Integrand size = 41, antiderivative size = 185 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {(4 A-B) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {2 (332 A-80 B+3 C) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(4 A-B) \tan (c+d x)}{a^4 d (1+\cos (c+d x))}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B-2 C) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3} \] Output:
-(4*A-B)*arctanh(sin(d*x+c))/a^4/d+2/105*(332*A-80*B+3*C)*tan(d*x+c)/a^4/d -1/105*(88*A-25*B-3*C)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-(4*A-B)*tan(d*x+c )/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*(1 2*A-5*B-2*C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(1190\) vs. \(2(185)=370\).
Time = 8.92 (sec) , antiderivative size = 1190, normalized size of antiderivative = 6.43 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a* Cos[c + d*x])^4,x]
Output:
((32*(4*A - B)*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2))/(d*(1 + Cos [c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (32*(4*A - B)*Cos[c/2 + (d*x)/2]^8*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x] )^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (4*Cos[c/2 + (d*x )/2]^2*Cos[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(A* Sin[c/2] - B*Sin[c/2] + C*Sin[c/2]))/(7*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (8*Cos[c/2 + (d*x)/2]^4*Cos[c + d*x]^2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(17*A*Sin[c/2] - 1 0*B*Sin[c/2] + 3*C*Sin[c/2]))/(35*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Co s[c + d*x] + C*Cos[2*c + 2*d*x])) + (16*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^ 2*Sec[c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(139*A*Sin[c/2] - 55*B* Sin[c/2] + 6*C*Sin[c/2]))/(105*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) + (4*Cos[c/2 + (d*x)/2]*Cos[c + d*x]^2*Sec[ c/2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] - B*Sin[(d*x) /2] + C*Sin[(d*x)/2]))/(7*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*Cos[c + d* x] + C*Cos[2*c + 2*d*x])) + (8*Cos[c/2 + (d*x)/2]^3*Cos[c + d*x]^2*Sec[c/2 ]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*(17*A*Sin[(d*x)/2] - 10*B*Sin[(d *x)/2] + 3*C*Sin[(d*x)/2]))/(35*d*(1 + Cos[c + d*x])^4*(2*A + C + 2*B*C...
Time = 1.54 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 3520, 3042, 3457, 3042, 3457, 3042, 3457, 3042, 3227, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(a (8 A-B+C)-a (4 A-4 B-3 C) \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (8 A-B+C)-a (4 A-4 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (52 A-10 B+3 C)-3 a^2 (12 A-5 B-2 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (52 A-10 B+3 C)-3 a^2 (12 A-5 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (a^3 (244 A-55 B+6 C)-2 a^3 (88 A-25 B-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (244 A-55 B+6 C)-2 a^3 (88 A-25 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \left (2 a^4 (332 A-80 B+3 C)-105 a^4 (4 A-B) \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {2 a^4 (332 A-80 B+3 C)-105 a^4 (4 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^4 (332 A-80 B+3 C) \int \sec ^2(c+d x)dx-105 a^4 (4 A-B) \int \sec (c+d x)dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^4 (332 A-80 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-105 a^4 (4 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {-\frac {2 a^4 (332 A-80 B+3 C) \int 1d(-\tan (c+d x))}{d}-105 a^4 (4 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {2 a^4 (332 A-80 B+3 C) \tan (c+d x)}{d}-105 a^4 (4 A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {2 a^4 (332 A-80 B+3 C) \tan (c+d x)}{d}-\frac {105 a^4 (4 A-B) \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {105 a^3 (4 A-B) \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B-3 C) \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B-2 C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]
Output:
-1/7*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + (-1/5*(a*(12* A - 5*B - 2*C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*((88*A - 2 5*B - 3*C)*Tan[c + d*x])/(d*(1 + Cos[c + d*x])^2) + ((-105*a^3*(4*A - B)*T an[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-105*a^4*(4*A - B)*ArcTanh[Sin[c + d*x]])/d + (2*a^4*(332*A - 80*B + 3*C)*Tan[c + d*x])/d)/a^2)/(3*a^2))/( 5*a^2))/(7*a^2)
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.69 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\frac {3360 \cos \left (d x +c \right ) \left (-\frac {B}{4}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3360 \cos \left (d x +c \right ) \left (-\frac {B}{4}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+559 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (\frac {15 \left (\frac {110 A}{13}-2 B +\frac {3 C}{26}\right ) \cos \left (2 d x +2 c \right )}{43}+\left (A -\frac {535 B}{2236}+\frac {6 C}{559}\right ) \cos \left (3 d x +3 c \right )+\frac {\left (83 A -20 B +\frac {3 C}{4}\right ) \cos \left (4 d x +4 c \right )}{559}+\frac {\left (2861 A -\frac {2645 B}{4}+54 C \right ) \cos \left (d x +c \right )}{559}+\frac {1672 A}{559}-\frac {370 B}{559}+\frac {87 C}{2236}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{840 d \,a^{4} \cos \left (d x +c \right )}\) | \(178\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B +\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-32 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (32 A -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{8 d \,a^{4}}\) | \(242\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B +\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-32 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (32 A -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{8 d \,a^{4}}\) | \(242\) |
| norman | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{56 a d}+\frac {\left (27 A -20 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{140 d a}-\frac {\left (65 A -15 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {\left (133 A -28 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d a}+\frac {\left (359 A -155 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{120 d a}+\frac {\left (937 A -475 B +153 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{840 d a}+\frac {\left (1447 A -460 B +33 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{210 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{3}}+\frac {\left (4 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}-\frac {\left (4 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}\) | \(285\) |
| risch | \(\frac {2 i \left (420 A \,{\mathrm e}^{8 i \left (d x +c \right )}-105 \,{\mathrm e}^{8 i \left (d x +c \right )} B +2940 A \,{\mathrm e}^{7 i \left (d x +c \right )}-735 B \,{\mathrm e}^{7 i \left (d x +c \right )}+9100 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}+16660 A \,{\mathrm e}^{5 i \left (d x +c \right )}-4165 B \,{\mathrm e}^{5 i \left (d x +c \right )}+210 C \,{\mathrm e}^{5 i \left (d x +c \right )}+20524 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4795 B \,{\mathrm e}^{4 i \left (d x +c \right )}+126 C \,{\mathrm e}^{4 i \left (d x +c \right )}+18788 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4445 B \,{\mathrm e}^{3 i \left (d x +c \right )}+252 C \,{\mathrm e}^{3 i \left (d x +c \right )}+11668 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2785 B \,{\mathrm e}^{2 i \left (d x +c \right )}+132 C \,{\mathrm e}^{2 i \left (d x +c \right )}+4228 A \,{\mathrm e}^{i \left (d x +c \right )}-1015 B \,{\mathrm e}^{i \left (d x +c \right )}+42 C \,{\mathrm e}^{i \left (d x +c \right )}+664 A -160 B +6 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}-\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}\) | \(386\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x,meth od=_RETURNVERBOSE)
Output:
1/840*(3360*cos(d*x+c)*(-1/4*B+A)*ln(tan(1/2*d*x+1/2*c)-1)-3360*cos(d*x+c) *(-1/4*B+A)*ln(tan(1/2*d*x+1/2*c)+1)+559*sec(1/2*d*x+1/2*c)^6*(15/43*(110/ 13*A-2*B+3/26*C)*cos(2*d*x+2*c)+(A-535/2236*B+6/559*C)*cos(3*d*x+3*c)+1/55 9*(83*A-20*B+3/4*C)*cos(4*d*x+4*c)+1/559*(2861*A-2645/4*B+54*C)*cos(d*x+c) +1672/559*A-370/559*B+87/2236*C)*tan(1/2*d*x+1/2*c))/d/a^4/cos(d*x+c)
Time = 0.10 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.88 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (332 \, A - 80 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2236 \, A - 535 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2636 \, A - 620 \, B + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (296 \, A - 65 \, B + 9 \, C\right )} \cos \left (d x + c\right ) + 105 \, A\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4, x, algorithm="fricas")
Output:
-1/210*(105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*cos(d*x + c)^4 + 6*(4* A - B)*cos(d*x + c)^3 + 4*(4*A - B)*cos(d*x + c)^2 + (4*A - B)*cos(d*x + c ))*log(sin(d*x + c) + 1) - 105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*cos (d*x + c)^4 + 6*(4*A - B)*cos(d*x + c)^3 + 4*(4*A - B)*cos(d*x + c)^2 + (4 *A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(332*A - 80*B + 3*C)*c os(d*x + c)^4 + (2236*A - 535*B + 24*C)*cos(d*x + c)^3 + (2636*A - 620*B + 39*C)*cos(d*x + c)^2 + 4*(296*A - 65*B + 9*C)*cos(d*x + c) + 105*A)*sin(d *x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))* *4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (177) = 354\).
Time = 0.05 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.22 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4, x, algorithm="maxima")
Output:
1/840*(A*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^ 2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d *x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos( d*x + c) + 1) + 1)/a^4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4 ) - 5*B*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/( cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^ 4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) + 3*C*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d* x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 )/d
Time = 0.20 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4, x, algorithm="giac")
Output:
-1/840*(840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 1680*A*tan(1/2*d*x + 1/2*c)/(( tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15* B*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 147*A*a ^24*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 63*C*a^24 *tan(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*t an(1/2*d*x + 1/2*c)^3 + 105*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*ta n(1/2*d*x + 1/2*c) - 1575*B*a^24*tan(1/2*d*x + 1/2*c) + 105*C*a^24*tan(1/2 *d*x + 1/2*c))/a^28)/d
Time = 0.34 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.36 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {5\,A-3\,B+C}{40\,a^4}+\frac {A-B+C}{20\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,A-3\,B+C}{12\,a^4}-\frac {2\,B-10\,A+2\,C}{24\,a^4}+\frac {A-B+C}{8\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (5\,A-3\,B+C\right )}{8\,a^4}-\frac {2\,B-10\,A+2\,C}{4\,a^4}+\frac {10\,A+2\,B-2\,C}{8\,a^4}+\frac {A-B+C}{2\,a^4}\right )}{d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-B\right )}{a^4\,d} \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a*cos(c + d*x))^4),x)
Output:
(tan(c/2 + (d*x)/2)^5*((5*A - 3*B + C)/(40*a^4) + (A - B + C)/(20*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((5*A - 3*B + C)/(12*a^4) - (2*B - 10*A + 2*C)/(2 4*a^4) + (A - B + C)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)*((3*(5*A - 3*B + C) )/(8*a^4) - (2*B - 10*A + 2*C)/(4*a^4) + (10*A + 2*B - 2*C)/(8*a^4) + (A - B + C)/(2*a^4)))/d - (2*A*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^ 2 - a^4)) + (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) - (2*atanh(tan(c /2 + (d*x)/2))*(4*A - B))/(a^4*d)
Time = 0.16 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.04 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +3360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} c +132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b +48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +658 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -280 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +4340 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -1190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -6825 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +1575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{840 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x)
Output:
(3360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 840*log(tan((c + d *x)/2) - 1)*tan((c + d*x)/2)**2*b - 3360*log(tan((c + d*x)/2) - 1)*a + 840 *log(tan((c + d*x)/2) - 1)*b - 3360*log(tan((c + d*x)/2) + 1)*tan((c + d*x )/2)**2*a + 840*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b + 3360*log (tan((c + d*x)/2) + 1)*a - 840*log(tan((c + d*x)/2) + 1)*b + 15*tan((c + d *x)/2)**9*a - 15*tan((c + d*x)/2)**9*b + 15*tan((c + d*x)/2)**9*c + 132*ta n((c + d*x)/2)**7*a - 90*tan((c + d*x)/2)**7*b + 48*tan((c + d*x)/2)**7*c + 658*tan((c + d*x)/2)**5*a - 280*tan((c + d*x)/2)**5*b + 42*tan((c + d*x) /2)**5*c + 4340*tan((c + d*x)/2)**3*a - 1190*tan((c + d*x)/2)**3*b - 6825* tan((c + d*x)/2)*a + 1575*tan((c + d*x)/2)*b - 105*tan((c + d*x)/2)*c)/(84 0*a**4*d*(tan((c + d*x)/2)**2 - 1))