3.792 \(\int \cos ^6(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=221 \[ -\frac{a \left (4 a^2 B+15 a b C+12 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (15 a^2 b C+4 a^3 B+14 a b^2 B+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right )+\frac{a^2 (5 a C+7 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a B \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[Out]

((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*x)/8 + ((4*a^3*B + 14*a*b^2*B + 15*a^2*b*C + 5*b^3*C)*Sin[c + d*
x])/(5*d) + ((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*(7*b*B + 5*a
*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a*B*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) - (a*
(4*a^2*B + 12*b^2*B + 15*a*b*C)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.548044, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4072, 4025, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{a \left (4 a^2 B+15 a b C+12 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (15 a^2 b C+4 a^3 B+14 a b^2 B+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right )+\frac{a^2 (5 a C+7 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a B \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*x)/8 + ((4*a^3*B + 14*a*b^2*B + 15*a^2*b*C + 5*b^3*C)*Sin[c + d*
x])/(5*d) + ((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*(7*b*B + 5*a
*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a*B*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) - (a*
(4*a^2*B + 12*b^2*B + 15*a*b*C)*Sin[c + d*x]^3)/(15*d)

Rule 4072

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

Rule 4025

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

Rule 4074

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_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (-a (7 b B+5 a C)-\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sec (c+d x)-b (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )+5 \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \sec (c+d x)+4 b^2 (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )+4 b^2 (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{4} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos (c+d x) \left (4 b^2 (2 a B+5 b C)+4 a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) x+\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{\operatorname{Subst}\left (\int \left (4 b^2 (2 a B+5 b C)+4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )-4 a \left (4 a^2 B+12 b^2 B+15 a b C\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac{1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) x+\frac{\left (4 a^3 B+14 a b^2 B+15 a^2 b C+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.667971, size = 176, normalized size = 0.8 \[ \frac{60 (c+d x) \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right )+10 a \left (5 a^2 B+12 a b C+12 b^2 B\right ) \sin (3 (c+d x))+60 \left (18 a^2 b C+5 a^3 B+18 a b^2 B+8 b^3 C\right ) \sin (c+d x)+120 \left (3 a^2 b B+a^3 C+3 a b^2 C+b^3 B\right ) \sin (2 (c+d x))+15 a^2 (a C+3 b B) \sin (4 (c+d x))+6 a^3 B \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(60*(9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*(c + d*x) + 60*(5*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 8*b^3*C)*
Sin[c + d*x] + 120*(3*a^2*b*B + b^3*B + a^3*C + 3*a*b^2*C)*Sin[2*(c + d*x)] + 10*a*(5*a^2*B + 12*b^2*B + 12*a*
b*C)*Sin[3*(c + d*x)] + 15*a^2*(3*b*B + a*C)*Sin[4*(c + d*x)] + 6*a^3*B*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.08, size = 227, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{B{a}^{3}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+{a}^{3}C \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,B{a}^{2}b \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{2}bC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +Ba{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,Ca{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +B{b}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +C{b}^{3}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/5*B*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^3*C*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+3*B*a^2*b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+a^2*b*C*(2+cos(d*x+c
)^2)*sin(d*x+c)+B*a*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+3*C*a*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+B*b^3*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+C*b^3*sin(d*x+c))

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Maxima [A]  time = 0.97372, size = 293, normalized size = 1.33 \begin{align*} \frac{32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} + 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 45 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 480 \, C b^{3} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c
) + 8*sin(2*d*x + 2*c))*C*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2*b - 480*(sin(
d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b^2 + 360*(2*d*x + 2*c + sin(
2*d*x + 2*c))*C*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^3 + 480*C*b^3*sin(d*x + c))/d

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Fricas [A]  time = 0.537056, size = 423, normalized size = 1.91 \begin{align*} \frac{15 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} d x +{\left (24 \, B a^{3} \cos \left (d x + c\right )^{4} + 64 \, B a^{3} + 240 \, C a^{2} b + 240 \, B a b^{2} + 120 \, C b^{3} + 30 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (4 \, B a^{3} + 15 \, C a^{2} b + 15 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*d*x + (24*B*a^3*cos(d*x + c)^4 + 64*B*a^3 + 240*C*a^2*b
+ 240*B*a*b^2 + 120*C*b^3 + 30*(C*a^3 + 3*B*a^2*b)*cos(d*x + c)^3 + 8*(4*B*a^3 + 15*C*a^2*b + 15*B*a*b^2)*cos
(d*x + c)^2 + 15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.27143, size = 907, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*(d*x + c) + 2*(120*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*C*
a^3*tan(1/2*d*x + 1/2*c)^9 - 225*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*b^3
*tan(1/2*d*x + 1/2*c)^9 + 160*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 30*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*B*a^2*b*tan(
1/2*d*x + 1/2*c)^7 + 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*C*a*b^2*tan
(1/2*d*x + 1/2*c)^7 - 120*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*B*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*b^3*tan(1/2
*d*x + 1/2*c)^5 + 160*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*B*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 + 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 360*C*a*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^3*tan(1/2*d*x + 1/
2*c) + 75*C*a^3*tan(1/2*d*x + 1/2*c) + 225*B*a^2*b*tan(1/2*d*x + 1/2*c) + 360*C*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*B*a*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*B*b^3*tan(1/2*d*x + 1/2*c) + 120*C*b^3
*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d