Integrand size = 31, antiderivative size = 104 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d} \] Output:
1/15*a^2*(3*A-2*B)*sec(d*x+c)^3/d+1/5*(A+B)*sec(d*x+c)^5*(a+a*sin(d*x+c))^ 2/d+1/5*a^2*(3*A-2*B)*tan(d*x+c)/d+1/15*a^2*(3*A-2*B)*tan(d*x+c)^3/d
Time = 0.34 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left ((6 A+B) \sec ^5(c+d x)+15 A \sec ^4(c+d x) \tan (c+d x)+5 B \sec ^3(c+d x) \tan ^2(c+d x)-5 (3 A-2 B) \sec ^2(c+d x) \tan ^3(c+d x)+2 (3 A-2 B) \tan ^5(c+d x)\right )}{15 d} \] Input:
Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^2*((6*A + B)*Sec[c + d*x]^5 + 15*A*Sec[c + d*x]^4*Tan[c + d*x] + 5*B*Se c[c + d*x]^3*Tan[c + d*x]^2 - 5*(3*A - 2*B)*Sec[c + d*x]^2*Tan[c + d*x]^3 + 2*(3*A - 2*B)*Tan[c + d*x]^5))/(15*d)
Time = 0.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3042, 3334, 3042, 3148, 3042, 4254, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3334 |
| \(\displaystyle \frac {1}{5} a (3 A-2 B) \int \sec ^4(c+d x) (\sin (c+d x) a+a)dx+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} a (3 A-2 B) \int \frac {\sin (c+d x) a+a}{\cos (c+d x)^4}dx+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{5} a (3 A-2 B) \left (a \int \sec ^4(c+d x)dx+\frac {a \sec ^3(c+d x)}{3 d}\right )+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} a (3 A-2 B) \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {a \sec ^3(c+d x)}{3 d}\right )+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} a (3 A-2 B) \left (\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d}+\frac {1}{5} a (3 A-2 B) \left (\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )\) |
Input:
Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(5*d) + (a*(3*A - 2*B)*((a *Sec[c + d*x]^3)/(3*d) - (a*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/5
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 1.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17
| method | result | size |
| parallelrisch | \(-\frac {2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\left (B -2 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \left (A -\frac {2 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\frac {\left (-3 A -\frac {4 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {2 A}{5}+\frac {B}{15}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\) | \(122\) |
| risch | \(-\frac {4 \left (-3 i A \,a^{2}+12 A \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i B \,a^{2}+10 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+15 i A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-10 i B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d}\) | \(126\) |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin \left (d x +c \right )^{4}}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{15}\right )+\frac {2 A \,a^{2}}{5 \cos \left (d x +c \right )^{5}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {a^{2} B}{5 \cos \left (d x +c \right )^{5}}}{d}\) | \(231\) |
| default | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin \left (d x +c \right )^{4}}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{15}\right )+\frac {2 A \,a^{2}}{5 \cos \left (d x +c \right )^{5}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {a^{2} B}{5 \cos \left (d x +c \right )^{5}}}{d}\) | \(231\) |
| norman | \(\frac {-\frac {12 A \,a^{2}+2 a^{2} B}{15 d}-\frac {2 \left (2 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}-\frac {2 \left (6 A \,a^{2}+5 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}-\frac {2 \left (18 A \,a^{2}+13 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d}-\frac {2 \left (30 A \,a^{2}+35 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}-\frac {2 \left (78 A \,a^{2}+73 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{15 d}-\frac {2 \left (186 A \,a^{2}+181 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {2 a^{2} \left (9 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a^{2} \left (9 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}-\frac {2 a^{2} \left (42 A +47 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}-\frac {2 a^{2} \left (111 A +136 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {2 a^{2} \left (111 A +136 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{15 d}-\frac {2 a^{2} \left (213 A +208 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}-\frac {2 a^{2} \left (213 A +208 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(444\) |
Input:
int(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-2*a^2*(tan(1/2*d*x+1/2*c)^5*A+(B-2*A)*tan(1/2*d*x+1/2*c)^4+2*(A-2/3*B)*ta n(1/2*d*x+1/2*c)^3+4/3*B*tan(1/2*d*x+1/2*c)^2+1/5*(-3*A-4/3*B)*tan(1/2*d*x +1/2*c)+2/5*A+1/15*B)/d/(tan(1/2*d*x+1/2*c)+1)/(tan(1/2*d*x+1/2*c)-1)^5
Time = 0.07 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {4 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, A - 3 \, B\right )} a^{2} - {\left (2 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
-1/15*(4*(3*A - 2*B)*a^2*cos(d*x + c)^2 - 3*(2*A - 3*B)*a^2 - (2*(3*A - 2* B)*a^2*cos(d*x + c)^2 - 3*(3*A - 2*B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)*sin(d*x + c) - 2*d*cos(d*x + c))
Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {6 \, A a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {3 \, B a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + (3* tan(d*x + c)^5 + 5*tan(d*x + c)^3)*A*a^2 + 2*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a^2 - (5*cos(d*x + c)^2 - 3)*B*a^2/cos(d*x + c)^5 + 6*A*a^2/cos (d*x + c)^5 + 3*B*a^2/cos(d*x + c)^5)/d
Time = 0.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\frac {15 \, {\left (A a^{2} - B a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {105 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 50 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 63 \, A a^{2} - 7 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
-1/60*(15*(A*a^2 - B*a^2)/(tan(1/2*d*x + 1/2*c) + 1) + (105*A*a^2*tan(1/2* d*x + 1/2*c)^4 + 15*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 360*A*a^2*tan(1/2*d*x + 1/2* c)^2 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 210*A*a^2*tan(1/2*d*x + 1/2*c) + 50*B*a^2*tan(1/2*d*x + 1/2*c) + 63*A*a^2 - 7*B*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^5)/d
Time = 34.75 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2\,a^2\,\left (\frac {5\,B\,\sin \left (c+d\,x\right )}{2}-\frac {15\,A\,\cos \left (c+d\,x\right )}{4}-\frac {5\,B\,\cos \left (c+d\,x\right )}{8}-\frac {15\,A\,\sin \left (c+d\,x\right )}{4}-\frac {5\,B}{2}-3\,A\,\cos \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\cos \left (3\,c+3\,d\,x\right )}{4}+2\,B\,\cos \left (2\,c+2\,d\,x\right )+\frac {B\,\cos \left (3\,c+3\,d\,x\right )}{8}+3\,A\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{2}-\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{2}\right )}{15\,d\,\left (\frac {\cos \left (3\,c+3\,d\,x\right )}{4}-\frac {5\,\cos \left (c+d\,x\right )}{4}+\sin \left (2\,c+2\,d\,x\right )\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^6,x)
Output:
(2*a^2*((5*B*sin(c + d*x))/2 - (15*A*cos(c + d*x))/4 - (5*B*cos(c + d*x))/ 8 - (15*A*sin(c + d*x))/4 - (5*B)/2 - 3*A*cos(2*c + 2*d*x) + (3*A*cos(3*c + 3*d*x))/4 + 2*B*cos(2*c + 2*d*x) + (B*cos(3*c + 3*d*x))/8 + 3*A*sin(2*c + 2*d*x) + (3*A*sin(3*c + 3*d*x))/4 + (B*sin(2*c + 2*d*x))/2 - (B*sin(3*c + 3*d*x))/2))/(15*d*(cos(3*c + 3*d*x)/4 - (5*cos(c + d*x))/4 + sin(2*c + 2 *d*x)))
Time = 0.17 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.80 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -3 \cos \left (d x +c \right ) a +2 \cos \left (d x +c \right ) b +12 \sin \left (d x +c \right )^{3} a -8 \sin \left (d x +c \right )^{3} b -24 \sin \left (d x +c \right )^{2} a +16 \sin \left (d x +c \right )^{2} b +6 \sin \left (d x +c \right ) a -4 \sin \left (d x +c \right ) b +12 a +2 b \right )}{30 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(a**2*( - 3*cos(c + d*x)*sin(c + d*x)**2*a + 2*cos(c + d*x)*sin(c + d*x)** 2*b + 6*cos(c + d*x)*sin(c + d*x)*a - 4*cos(c + d*x)*sin(c + d*x)*b - 3*co s(c + d*x)*a + 2*cos(c + d*x)*b + 12*sin(c + d*x)**3*a - 8*sin(c + d*x)**3 *b - 24*sin(c + d*x)**2*a + 16*sin(c + d*x)**2*b + 6*sin(c + d*x)*a - 4*si n(c + d*x)*b + 12*a + 2*b))/(30*cos(c + d*x)*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1))