\frac {17 a^5}{64 (\sin (c+d x) a+a)^5}-\frac {a^4}{8 (a-a \sin (c+d x))^4}-\frac {125 a^4}{128 (\sin (c+d x) a+a)^4}+\frac {109 a^3}{256 (a-a \sin (c+d x))^3}+\frac {515 a^3}{256 (\sin (c+d x) a+a)^3}-\frac {203 a^2}{256 (a-a \sin (c+d x))^2}-\frac {5 a^2}{2 (\sin (c+d x) a+a)^2}+\frac {437 a}{512 (a-a \sin (c+d x))}+\frac {949 a}{512 (\sin (c+d x) a+a)}-1\right )d(a \sin (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^6}{160 (a \sin (c+d x)+a)^5}+\frac {a^5}{256 (a-a \sin (c+d x))^4}-\frac {17 a^5}{256 (a \sin (c+d x)+a)^4}-\frac {a^4}{24 (a-a \sin (c+d x))^3}+\frac {125 a^4}{384 (a \sin (c+d x)+a)^3}+\frac {109 a^3}{512 (a-a \sin (c+d x))^2}-\frac {515 a^3}{512 (a \sin (c+d x)+a)^2}-\frac {203 a^2}{256 (a-a \sin (c+d x))}+\frac {5 a^2}{2 (a \sin (c+d x)+a)}-a \sin (c+d x)-\frac {437}{512} a \log (a-a \sin (c+d x))+\frac {949}{512} a \log (a \sin (c+d x)+a)}{a^2 d}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]
Output:
((-437*a*Log[a - a*Sin[c + d*x]])/512 + (949*a*Log[a + a*Sin[c + d*x]])/51 2 - a*Sin[c + d*x] + a^5/(256*(a - a*Sin[c + d*x])^4) - a^4/(24*(a - a*Sin [c + d*x])^3) + (109*a^3)/(512*(a - a*Sin[c + d*x])^2) - (203*a^2)/(256*(a - a*Sin[c + d*x])) + a^6/(160*(a + a*Sin[c + d*x])^5) - (17*a^5)/(256*(a + a*Sin[c + d*x])^4) + (125*a^4)/(384*(a + a*Sin[c + d*x])^3) - (515*a^3)/ (512*(a + a*Sin[c + d*x])^2) + (5*a^2)/(2*(a + a*Sin[c + d*x])))/(a^2*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 = 6.42 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.56
| method | result | size |
| derivativedivides | \(\frac {-\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{24 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {109}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {203}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {437 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {17}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {125}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {515}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{2 \left (1+\sin \left (d x +c \right )\right )}+\frac {949 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(147\) |
| default | \(\frac {-\sin \left (d x +c \right )+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{24 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {109}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {203}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {437 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {17}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {125}{384 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {515}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{2 \left (1+\sin \left (d x +c \right )\right )}+\frac {949 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(147\) |
| risch | \(-\frac {i x}{a}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d a}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}-\frac {2 i c}{a d}+\frac {i \left (12645 \,{\mathrm e}^{i \left (d x +c \right )}-16594 i {\mathrm e}^{8 i \left (d x +c \right )}+149124 \,{\mathrm e}^{13 i \left (d x +c \right )}+45900 \,{\mathrm e}^{15 i \left (d x +c \right )}+12645 \,{\mathrm e}^{17 i \left (d x +c \right )}+207028 \,{\mathrm e}^{11 i \left (d x +c \right )}+292910 \,{\mathrm e}^{9 i \left (d x +c \right )}+45900 \,{\mathrm e}^{3 i \left (d x +c \right )}+21090 i {\mathrm e}^{14 i \left (d x +c \right )}-37738 i {\mathrm e}^{6 i \left (d x +c \right )}+37738 i {\mathrm e}^{12 i \left (d x +c \right )}+6090 i {\mathrm e}^{16 i \left (d x +c \right )}-6090 i {\mathrm e}^{2 i \left (d x +c \right )}+16594 i {\mathrm e}^{10 i \left (d x +c \right )}-21090 i {\mathrm e}^{4 i \left (d x +c \right )}+207028 \,{\mathrm e}^{7 i \left (d x +c \right )}+149124 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{1920 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {437 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 a d}+\frac {949 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) | \(330\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^9/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-sin(d*x+c)+1/256/(sin(d*x+c)-1)^4+1/24/(sin(d*x+c)-1)^3+109/512/(s in(d*x+c)-1)^2+203/256/(sin(d*x+c)-1)-437/512*ln(sin(d*x+c)-1)+1/160/(1+si n(d*x+c))^5-17/256/(1+sin(d*x+c))^4+125/384/(1+sin(d*x+c))^3-515/512/(1+si n(d*x+c))^2+5/2/(1+sin(d*x+c))+949/512*ln(1+sin(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {7680 \, \cos \left (d x + c\right )^{10} + 17610 \, \cos \left (d x + c\right )^{8} - 27630 \, \cos \left (d x + c\right )^{6} + 15828 \, \cos \left (d x + c\right )^{4} - 5584 \, \cos \left (d x + c\right )^{2} + 14235 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6555 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3840 \, \cos \left (d x + c\right )^{8} + 3045 \, \cos \left (d x + c\right )^{6} - 1170 \, \cos \left (d x + c\right )^{4} + 344 \, \cos \left (d x + c\right )^{2} - 48\right )} \sin \left (d x + c\right ) + 864}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/7680*(7680*cos(d*x + c)^10 + 17610*cos(d*x + c)^8 - 27630*cos(d*x + c)^6 + 15828*cos(d*x + c)^4 - 5584*cos(d*x + c)^2 + 14235*(cos(d*x + c)^8*sin( d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) - 6555*(cos(d*x + c)^8*si n(d*x + c) + cos(d*x + c)^8)*log(-sin(d*x + c) + 1) - 2*(3840*cos(d*x + c) ^8 + 3045*cos(d*x + c)^6 - 1170*cos(d*x + c)^4 + 344*cos(d*x + c)^2 - 48)* sin(d*x + c) + 864)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)
Timed out. \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**9/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (12645 \, \sin \left (d x + c\right )^{8} + 3045 \, \sin \left (d x + c\right )^{7} - 36765 \, \sin \left (d x + c\right )^{6} - 7965 \, \sin \left (d x + c\right )^{5} + 42339 \, \sin \left (d x + c\right )^{4} + 7139 \, \sin \left (d x + c\right )^{3} - 22171 \, \sin \left (d x + c\right )^{2} - 2171 \, \sin \left (d x + c\right ) + 4384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {14235 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {6555 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {7680 \, \sin \left (d x + c\right )}{a}}{7680 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/7680*(2*(12645*sin(d*x + c)^8 + 3045*sin(d*x + c)^7 - 36765*sin(d*x + c) ^6 - 7965*sin(d*x + c)^5 + 42339*sin(d*x + c)^4 + 7139*sin(d*x + c)^3 - 22 171*sin(d*x + c)^2 - 2171*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) + 14235*log(sin(d*x + c) + 1)/a - 6555*log(sin(d*x + c) - 1)/a - 7680*sin(d*x + c)/a)/d
Time = 0.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {949 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a d} - \frac {437 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a d} - \frac {\sin \left (d x + c\right )}{a d} + \frac {12645 \, \sin \left (d x + c\right )^{8} + 3045 \, \sin \left (d x + c\right )^{7} - 36765 \, \sin \left (d x + c\right )^{6} - 7965 \, \sin \left (d x + c\right )^{5} + 42339 \, \sin \left (d x + c\right )^{4} + 7139 \, \sin \left (d x + c\right )^{3} - 22171 \, \sin \left (d x + c\right )^{2} - 2171 \, \sin \left (d x + c\right ) + 4384}{3840 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
949/512*log(abs(sin(d*x + c) + 1))/(a*d) - 437/512*log(abs(sin(d*x + c) - 1))/(a*d) - sin(d*x + c)/(a*d) + 1/3840*(12645*sin(d*x + c)^8 + 3045*sin(d *x + c)^7 - 36765*sin(d*x + c)^6 - 7965*sin(d*x + c)^5 + 42339*sin(d*x + c )^4 + 7139*sin(d*x + c)^3 - 22171*sin(d*x + c)^2 - 2171*sin(d*x + c) + 438 4)/(a*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^4)
Time = 34.35 (sec) , antiderivative size = 595, normalized size of antiderivative = 2.26 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^9)/(a + a*sin(c + d*x)),x)
Output:
(949*log(tan(c/2 + (d*x)/2) + 1))/(256*a*d) - (437*log(tan(c/2 + (d*x)/2) - 1))/(256*a*d) - ((693*tan(c/2 + (d*x)/2))/128 + (565*tan(c/2 + (d*x)/2)^ 2)/64 - (4439*tan(c/2 + (d*x)/2)^3)/128 - (963*tan(c/2 + (d*x)/2)^4)/16 + (7091*tan(c/2 + (d*x)/2)^5)/80 + (40031*tan(c/2 + (d*x)/2)^6)/240 - (12829 *tan(c/2 + (d*x)/2)^7)/120 - (17969*tan(c/2 + (d*x)/2)^8)/80 + (39491*tan( c/2 + (d*x)/2)^9)/960 + (49513*tan(c/2 + (d*x)/2)^10)/480 + (39491*tan(c/2 + (d*x)/2)^11)/960 - (17969*tan(c/2 + (d*x)/2)^12)/80 - (12829*tan(c/2 + (d*x)/2)^13)/120 + (40031*tan(c/2 + (d*x)/2)^14)/240 + (7091*tan(c/2 + (d* x)/2)^15)/80 - (963*tan(c/2 + (d*x)/2)^16)/16 - (4439*tan(c/2 + (d*x)/2)^1 7)/128 + (565*tan(c/2 + (d*x)/2)^18)/64 + (693*tan(c/2 + (d*x)/2)^19)/128) /(d*(a + 2*a*tan(c/2 + (d*x)/2) - 6*a*tan(c/2 + (d*x)/2)^2 - 14*a*tan(c/2 + (d*x)/2)^3 + 13*a*tan(c/2 + (d*x)/2)^4 + 40*a*tan(c/2 + (d*x)/2)^5 - 8*a *tan(c/2 + (d*x)/2)^6 - 56*a*tan(c/2 + (d*x)/2)^7 - 14*a*tan(c/2 + (d*x)/2 )^8 + 28*a*tan(c/2 + (d*x)/2)^9 + 28*a*tan(c/2 + (d*x)/2)^10 + 28*a*tan(c/ 2 + (d*x)/2)^11 - 14*a*tan(c/2 + (d*x)/2)^12 - 56*a*tan(c/2 + (d*x)/2)^13 - 8*a*tan(c/2 + (d*x)/2)^14 + 40*a*tan(c/2 + (d*x)/2)^15 + 13*a*tan(c/2 + (d*x)/2)^16 - 14*a*tan(c/2 + (d*x)/2)^17 - 6*a*tan(c/2 + (d*x)/2)^18 + 2*a *tan(c/2 + (d*x)/2)^19 + a*tan(c/2 + (d*x)/2)^20)) - log(tan(c/2 + (d*x)/2 )^2 + 1)/(a*d)
Time = 0.24 (sec) , antiderivative size = 2608, normalized size of antiderivative = 9.92 \[ \int \frac {\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^9/(a+a*sin(d*x+c)),x)
Output:
( - 480*cos(c + d*x)*sin(c + d*x)**9*tan(c + d*x)**7 + 1584*cos(c + d*x)*s in(c + d*x)**9*tan(c + d*x)**5 - 5052*cos(c + d*x)*sin(c + d*x)**9*tan(c + d*x)**3 + 32154*cos(c + d*x)*sin(c + d*x)**9*tan(c + d*x) - 480*cos(c + d *x)*sin(c + d*x)**8*tan(c + d*x)**7 + 1584*cos(c + d*x)*sin(c + d*x)**8*ta n(c + d*x)**5 - 5052*cos(c + d*x)*sin(c + d*x)**8*tan(c + d*x)**3 + 32154* cos(c + d*x)*sin(c + d*x)**8*tan(c + d*x) + 1920*cos(c + d*x)*sin(c + d*x) **7*tan(c + d*x)**7 - 6336*cos(c + d*x)*sin(c + d*x)**7*tan(c + d*x)**5 + 20208*cos(c + d*x)*sin(c + d*x)**7*tan(c + d*x)**3 - 128616*cos(c + d*x)*s in(c + d*x)**7*tan(c + d*x) + 1920*cos(c + d*x)*sin(c + d*x)**6*tan(c + d* x)**7 - 6336*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)**5 + 20208*cos(c + d*x)*sin(c + d*x)**6*tan(c + d*x)**3 - 128616*cos(c + d*x)*sin(c + d*x)**6 *tan(c + d*x) - 2880*cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x)**7 + 9504*c os(c + d*x)*sin(c + d*x)**5*tan(c + d*x)**5 - 30312*cos(c + d*x)*sin(c + d *x)**5*tan(c + d*x)**3 + 192924*cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x) - 2880*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**7 + 9504*cos(c + d*x)*si n(c + d*x)**4*tan(c + d*x)**5 - 30312*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**3 + 192924*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x) + 1920*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x)**7 - 6336*cos(c + d*x)*sin(c + d*x)**3* tan(c + d*x)**5 + 20208*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x)**3 - 128 616*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x) + 1920*cos(c + d*x)*sin(c...