Integrand size = 31, antiderivative size = 162 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(5 A+B) \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {3 A}{16 d (a+a \sin (c+d x))}-\frac {a^5 (A-B)}{24 d \left (a^2+a^2 \sin (c+d x)\right )^3}+\frac {a^5 (A+B)}{32 d \left (a^3-a^3 \sin (c+d x)\right )^2}-\frac {a^5 (3 A-B)}{32 d \left (a^3+a^3 \sin (c+d x)\right )^2} \] Output:
1/16*(5*A+B)*arctanh(sin(d*x+c))/a/d+1/16*(2*A+B)/d/(a-a*sin(d*x+c))-3/16* A/d/(a+a*sin(d*x+c))-1/24*a^5*(A-B)/d/(a^2+a^2*sin(d*x+c))^3+1/32*a^5*(A+B )/d/(a^3-a^3*sin(d*x+c))^2-1/32*a^5*(3*A-B)/d/(a^3+a^3*sin(d*x+c))^2
Time = 0.62 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {6 (5 A+B) \text {arctanh}(\sin (c+d x))+\frac {3 (A+B)}{(-1+\sin (c+d x))^2}-\frac {6 (2 A+B)}{-1+\sin (c+d x)}-\frac {4 (A-B)}{(1+\sin (c+d x))^3}+\frac {-9 A+3 B}{(1+\sin (c+d x))^2}-\frac {18 A}{1+\sin (c+d x)}}{96 a d} \] Input:
Integrate[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(6*(5*A + B)*ArcTanh[Sin[c + d*x]] + (3*(A + B))/(-1 + Sin[c + d*x])^2 - ( 6*(2*A + B))/(-1 + Sin[c + d*x]) - (4*(A - B))/(1 + Sin[c + d*x])^3 + (-9* A + 3*B)/(1 + Sin[c + d*x])^2 - (18*A)/(1 + Sin[c + d*x]))/(96*a*d)
Time = 0.41 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^5 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^4 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^4 \int \left (\frac {3 A}{16 a^4 (\sin (c+d x) a+a)^2}+\frac {5 A+B}{16 a^4 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {2 A+B}{16 a^4 (a-a \sin (c+d x))^2}+\frac {A+B}{16 a^3 (a-a \sin (c+d x))^3}+\frac {3 A-B}{16 a^3 (\sin (c+d x) a+a)^3}+\frac {A-B}{8 a^2 (\sin (c+d x) a+a)^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 \left (\frac {(5 A+B) \text {arctanh}(\sin (c+d x))}{16 a^5}+\frac {2 A+B}{16 a^4 (a-a \sin (c+d x))}-\frac {3 A}{16 a^4 (a \sin (c+d x)+a)}+\frac {A+B}{32 a^3 (a-a \sin (c+d x))^2}-\frac {3 A-B}{32 a^3 (a \sin (c+d x)+a)^2}-\frac {A-B}{24 a^2 (a \sin (c+d x)+a)^3}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(a^4*(((5*A + B)*ArcTanh[Sin[c + d*x]])/(16*a^5) + (A + B)/(32*a^3*(a - a* Sin[c + d*x])^2) + (2*A + B)/(16*a^4*(a - a*Sin[c + d*x])) - (A - B)/(24*a ^2*(a + a*Sin[c + d*x])^3) - (3*A - B)/(32*a^3*(a + a*Sin[c + d*x])^2) - ( 3*A)/(16*a^4*(a + a*Sin[c + d*x]))))/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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.46 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {-\frac {3 A}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{8}-\frac {B}{8}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {5 A}{32}+\frac {B}{32}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {5 A}{32}-\frac {B}{32}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{16}-\frac {B}{16}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{8}+\frac {B}{16}}{\sin \left (d x +c \right )-1}}{d a}\) | \(132\) |
| default | \(\frac {-\frac {3 A}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{8}-\frac {B}{8}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {5 A}{32}+\frac {B}{32}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {5 A}{32}-\frac {B}{32}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{16}-\frac {B}{16}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{8}+\frac {B}{16}}{\sin \left (d x +c \right )-1}}{d a}\) | \(132\) |
| parallelrisch | \(\frac {-45 \left (2+\frac {\sin \left (5 d x +5 c \right )}{3}+\sin \left (3 d x +3 c \right )+\frac {2 \sin \left (d x +c \right )}{3}+\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {8 \cos \left (2 d x +2 c \right )}{3}\right ) \left (A +\frac {B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \left (2+\frac {\sin \left (5 d x +5 c \right )}{3}+\sin \left (3 d x +3 c \right )+\frac {2 \sin \left (d x +c \right )}{3}+\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {8 \cos \left (2 d x +2 c \right )}{3}\right ) \left (A +\frac {B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-16 A -80 B \right ) \cos \left (2 d x +2 c \right )+\left (-14 A -22 B \right ) \cos \left (4 d x +4 c \right )+\left (84 A -12 B \right ) \sin \left (3 d x +3 c \right )+\left (8 A -8 B \right ) \sin \left (5 d x +5 c \right )+\left (236 A +28 B \right ) \sin \left (d x +c \right )+30 A +102 B}{48 a d \left (6+\sin \left (5 d x +5 c \right )+3 \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right )+2 \cos \left (4 d x +4 c \right )+8 \cos \left (2 d x +2 c \right )\right )}\) | \(294\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (22 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+15 A \,{\mathrm e}^{8 i \left (d x +c \right )}+110 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{8 i \left (d x +c \right )}-6 i B \,{\mathrm e}^{i \left (d x +c \right )}+40 A \,{\mathrm e}^{6 i \left (d x +c \right )}-110 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+8 B \,{\mathrm e}^{6 i \left (d x +c \right )}+30 i A \,{\mathrm e}^{7 i \left (d x +c \right )}+18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-30 i A \,{\mathrm e}^{i \left (d x +c \right )}-150 B \,{\mathrm e}^{4 i \left (d x +c \right )}-22 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+6 i B \,{\mathrm e}^{7 i \left (d x +c \right )}+8 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A +3 B \right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{16 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{16 d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{16 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{16 d a}\) | \(371\) |
| norman | \(\frac {\frac {\left (41 A +37 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d a}+\frac {\left (41 A +37 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d a}-\frac {\left (A -19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{6 d a}+\frac {\left (3 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 d a}+\frac {\left (3 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 d a}+\frac {\left (2 A +10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {\left (2 A +10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}+\frac {\left (11 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\left (11 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d a}+\frac {\left (43 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d a}+\frac {\left (43 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {\left (5 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d a}+\frac {\left (5 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d a}\) | \(377\) |
Input:
int(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d/a*(-3/16*A/(1+sin(d*x+c))-1/3*(1/8*A-1/8*B)/(1+sin(d*x+c))^3-1/2*(3/16 *A-1/16*B)/(1+sin(d*x+c))^2+(5/32*A+1/32*B)*ln(1+sin(d*x+c))+(-5/32*A-1/32 *B)*ln(sin(d*x+c)-1)-1/2*(-1/16*A-1/16*B)/(sin(d*x+c)-1)^2-(1/8*A+1/16*B)/ (sin(d*x+c)-1))
Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {6 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 2 \, B\right )} \sin \left (d x + c\right ) - 4 \, A - 20 \, B}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \] Input:
integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/96*(6*(5*A + B)*cos(d*x + c)^4 - 2*(5*A + B)*cos(d*x + c)^2 - 3*((5*A + B)*cos(d*x + c)^4*sin(d*x + c) + (5*A + B)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 3*((5*A + B)*cos(d*x + c)^4*sin(d*x + c) + (5*A + B)*cos(d*x + c )^4)*log(-sin(d*x + c) + 1) - 2*(3*(5*A + B)*cos(d*x + c)^2 + 10*A + 2*B)* sin(d*x + c) - 4*A - 20*B)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)
\[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
(Integral(A*sec(c + d*x)**5/(sin(c + d*x) + 1), x) + Integral(B*sin(c + d* x)*sec(c + d*x)**5/(sin(c + d*x) + 1), x))/a
Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{2} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right ) + 8 \, A - 8 \, B\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a}}{96 \, d} \] Input:
integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="max ima")
Output:
1/96*(3*(5*A + B)*log(sin(d*x + c) + 1)/a - 3*(5*A + B)*log(sin(d*x + c) - 1)/a - 2*(3*(5*A + B)*sin(d*x + c)^4 + 3*(5*A + B)*sin(d*x + c)^3 - 5*(5* A + B)*sin(d*x + c)^2 - 5*(5*A + B)*sin(d*x + c) + 8*A - 8*B)/(a*sin(d*x + c)^5 + a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin (d*x + c) + a))/d
Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {{\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, a d} - \frac {{\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, a d} - \frac {3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{2} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right ) + 8 \, A - 8 \, B}{48 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="gia c")
Output:
1/32*(5*A + B)*log(abs(sin(d*x + c) + 1))/(a*d) - 1/32*(5*A + B)*log(abs(s in(d*x + c) - 1))/(a*d) - 1/48*(3*(5*A + B)*sin(d*x + c)^4 + 3*(5*A + B)*s in(d*x + c)^3 - 5*(5*A + B)*sin(d*x + c)^2 - 5*(5*A + B)*sin(d*x + c) + 8* A - 8*B)/(a*d*(sin(d*x + c) + 1)^3*(sin(d*x + c) - 1)^2)
Time = 34.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A+B\right )}{16\,a\,d}-\frac {\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A}{6}-\frac {B}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)^5*(a + a*sin(c + d*x))),x)
Output:
(atanh(sin(c + d*x))*(5*A + B))/(16*a*d) - (A/6 - B/6 - sin(c + d*x)*((25* A)/48 + (5*B)/48) + sin(c + d*x)^3*((5*A)/16 + B/16) + sin(c + d*x)^4*((5* A)/16 + B/16) - sin(c + d*x)^2*((25*A)/48 + (5*B)/48))/(d*(a + a*sin(c + d *x) - 2*a*sin(c + d*x)^2 - 2*a*sin(c + d*x)^3 + a*sin(c + d*x)^4 + a*sin(c + d*x)^5))
Time = 0.18 (sec) , antiderivative size = 661, normalized size of antiderivative = 4.08 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a - 3*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**5*b - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b + 3 0*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 6*log(tan((c + d*x)/2) - 1 )*sin(c + d*x)**2*b - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a - 3*log( tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 15*log(tan((c + d*x)/2) - 1)*a - 3* log(tan((c + d*x)/2) - 1)*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 *a + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*b + 15*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**4*a + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*b - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b + 15*log(ta n((c + d*x)/2) + 1)*a + 3*log(tan((c + d*x)/2) + 1)*b - 25*sin(c + d*x)**5 *a - 5*sin(c + d*x)**5*b - 40*sin(c + d*x)**4*a - 8*sin(c + d*x)**4*b + 35 *sin(c + d*x)**3*a + 7*sin(c + d*x)**3*b + 75*sin(c + d*x)**2*a + 15*sin(c + d*x)**2*b - 33*a + 3*b)/(48*a*d*(sin(c + d*x)**5 + sin(c + d*x)**4 - 2* sin(c + d*x)**3 - 2*sin(c + d*x)**2 + sin(c + d*x) + 1))