Integrand size = 29, antiderivative size = 112 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (3 A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^5 (A+B)}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac {a^5 A}{4 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {a^5 (A-B)}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
1/8*a*(3*A-B)*arctanh(sin(d*x+c))/d+1/8*a^5*(A+B)/d/(a^2-a^2*sin(d*x+c))^2 +1/4*a^5*A/d/(a^4-a^4*sin(d*x+c))-1/8*a^5*(A-B)/d/(a^4+a^4*sin(d*x+c))
Time = 0.94 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.60 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left ((3 A-B) \text {arctanh}(\sin (c+d x))+\frac {A+B}{(-1+\sin (c+d x))^2}-\frac {2 A}{-1+\sin (c+d x)}+\frac {-A+B}{1+\sin (c+d x)}\right )}{8 d} \] Input:
Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*((3*A - B)*ArcTanh[Sin[c + d*x]] + (A + B)/(-1 + Sin[c + d*x])^2 - (2*A )/(-1 + Sin[c + d*x]) + (-A + B)/(1 + Sin[c + d*x])))/(8*d)
Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87, 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, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a) (A+B \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^4 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^4 \int \left (\frac {A}{4 a^2 (a-a \sin (c+d x))^2}+\frac {3 A-B}{8 a^2 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {A-B}{8 a^2 (\sin (c+d x) a+a)^2}+\frac {A+B}{4 a (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 \left (\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{8 a^3}-\frac {A-B}{8 a^2 (a \sin (c+d x)+a)}+\frac {A}{4 a^2 (a-a \sin (c+d x))}+\frac {A+B}{8 a (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a^4*(((3*A - B)*ArcTanh[Sin[c + d*x]])/(8*a^3) + (A + B)/(8*a*(a - a*Sin[ c + d*x])^2) + A/(4*a^2*(a - a*Sin[c + d*x])) - (A - B)/(8*a^2*(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 = 1.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {\frac {a A}{4 \cos \left (d x +c \right )^{4}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(141\) |
| default | \(\frac {\frac {a A}{4 \cos \left (d x +c \right )^{4}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a B}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(141\) |
| parallelrisch | \(-\frac {3 \left (\left (A -\frac {B}{3}\right ) \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -\frac {B}{3}\right ) \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {A}{3}+B \right ) \cos \left (2 d x +2 c \right )+\frac {\left (-A -B \right ) \sin \left (3 d x +3 c \right )}{3}+\frac {\left (-7 A +B \right ) \sin \left (d x +c \right )}{3}+\frac {A}{3}-B \right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\) | \(191\) |
| risch | \(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (-6 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-B \,{\mathrm e}^{4 i \left (d x +c \right )}+6 i A \,{\mathrm e}^{i \left (d x +c \right )}+2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i B \,{\mathrm e}^{i \left (d x +c \right )}+10 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A -B \right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\) | \(245\) |
| norman | \(\frac {\frac {\left (4 a A +4 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (4 a A +4 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {4 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {a \left (5 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (5 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a \left (7 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {a \left (7 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {a \left (13 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {a \left (13 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {a \left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(339\) |
Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d*(1/4*a*A/cos(d*x+c)^4+a*B*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c )^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+a*A*(-(-1/4 *sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+1/ 4*a*B/cos(d*x+c)^4)
Time = 0.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.62 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \, {\left (A - 3 \, B\right )} a - {\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/16*(2*(3*A - B)*a*cos(d*x + c)^2 + 2*(3*A - B)*a*sin(d*x + c) - 2*(A - 3*B)*a - ((3*A - B)*a*cos(d*x + c)^2*sin(d*x + c) - (3*A - B)*a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((3*A - B)*a*cos(d*x + c)^2*sin(d*x + c) - ( 3*A - B)*a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2*sin(d *x + c) - d*cos(d*x + c)^2)
\[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=a \left (\int A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x)**5, x) + Integral(A*sin(c + d*x)*sec(c + d*x)** 5, x) + Integral(B*sin(c + d*x)*sec(c + d*x)**5, x) + Integral(B*sin(c + d *x)**2*sec(c + d*x)**5, x))
Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, A - B\right )} a \sin \left (d x + c\right )^{2} - {\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \, {\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/16*((3*A - B)*a*log(sin(d*x + c) + 1) - (3*A - B)*a*log(sin(d*x + c) - 1 ) - 2*((3*A - B)*a*sin(d*x + c)^2 - (3*A - B)*a*sin(d*x + c) - 2*(A + B)*a )/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, A a - B a\right )} \sin \left (d x + c\right )^{2} - 2 \, A a - 2 \, B a - {\left (3 \, A a - B a\right )} \sin \left (d x + c\right )}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/16*(3*A*a - B*a)*log(abs(sin(d*x + c) + 1))/d - 1/16*(3*A*a - B*a)*log(a bs(sin(d*x + c) - 1))/d - 1/8*((3*A*a - B*a)*sin(d*x + c)^2 - 2*A*a - 2*B* a - (3*A*a - B*a)*sin(d*x + c))/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2 )
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A-B\right )}{8\,d}-\frac {\left (\frac {B\,a}{8}-\frac {3\,A\,a}{8}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {3\,A\,a}{8}-\frac {B\,a}{8}\right )\,\sin \left (c+d\,x\right )+\frac {A\,a}{4}+\frac {B\,a}{4}}{d\,\left (-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x)^5,x)
Output:
(a*atanh(sin(c + d*x))*(3*A - B))/(8*d) - ((A*a)/4 + (B*a)/4 + sin(c + d*x )*((3*A*a)/8 - (B*a)/8) - sin(c + d*x)^2*((3*A*a)/8 - (B*a)/8))/(d*(sin(c + d*x) + sin(c + d*x)^2 - sin(c + d*x)^3 - 1))
Time = 0.19 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.65 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +3 \sin \left (d x +c \right )^{3} a -\sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right )^{2} a +2 \sin \left (d x +c \right )^{2} b +5 a +b \right )}{8 d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(a*( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a + log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**3*b + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*log(tan((c + d*x)/2) - 1) *sin(c + d*x)*a - log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 3*log(tan((c + d*x)/2) - 1)*a + log(tan((c + d*x)/2) - 1)*b + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*b - 3*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + log(tan((c + d *x)/2) + 1)*sin(c + d*x)*b + 3*log(tan((c + d*x)/2) + 1)*a - log(tan((c + d*x)/2) + 1)*b + 3*sin(c + d*x)**3*a - sin(c + d*x)**3*b - 6*sin(c + d*x)* *2*a + 2*sin(c + d*x)**2*b + 5*a + b))/(8*d*(sin(c + d*x)**3 - sin(c + d*x )**2 - sin(c + d*x) + 1))