Integrand size = 29, antiderivative size = 86 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (A+B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+2 B) \tan (c+d x)}{3 d}+\frac {a (A+B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*a*(A+B)*arctanh(sin(d*x+c))/d+1/3*a*(3*A+2*B)*tan(d*x+c)/d+1/2*a*(A+B) *sec(d*x+c)*tan(d*x+c)/d+1/3*a*B*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.65 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (3 (A+B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 (A+B)+3 (A+B) \sec (c+d x)+2 B \tan ^2(c+d x)\right )\right )}{6 d} \] Input:
Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(3*(A + B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + B) + 3*(A + B)* Sec[c + d*x] + 2*B*Tan[c + d*x]^2)))/(6*d)
Time = 0.58 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4485, 3042, 4274, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{3} \int \sec ^2(c+d x) (a (3 A+2 B)+3 a (A+B) \sec (c+d x))dx+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (3 A+2 B)+3 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \int \sec ^3(c+d x)dx+a (3 A+2 B) \int \sec ^2(c+d x)dx\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {a (3 A+2 B) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (3 a (A+B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+2 B) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
Input:
Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*B*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a*(3*A + 2*B)*Tan[c + d*x])/d + 3*a*(A + B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/( 2*d)))/3
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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Time = 0.70 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94
| method | result | size |
| parts | \(\frac {\left (a A +B a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}\) | \(81\) |
| derivativedivides | \(\frac {a A \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(105\) |
| default | \(\frac {a A \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(105\) |
| norman | \(\frac {-\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(121\) |
| parallelrisch | \(\frac {a \left (-\frac {3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (A +B \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {2 B}{3}\right ) \sin \left (3 d x +3 c \right )+\sin \left (d x +c \right ) \left (A +2 B \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(137\) |
| risch | \(-\frac {i a \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A -3 B \,{\mathrm e}^{i \left (d x +c \right )}-6 A -4 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\) | \(193\) |
Input:
int(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
(A*a+B*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*a/ d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*A/d*tan(d*x+c)
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A + 2 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, B a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/12*(3*(A + B)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + B)*a*cos(d *x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*A + 2*B)*a*cos(d*x + c)^2 + 3*( A + B)*a*cos(d*x + c) + 2*B*a)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integ ral(B*sec(c + d*x)**3, x) + Integral(B*sec(c + d*x)**4, x))
Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.48 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a - 3*A*a*(2*sin(d*x + c)/(sin (d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*a* (2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*A*a*tan(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.79 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/6*(3*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a + B*a)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*t an(1/2*d*x + 1/2*c)^5 - 12*A*a*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*tan(1/2*d*x + 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c))/(tan (1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 13.77 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+B\right )}{d}-\frac {\left (A\,a+B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a-\frac {4\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x)))/cos(c + d*x)^2,x)
Output:
(a*atanh(tan(c/2 + (d*x)/2))*(A + B))/d - (tan(c/2 + (d*x)/2)*(3*A*a + 3*B *a) + tan(c/2 + (d*x)/2)^5*(A*a + B*a) - tan(c/2 + (d*x)/2)^3*(4*A*a + (4* B*a)/3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + ( d*x)/2)^6 - 1))
Time = 0.15 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.45 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +6 \sin \left (d x +c \right )^{3} a +4 \sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right ) a -6 \sin \left (d x +c \right ) b \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(a*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 3*cos( c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*cos(c + d*x)*log( tan((c + d*x)/2) - 1)*a + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 3*c os(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 3*cos(c + d*x)*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*a - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 3*cos(c + d*x )*sin(c + d*x)*a - 3*cos(c + d*x)*sin(c + d*x)*b + 6*sin(c + d*x)**3*a + 4 *sin(c + d*x)**3*b - 6*sin(c + d*x)*a - 6*sin(c + d*x)*b))/(6*cos(c + d*x) *d*(sin(c + d*x)**2 - 1))