\(\int \cos ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [891]

Optimal result

Integrand size = 41, antiderivative size = 274 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac {b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac {b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d} \]

[Out]

1/2*a^2*(12*A*b^2+8*B*a*b+a^2*(A+2*C))*x+1/2*b*(12*B*a^2*b+B*b^3+8*a^3*C+4*a*b^2*(2*A+C))*arctanh(sin(d*x+c))/
d+(2*A*b+B*a)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+b*sec(d*x+c))^4*sin(d*x+c)/d-1/6*b*(12*B*a^3
-24*B*a*b^2+a^2*b*(39*A-34*C)-2*b^3*(3*A+2*C))*tan(d*x+c)/d-1/6*b^2*(6*B*a^2-3*B*b^2+2*a*b*(9*A-4*C))*sec(d*x+
c)*tan(d*x+c)/d-1/6*b*(15*A*b+6*B*a-2*C*b)*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4179, 4141, 4133, 3855, 3852, 8} \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^2 \tan (c+d x) \sec (c+d x) \left (6 a^2 B+2 a b (9 A-4 C)-3 b^2 B\right )}{6 d}+\frac {1}{2} a^2 x \left (a^2 (A+2 C)+8 a b B+12 A b^2\right )+\frac {b \left (8 a^3 C+12 a^2 b B+4 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b \tan (c+d x) \left (12 a^3 B+a^2 b (39 A-34 C)-24 a b^2 B-2 b^3 (3 A+2 C)\right )}{6 d}-\frac {b \tan (c+d x) (6 a B+15 A b-2 b C) (a+b \sec (c+d x))^2}{6 d}+\frac {(a B+2 A b) \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

[In]

Int[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(12*A*b^2 + 8*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(12*a^2*b*B + b^3*B + 8*a^3*C + 4*a*b^2*(2*A + C))*ArcTanh
[Sin[c + d*x]])/(2*d) + ((2*A*b + a*B)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c +
d*x])^4*Sin[c + d*x])/(2*d) - (b*(12*a^3*B - 24*a*b^2*B + a^2*b*(39*A - 34*C) - 2*b^3*(3*A + 2*C))*Tan[c + d*
x])/(6*d) - (b^2*(6*a^2*B - 3*b^2*B + 2*a*b*(9*A - 4*C))*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(15*A*b + 6*a*B
- 2*b*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4133

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b
*(2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 4141

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4179

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (2 (2 A b+a B)+(2 b B+a (A+2 C)) \sec (c+d x)-b (3 A-2 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac {1}{2} \int (a+b \sec (c+d x))^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)-2 b (a A-b B-2 a C) \sec (c+d x)-b (15 A b+6 a B-2 b C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \sec (c+d x)) \left (3 a \left (12 A b^2+8 a b B+a^2 (A+2 C)\right )+b \left (18 a b B-3 a^2 (A-6 C)+2 b^2 (3 A+2 C)\right ) \sec (c+d x)-2 b \left (18 a A b+6 a^2 B-3 b^2 B-8 a b C\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{12} \int \left (6 a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right )+6 b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \sec (c+d x)-2 b \left (12 a^3 B-24 a b^2 B-2 b^3 (3 A+2 C)+a^2 (39 A b-34 b C)\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{2} \left (b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{6} \left (b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac {b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {\left (b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d} \\ & = \frac {1}{2} a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) x+\frac {b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 A b+a B) (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {b \left (12 a^3 B-24 a b^2 B+a^2 b (39 A-34 C)-2 b^3 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac {b^2 \left (6 a^2 B-3 b^2 B+2 a b (9 A-4 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (15 A b+6 a B-2 b C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.50 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.27 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sec ^3(c+d x) \left (36 a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) (c+d x) \cos (c+d x)+12 a^2 \left (12 A b^2+8 a b B+a^2 (A+2 C)\right ) (c+d x) \cos (3 (c+d x))-48 b \left (12 a^2 b B+b^3 B+8 a^3 C+4 a b^2 (2 A+C)\right ) \cos ^3(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \left (9 a^4 A+24 A b^4+96 a b^3 B+144 a^2 b^2 C+32 b^4 C+12 \left (12 a^3 A b+3 a^4 B+2 b^4 B+8 a b^3 C\right ) \cos (c+d x)+4 \left (3 a^4 A+6 A b^4+24 a b^3 B+36 a^2 b^2 C+4 b^4 C\right ) \cos (2 (c+d x))+48 a^3 A b \cos (3 (c+d x))+12 a^4 B \cos (3 (c+d x))+3 a^4 A \cos (4 (c+d x))\right ) \sin (c+d x)\right )}{96 d} \]

[In]

Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Sec[c + d*x]^3*(36*a^2*(12*A*b^2 + 8*a*b*B + a^2*(A + 2*C))*(c + d*x)*Cos[c + d*x] + 12*a^2*(12*A*b^2 + 8*a*b
*B + a^2*(A + 2*C))*(c + d*x)*Cos[3*(c + d*x)] - 48*b*(12*a^2*b*B + b^3*B + 8*a^3*C + 4*a*b^2*(2*A + C))*Cos[c
+ d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(9*a^4*A +
24*A*b^4 + 96*a*b^3*B + 144*a^2*b^2*C + 32*b^4*C + 12*(12*a^3*A*b + 3*a^4*B + 2*b^4*B + 8*a*b^3*C)*Cos[c + d*
x] + 4*(3*a^4*A + 6*A*b^4 + 24*a*b^3*B + 36*a^2*b^2*C + 4*b^4*C)*Cos[2*(c + d*x)] + 48*a^3*A*b*Cos[3*(c + d*x)
] + 12*a^4*B*Cos[3*(c + d*x)] + 3*a^4*A*Cos[4*(c + d*x)])*Sin[c + d*x]))/(96*d)

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {a^{4} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )+a^{4} C \left (d x +c \right )+4 A \,a^{3} b \sin \left (d x +c \right )+4 B \,a^{3} b \left (d x +c \right )+4 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \left (d x +c \right )+6 B \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{2} b^{2} \tan \left (d x +c \right )+4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B a \,b^{3} \tan \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b^{4}+B \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)
default	\(\frac {a^{4} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )+a^{4} C \left (d x +c \right )+4 A \,a^{3} b \sin \left (d x +c \right )+4 B \,a^{3} b \left (d x +c \right )+4 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \left (d x +c \right )+6 B \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{2} b^{2} \tan \left (d x +c \right )+4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B a \,b^{3} \tan \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b^{4}+B \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)
parallelrisch	\(\frac {-32 \left (a^{3} C +\frac {3 B \,a^{2} b}{2}+a \left (A +\frac {C}{2}\right ) b^{2}+\frac {B \,b^{3}}{8}\right ) b \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+32 \left (a^{3} C +\frac {3 B \,a^{2} b}{2}+a \left (A +\frac {C}{2}\right ) b^{2}+\frac {B \,b^{3}}{8}\right ) b \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 d \,a^{2} x \left (12 A \,b^{2}+8 B a b +a^{2} \left (A +2 C \right )\right ) \cos \left (3 d x +3 c \right )+\left (3 a^{4} A +48 C \,a^{2} b^{2}+32 B a \,b^{3}+8 b^{4} \left (A +\frac {2 C}{3}\right )\right ) \sin \left (3 d x +3 c \right )+8 \left (4 A \,a^{3} b +B \,a^{4}+B \,b^{4}+4 C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+4 \left (4 A \,a^{3} b +B \,a^{4}\right ) \sin \left (4 d x +4 c \right )+a^{4} A \sin \left (5 d x +5 c \right )+12 d \,a^{2} x \left (12 A \,b^{2}+8 B a b +a^{2} \left (A +2 C \right )\right ) \cos \left (d x +c \right )+2 \sin \left (d x +c \right ) \left (a^{4} A +24 C \,a^{2} b^{2}+16 B a \,b^{3}+4 b^{4} \left (A +2 C \right )\right )}{8 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(379\)
risch	\(\frac {a^{4} A x}{2}+6 A \,a^{2} b^{2} x +4 B \,a^{3} b x +a^{4} x C +\frac {2 i A \,a^{3} b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i b^{2} \left (3 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-48 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-72 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 C b a \,{\mathrm e}^{i \left (d x +c \right )}-6 A \,b^{2}-24 B a b -36 C \,a^{2}-4 C \,b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 d}-\frac {2 i A \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{4}}{2 d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,a^{2} b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{4}}{2 d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(588\)
norman	\(\frac {\left (-6 A \,a^{2} b^{2}-\frac {1}{2} a^{4} A -4 B \,a^{3} b -a^{4} C \right ) x +\left (-30 A \,a^{2} b^{2}-\frac {5}{2} a^{4} A -20 B \,a^{3} b -5 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-18 A \,a^{2} b^{2}-\frac {3}{2} a^{4} A -12 B \,a^{3} b -3 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-6 A \,a^{2} b^{2}-\frac {1}{2} a^{4} A -4 B \,a^{3} b -a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (6 A \,a^{2} b^{2}+\frac {1}{2} a^{4} A +4 B \,a^{3} b +a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (6 A \,a^{2} b^{2}+\frac {1}{2} a^{4} A +4 B \,a^{3} b +a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (18 A \,a^{2} b^{2}+\frac {3}{2} a^{4} A +12 B \,a^{3} b +3 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (30 A \,a^{2} b^{2}+\frac {5}{2} a^{4} A +20 B \,a^{3} b +5 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {4 \left (15 a^{4} A -6 A \,b^{4}-24 B a \,b^{3}-36 C \,a^{2} b^{2}-2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {2 \left (9 a^{4} A -48 A \,a^{3} b +6 A \,b^{4}-12 B \,a^{4}+24 B a \,b^{3}+36 C \,a^{2} b^{2}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {2 \left (9 a^{4} A +48 A \,a^{3} b +6 A \,b^{4}+12 B \,a^{4}+24 B a \,b^{3}+36 C \,a^{2} b^{2}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (a^{4} A -8 A \,a^{3} b +2 A \,b^{4}-2 B \,a^{4}+8 B a \,b^{3}-B \,b^{4}+12 C \,a^{2} b^{2}-4 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {\left (a^{4} A +8 A \,a^{3} b +2 A \,b^{4}+2 B \,a^{4}+8 B a \,b^{3}+B \,b^{4}+12 C \,a^{2} b^{2}+4 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (15 a^{4} A -40 A \,a^{3} b -2 A \,b^{4}-10 B \,a^{4}-8 B a \,b^{3}+3 B \,b^{4}-12 C \,a^{2} b^{2}+12 C a \,b^{3}-2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (15 a^{4} A +40 A \,a^{3} b -2 A \,b^{4}+10 B \,a^{4}-8 B a \,b^{3}-3 B \,b^{4}-12 C \,a^{2} b^{2}-12 C a \,b^{3}-2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\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 )^{5}}-\frac {b \left (8 a A \,b^{2}+12 B \,a^{2} b +B \,b^{3}+8 a^{3} C +4 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (8 a A \,b^{2}+12 B \,a^{2} b +B \,b^{3}+8 a^{3} C +4 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(941\)

[In]

int(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(a^4*A*(1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+B*a^4*sin(d*x+c)+a^4*C*(d*x+c)+4*A*a^3*b*sin(d*x+c)+4*B*a
^3*b*(d*x+c)+4*a^3*b*C*ln(sec(d*x+c)+tan(d*x+c))+6*A*a^2*b^2*(d*x+c)+6*B*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+6*C
*a^2*b^2*tan(d*x+c)+4*a*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+4*B*a*b^3*tan(d*x+c)+4*C*a*b^3*(1/2*sec(d*x+c)*tan(d*x
+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+A*tan(d*x+c)*b^4+B*b^4*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x
+c)))-C*b^4*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.98 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 4 \, {\left (2 \, A + C\right )} a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 4 \, {\left (2 \, A + C\right )} a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 2 \, C b^{4} + 6 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (18 \, C a^{2} b^{2} + 12 \, B a b^{3} + {\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*((A + 2*C)*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*d*x*cos(d*x + c)^3 + 3*(8*C*a^3*b + 12*B*a^2*b^2 + 4*(2*A +
C)*a*b^3 + B*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(8*C*a^3*b + 12*B*a^2*b^2 + 4*(2*A + C)*a*b^3 + B*
b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*A*a^4*cos(d*x + c)^4 + 2*C*b^4 + 6*(B*a^4 + 4*A*a^3*b)*cos(d
*x + c)^3 + 2*(18*C*a^2*b^2 + 12*B*a*b^3 + (3*A + 2*C)*b^4)*cos(d*x + c)^2 + 3*(4*C*a*b^3 + B*b^4)*cos(d*x + c
))*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.22 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} C a^{4} + 48 \, {\left (d x + c\right )} B a^{3} b + 72 \, {\left (d x + c\right )} A a^{2} b^{2} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{4} - 12 \, C a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4} \sin \left (d x + c\right ) + 48 \, A a^{3} b \sin \left (d x + c\right ) + 72 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 48 \, B a b^{3} \tan \left (d x + c\right ) + 12 \, A b^{4} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 12*(d*x + c)*C*a^4 + 48*(d*x + c)*B*a^3*b + 72*(d*x + c)*A*a^
2*b^2 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^4 - 12*C*a*b^3*(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*b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^2*b^2*(log(sin(
d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^4
*sin(d*x + c) + 48*A*a^3*b*sin(d*x + c) + 72*C*a^2*b^2*tan(d*x + c) + 48*B*a*b^3*tan(d*x + c) + 12*A*b^4*tan(d
*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (262) = 524\).

Time = 0.39 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.01 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (A a^{4} + 2 \, C a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3} + 4 \, C a b^{3} + B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, C a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3} + 4 \, C a b^{3} + B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a^{3} b \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 )}^{2}} - \frac {2 \, {\left (36 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{4} \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} \]

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(A*a^4 + 2*C*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*(d*x + c) + 3*(8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3 + 4*C*
a*b^3 + B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3 + 4*C*a*b^3 + B*b^
4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(A*a^4*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 8*A
*a^3*b*tan(1/2*d*x + 1/2*c)^3 - A*a^4*tan(1/2*d*x + 1/2*c) - 2*B*a^4*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 2*(36*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*B*a*b^3*tan(1/2*d*x
+ 1/2*c)^5 - 12*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^4*tan(1/2*d*x + 1/2*c
)^5 + 6*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 48*B*a*b^3*tan(1/2*d*x + 1/2*c)^3
- 12*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 36*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 24*B
*a*b^3*tan(1/2*d*x + 1/2*c) + 12*C*a*b^3*tan(1/2*d*x + 1/2*c) + 6*A*b^4*tan(1/2*d*x + 1/2*c) + 3*B*b^4*tan(1/2
*d*x + 1/2*c) + 6*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 21.25 (sec) , antiderivative size = 4849, normalized size of antiderivative = 17.70 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

int(cos(c + d*x)^2*(a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(4*A*a^4 + 4*B*a^4 - 2*B*b^4 - (8*C*b^4)/3 + 16*A*a^3*b - 8*C*a*b^3) + tan(c/2 + (d*x)/2
)^7*(4*A*a^4 - 4*B*a^4 + 2*B*b^4 - (8*C*b^4)/3 - 16*A*a^3*b + 8*C*a*b^3) - tan(c/2 + (d*x)/2)^9*(A*a^4 + 2*A*b
^4 - 2*B*a^4 - B*b^4 + 2*C*b^4 + 12*C*a^2*b^2 - 8*A*a^3*b + 8*B*a*b^3 - 4*C*a*b^3) - tan(c/2 + (d*x)/2)*(A*a^4
+ 2*A*b^4 + 2*B*a^4 + B*b^4 + 2*C*b^4 + 12*C*a^2*b^2 + 8*A*a^3*b + 8*B*a*b^3 + 4*C*a*b^3) + tan(c/2 + (d*x)/2
)^5*(4*A*b^4 - 6*A*a^4 - (4*C*b^4)/3 + 24*C*a^2*b^2 + 16*B*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x
)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) + (atan(((((B*b^4)/2 + 6*
B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*b^2 + 192*B*a^2*b^2
+ 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*
a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*
b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*
C*a*b^7 + 256*B*C*a^7*b + 1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a
^6*b^2 + 896*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b)*1i
- (((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*
b^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^8 +
8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4
*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 12
8*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^7*b + 1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*
a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*
b^3 + 4*C*a^3*b)*1i)/(6144*A^3*a^4*b^8 - (((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b)*(16*A*
a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*b^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3
*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^
6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6
*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^7*b + 1536*A*B*a^3*b^5 + 1536*A*B
*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*((B*b^4
)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b) - 256*C^3*a^11*b - (((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^
3 + 2*C*a*b^3 + 4*C*a^3*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*b^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128
*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^
6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^
6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^
7*b + 1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3
*b^5 + 1536*B*C*a^5*b^3))*((B*b^4)/2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 2*C*a*b^3 + 4*C*a^3*b) - 9216*A^3*a^5*b^7 + 5
12*A^3*a^6*b^6 - 1536*A^3*a^7*b^5 - 64*A^3*a^9*b^3 + 64*B^3*a^3*b^9 + 1536*B^3*a^5*b^7 - 512*B^3*a^6*b^6 + 921
6*B^3*a^7*b^5 - 6144*B^3*a^8*b^4 + 256*C^3*a^6*b^6 + 1024*C^3*a^8*b^4 - 128*C^3*a^9*b^3 + 1024*C^3*a^10*b^2 -
256*A*C^2*a^11*b - 64*A^2*C*a^11*b + 96*A*B^2*a^2*b^10 + 3336*A*B^2*a^4*b^8 - 1536*A*B^2*a^5*b^7 + 26304*A*B^2
*a^6*b^6 - 22656*A*B^2*a^7*b^5 + 1152*A*B^2*a^8*b^4 - 1536*A*B^2*a^9*b^3 + 1536*A^2*B*a^3*b^9 - 1152*A^2*B*a^4
*b^8 + 22656*A^2*B*a^5*b^7 - 26304*A^2*B*a^6*b^6 + 1536*A^2*B*a^7*b^5 - 3336*A^2*B*a^8*b^4 - 96*A^2*B*a^10*b^2
+ 1536*A*C^2*a^4*b^8 + 7296*A*C^2*a^6*b^6 - 1536*A*C^2*a^7*b^5 + 8704*A*C^2*a^8*b^4 - 3456*A*C^2*a^9*b^3 + 51
2*A*C^2*a^10*b^2 + 6144*A^2*C*a^4*b^8 - 4608*A^2*C*a^5*b^7 + 13824*A^2*C*a^6*b^6 - 13056*A^2*C*a^7*b^5 + 1024*
A^2*C*a^8*b^4 - 1824*A^2*C*a^9*b^3 + 1152*B*C^2*a^5*b^7 + 5888*B*C^2*a^7*b^5 - 1056*B*C^2*a^8*b^4 + 7168*B*C^2
*a^9*b^3 - 2432*B*C^2*a^10*b^2 + 528*B^2*C*a^4*b^8 + 7552*B^2*C*a^6*b^6 - 2304*B^2*C*a^7*b^5 + 14592*B^2*C*a^8
*b^4 - 7168*B^2*C*a^9*b^3 + 768*A*B*C*a^3*b^9 + 15168*A*B*C*a^5*b^7 - 6528*A*B*C*a^6*b^6 + 30592*A*B*C*a^7*b^5
- 19488*A*B*C*a^8*b^4 + 1536*A*B*C*a^9*b^3 - 1408*A*B*C*a^10*b^2))*(B*b^4*1i + B*a^2*b^2*12i + A*a*b^3*8i + C
*a*b^3*4i + C*a^3*b*8i))/d + (a^2*atan(((a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2
*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2
*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256
*B*C*a^7*b + 1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*
B*C*a^3*b^5 + 1536*B*C*a^5*b^3) - (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4
+ 192*A*a^2*b^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b
^2 + 2*C*a^2 + 8*B*a*b))/2 + (a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 +
1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^6 +
512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^7*b
+ 1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3*b^5
+ 1536*B*C*a^5*b^3) + (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2
*b^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^
2 + 8*B*a*b))/2)/((a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^
4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4
*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^7*b + 1536*A*B*
a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3*b^5 + 1536*B*C
*a^5*b^3) - (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*b^2 + 192*
B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b
)*1i)/2 - 256*C^3*a^11*b - (a^2*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 8*B^2*b^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 11
52*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 192*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 512*B^2*a^6*b^2 + 128*C^2*a^2*b^6 + 51
2*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 128*A*B*a*b^7 + 128*A*B*a^7*b + 64*B*C*a*b^7 + 256*B*C*a^7*b +
1536*A*B*a^3*b^5 + 1536*A*B*a^5*b^3 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2 + 896*B*C*a^3*b^5 +
1536*B*C*a^5*b^3) + (a^2*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b)*(16*A*a^4 + 16*B*b^4 + 32*C*a^4 + 192*A*a^2*b
^2 + 192*B*a^2*b^2 + 128*A*a*b^3 + 128*B*a^3*b + 64*C*a*b^3 + 128*C*a^3*b)*1i)/2)*(A*a^2 + 12*A*b^2 + 2*C*a^2
+ 8*B*a*b)*1i)/2 + 6144*A^3*a^4*b^8 - 9216*A^3*a^5*b^7 + 512*A^3*a^6*b^6 - 1536*A^3*a^7*b^5 - 64*A^3*a^9*b^3 +
64*B^3*a^3*b^9 + 1536*B^3*a^5*b^7 - 512*B^3*a^6*b^6 + 9216*B^3*a^7*b^5 - 6144*B^3*a^8*b^4 + 256*C^3*a^6*b^6 +
1024*C^3*a^8*b^4 - 128*C^3*a^9*b^3 + 1024*C^3*a^10*b^2 - 256*A*C^2*a^11*b - 64*A^2*C*a^11*b + 96*A*B^2*a^2*b^
10 + 3336*A*B^2*a^4*b^8 - 1536*A*B^2*a^5*b^7 + 26304*A*B^2*a^6*b^6 - 22656*A*B^2*a^7*b^5 + 1152*A*B^2*a^8*b^4
- 1536*A*B^2*a^9*b^3 + 1536*A^2*B*a^3*b^9 - 1152*A^2*B*a^4*b^8 + 22656*A^2*B*a^5*b^7 - 26304*A^2*B*a^6*b^6 + 1
536*A^2*B*a^7*b^5 - 3336*A^2*B*a^8*b^4 - 96*A^2*B*a^10*b^2 + 1536*A*C^2*a^4*b^8 + 7296*A*C^2*a^6*b^6 - 1536*A*
C^2*a^7*b^5 + 8704*A*C^2*a^8*b^4 - 3456*A*C^2*a^9*b^3 + 512*A*C^2*a^10*b^2 + 6144*A^2*C*a^4*b^8 - 4608*A^2*C*a
^5*b^7 + 13824*A^2*C*a^6*b^6 - 13056*A^2*C*a^7*b^5 + 1024*A^2*C*a^8*b^4 - 1824*A^2*C*a^9*b^3 + 1152*B*C^2*a^5*
b^7 + 5888*B*C^2*a^7*b^5 - 1056*B*C^2*a^8*b^4 + 7168*B*C^2*a^9*b^3 - 2432*B*C^2*a^10*b^2 + 528*B^2*C*a^4*b^8 +
7552*B^2*C*a^6*b^6 - 2304*B^2*C*a^7*b^5 + 14592*B^2*C*a^8*b^4 - 7168*B^2*C*a^9*b^3 + 768*A*B*C*a^3*b^9 + 1516
8*A*B*C*a^5*b^7 - 6528*A*B*C*a^6*b^6 + 30592*A*B*C*a^7*b^5 - 19488*A*B*C*a^8*b^4 + 1536*A*B*C*a^9*b^3 - 1408*A
*B*C*a^10*b^2))*(A*a^2 + 12*A*b^2 + 2*C*a^2 + 8*B*a*b))/d