Integrand size = 19, antiderivative size = 151 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {4 (b \cos (c+d x)-a \sin (c+d x))}{15 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {8 \sin (c+d x)}{15 a \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))} \]
1/5*(-b*cos(d*x+c)+a*sin(d*x+c))/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^5 -4/15*(b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c) )^3+8/15*sin(d*x+c)/a/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c))
Time = 0.51 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=\frac {-10 a b \left (a^2+b^2\right ) \cos (3 (c+d x))+\left (-4 a^3 b+4 a b^3\right ) \cos (5 (c+d x))+10 a^4 \sin (c+d x)+20 a^2 b^2 \sin (c+d x)+10 b^4 \sin (c+d x)+5 a^4 \sin (3 (c+d x))-5 b^4 \sin (3 (c+d x))+a^4 \sin (5 (c+d x))-6 a^2 b^2 \sin (5 (c+d x))+b^4 \sin (5 (c+d x))}{30 a \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^5} \]
(-10*a*b*(a^2 + b^2)*Cos[3*(c + d*x)] + (-4*a^3*b + 4*a*b^3)*Cos[5*(c + d* x)] + 10*a^4*Sin[c + d*x] + 20*a^2*b^2*Sin[c + d*x] + 10*b^4*Sin[c + d*x] + 5*a^4*Sin[3*(c + d*x)] - 5*b^4*Sin[3*(c + d*x)] + a^4*Sin[5*(c + d*x)] - 6*a^2*b^2*Sin[5*(c + d*x)] + b^4*Sin[5*(c + d*x)])/(30*a*(a^2 + b^2)^2*d* (a*Cos[c + d*x] + b*Sin[c + d*x])^5)
Time = 0.45 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3555, 3042, 3555, 3042, 3554}
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.
| \(\displaystyle \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6}dx\) |
| \(\Big \downarrow \) 3555 |
| \(\displaystyle \frac {4 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^4}dx}{5 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^4}dx}{5 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5}\) |
| \(\Big \downarrow \) 3555 |
| \(\displaystyle \frac {4 \left (\frac {2 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^2}dx}{3 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3}\right )}{5 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {2 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^2}dx}{3 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3}\right )}{5 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5}\) |
| \(\Big \downarrow \) 3554 |
| \(\displaystyle \frac {4 \left (\frac {2 \sin (c+d x)}{3 a d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}-\frac {b \cos (c+d x)-a \sin (c+d x)}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3}\right )}{5 \left (a^2+b^2\right )}-\frac {b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5}\) |
-1/5*(b*Cos[c + d*x] - a*Sin[c + d*x])/((a^2 + b^2)*d*(a*Cos[c + d*x] + b* Sin[c + d*x])^5) + (4*(-1/3*(b*Cos[c + d*x] - a*Sin[c + d*x])/((a^2 + b^2) *d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (2*Sin[c + d*x])/(3*a*(a^2 + b^2 )*d*(a*Cos[c + d*x] + b*Sin[c + d*x]))))/(5*(a^2 + b^2))
3.3.31.3.1 Defintions of rubi rules used
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-2), x _Symbol] :> Simp[Sin[c + d*x]/(a*d*(a*Cos[c + d*x] + b*Sin[c + d*x])), x] / ; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(b*Cos[c + d*x] - a*Sin[c + d*x])*((a*Cos[c + d*x] + b*Sin [c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[(n + 2)/((n + 1)*(a^ 2 + b^2)) Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1] && NeQ[n, -2]
Time = 2.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (a^{2}+b^{2}\right ) a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{4}}-\frac {6 a^{2}+2 b^{2}}{3 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {a^{4}+2 a^{2} b^{2}+b^{4}}{5 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{5}}-\frac {1}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}}{d}\) | \(125\) |
| default | \(\frac {\frac {\left (a^{2}+b^{2}\right ) a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{4}}-\frac {6 a^{2}+2 b^{2}}{3 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {a^{4}+2 a^{2} b^{2}+b^{4}}{5 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{5}}-\frac {1}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}}{d}\) | \(125\) |
| risch | \(\frac {16 i \left (-20 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+10 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-10 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 i b a +a^{2}-b^{2}\right )}{15 d \left (-i b +a \right )^{3} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )} a +i b +a \right )^{5}}\) | \(131\) |
| norman | \(\frac {\frac {1}{5 b d}+\frac {2 \left (32 a^{3}-16 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d \,a^{4}}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{5 b d}+\frac {\left (6 a^{3}-16 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d \,a^{3} b}-\frac {\left (6 a^{3}-16 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d \,a^{3} b}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d b}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{5}}\) | \(235\) |
| parallelrisch | \(-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{4}-60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a^{3} b -20 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{2} b^{2}+140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{3} b -120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \,b^{3}+58 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{4}-224 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} b^{2}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{4}-140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{3} b +120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a \,b^{3}-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{4}+120 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2} b^{2}+60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} b +15 a^{4}\right )}{15 a^{5} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{5}}\) | \(292\) |
1/d*((a^2+b^2)*a/b^5/(a+b*tan(d*x+c))^4-1/3*(6*a^2+2*b^2)/b^5/(a+b*tan(d*x +c))^3-1/5*(a^4+2*a^2*b^2+b^4)/b^5/(a+b*tan(d*x+c))^5-1/b^5/(a+b*tan(d*x+c ))+2*a/b^5/(a+b*tan(d*x+c))^2)
Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (145) = 290\).
Time = 0.28 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.92 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=-\frac {8 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left ({\left (a^{11} - 7 \, a^{9} b^{2} - 22 \, a^{7} b^{4} - 14 \, a^{5} b^{6} + 5 \, a^{3} b^{8} + 5 \, a b^{10}\right )} d \cos \left (d x + c\right )^{5} + 10 \, {\left (a^{9} b^{2} + 2 \, a^{7} b^{4} - 2 \, a^{3} b^{8} - a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 5 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left ({\left (5 \, a^{10} b + 5 \, a^{8} b^{3} - 14 \, a^{6} b^{5} - 22 \, a^{4} b^{7} - 7 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 12 \, a^{4} b^{7} + 2 \, a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]
-1/15*(8*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^5 - 20*(a^4*b - 6*a^2*b ^3 + b^5)*cos(d*x + c)^3 - 5*(a^4*b + 6*a^2*b^3 - 3*b^5)*cos(d*x + c) - (3 *a^5 + 10*a^3*b^2 + 15*a*b^4 + 8*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c) ^4 + 4*(a^5 + 10*a^3*b^2 - 15*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 7*a^9*b^2 - 22*a^7*b^4 - 14*a^5*b^6 + 5*a^3*b^8 + 5*a*b^10)*d*cos(d*x + c)^5 + 10*(a^9*b^2 + 2*a^7*b^4 - 2*a^3*b^8 - a*b^10)*d*cos(d*x + c)^3 + 5* (a^7*b^4 + 3*a^5*b^6 + 3*a^3*b^8 + a*b^10)*d*cos(d*x + c) + ((5*a^10*b + 5 *a^8*b^3 - 14*a^6*b^5 - 22*a^4*b^7 - 7*a^2*b^9 + b^11)*d*cos(d*x + c)^4 + 2*(5*a^8*b^3 + 14*a^6*b^5 + 12*a^4*b^7 + 2*a^2*b^9 - b^11)*d*cos(d*x + c)^ 2 + (a^6*b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11)*d)*sin(d*x + c))
Timed out. \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.15 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=-\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4} + 10 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} + 5 \, {\left (3 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{15 \, {\left (b^{10} \tan \left (d x + c\right )^{5} + 5 \, a b^{9} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{8} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{7} \tan \left (d x + c\right )^{2} + 5 \, a^{4} b^{6} \tan \left (d x + c\right ) + a^{5} b^{5}\right )} d} \]
-1/15*(15*b^4*tan(d*x + c)^4 + 30*a*b^3*tan(d*x + c)^3 + 3*a^4 + a^2*b^2 + 3*b^4 + 10*(3*a^2*b^2 + b^4)*tan(d*x + c)^2 + 5*(3*a^3*b + a*b^3)*tan(d*x + c))/((b^10*tan(d*x + c)^5 + 5*a*b^9*tan(d*x + c)^4 + 10*a^2*b^8*tan(d*x + c)^3 + 10*a^3*b^7*tan(d*x + c)^2 + 5*a^4*b^6*tan(d*x + c) + a^5*b^5)*d)
Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=-\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 10 \, b^{4} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 5 \, a b^{3} \tan \left (d x + c\right ) + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4}}{15 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{5} b^{5} d} \]
-1/15*(15*b^4*tan(d*x + c)^4 + 30*a*b^3*tan(d*x + c)^3 + 30*a^2*b^2*tan(d* x + c)^2 + 10*b^4*tan(d*x + c)^2 + 15*a^3*b*tan(d*x + c) + 5*a*b^3*tan(d*x + c) + 3*a^4 + a^2*b^2 + 3*b^4)/((b*tan(d*x + c) + a)^5*b^5*d)
Time = 31.23 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.11 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{a}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2-6\,b^2\right )}{3\,a^3}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2-6\,b^2\right )}{3\,a^3}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2}-\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (29\,a^4-112\,a^2\,b^2+24\,b^4\right )}{15\,a^5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (60\,a^4\,b-160\,a^2\,b^3+32\,b^5\right )+a^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^5-40\,a^3\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (5\,a^5-40\,a^3\,b^2\right )+10\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\right )} \]
((2*tan(c/2 + (d*x)/2)^9)/a + (2*tan(c/2 + (d*x)/2))/a - (8*tan(c/2 + (d*x )/2)^4*(7*a^2*b - 6*b^3))/(3*a^4) + (8*tan(c/2 + (d*x)/2)^6*(7*a^2*b - 6*b ^3))/(3*a^4) - (8*tan(c/2 + (d*x)/2)^3*(a^2 - 6*b^2))/(3*a^3) - (8*tan(c/2 + (d*x)/2)^7*(a^2 - 6*b^2))/(3*a^3) + (8*b*tan(c/2 + (d*x)/2)^2)/a^2 - (8 *b*tan(c/2 + (d*x)/2)^8)/a^2 + (4*tan(c/2 + (d*x)/2)^5*(29*a^4 + 24*b^4 - 112*a^2*b^2))/(15*a^5))/(d*(tan(c/2 + (d*x)/2)^4*(80*a*b^4 + 10*a^5 - 120* a^3*b^2) - tan(c/2 + (d*x)/2)^3*(40*a^4*b - 80*a^2*b^3) - tan(c/2 + (d*x)/ 2)^7*(40*a^4*b - 80*a^2*b^3) - a^5*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x) /2)^6*(80*a*b^4 + 10*a^5 - 120*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(60*a^4*b + 32*b^5 - 160*a^2*b^3) + a^5 - tan(c/2 + (d*x)/2)^2*(5*a^5 - 40*a^3*b^2) + tan(c/2 + (d*x)/2)^8*(5*a^5 - 40*a^3*b^2) + 10*a^4*b*tan(c/2 + (d*x)/2) + 10*a^4*b*tan(c/2 + (d*x)/2)^9))