3.646 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=226 \[ \frac{\left (2 a^2 b^2 (5 A+3 C)+a^4 C+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac{a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac{a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]

[Out]

(a*b*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((a^4*C + 2*a^2*b^2*(5*A + 3*C) + 2*b^4*(5*A + 4*C))*Tan[c + d
*x])/(15*b^2*d) + (a*(20*A*b^2 + 2*a^2*C + 13*b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(60*b*d) + ((a^2*C + 2*b^2*(5*
A + 4*C))*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(30*b^2*d) - (a*C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(10*b^2*
d) + (C*Sec[c + d*x]*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*b*d)

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Rubi [A]  time = 0.502938, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4093, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{\left (2 a^2 b^2 (5 A+3 C)+a^4 C+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac{a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac{a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac{C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((a^4*C + 2*a^2*b^2*(5*A + 3*C) + 2*b^4*(5*A + 4*C))*Tan[c + d
*x])/(15*b^2*d) + (a*(20*A*b^2 + 2*a^2*C + 13*b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(60*b*d) + ((a^2*C + 2*b^2*(5*
A + 4*C))*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(30*b^2*d) - (a*C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(10*b^2*
d) + (C*Sec[c + d*x]*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*b*d)

Rule 4093

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))
^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[
1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - 2*a
*C*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4002

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

Rule 3997

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] + Dist[1/(n + 1), Int[(d*C
sc[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]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (-2 a b C+2 \left (a^2 C+2 b^2 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (2 b \left (20 A b^2-a^2 C+16 b^2 C\right )+2 a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=\frac{a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) \left (30 a b^3 (4 A+3 C)+8 \left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac{a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac{1}{4} (a b (4 A+3 C)) \int \sec (c+d x) \, dx+\frac{\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{15 b^2}\\ &=\frac{a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}-\frac{\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 b^2 d}\\ &=\frac{a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac{a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac{\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac{C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 2.29767, size = 275, normalized size = 1.22 \[ -\frac{\sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (60 a b (4 A+3 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sin (c+d x) \left (24 \left (5 a^2 (A+C)+b^2 (5 A+4 C)\right ) \cos (2 (c+d x))+30 a^2 A \cos (4 (c+d x))+90 a^2 A+20 a^2 C \cos (4 (c+d x))+100 a^2 C+15 a b (12 A+17 C) \cos (c+d x)+60 a A b \cos (3 (c+d x))+45 a b C \cos (3 (c+d x))+20 A b^2 \cos (4 (c+d x))+100 A b^2+16 b^2 C \cos (4 (c+d x))+128 b^2 C\right )\right )}{120 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C + A*Cos[c + d*x]^2)*Sec[c + d*x]^5*(60*a*b*(4*A + 3*C)*Cos[c + d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*
x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - (90*a^2*A + 100*A*b^2 + 100*a^2*C + 128*b^2*C + 15*a*b*(1
2*A + 17*C)*Cos[c + d*x] + 24*(5*a^2*(A + C) + b^2*(5*A + 4*C))*Cos[2*(c + d*x)] + 60*a*A*b*Cos[3*(c + d*x)] +
45*a*b*C*Cos[3*(c + d*x)] + 30*a^2*A*Cos[4*(c + d*x)] + 20*A*b^2*Cos[4*(c + d*x)] + 20*a^2*C*Cos[4*(c + d*x)]
+ 16*b^2*C*Cos[4*(c + d*x)])*Sin[c + d*x]))/(120*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.044, size = 257, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Aab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{Aab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{abC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,abC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{2\,A{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{8\,{b}^{2}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*a^2*A*tan(d*x+c)+2/3/d*a^2*C*tan(d*x+c)+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+1/d*A*a*b*sec(d*x+c)*tan(d*x+c
)+1/d*A*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a*b*C*tan(d*x+c)*sec(d*x+c)^3+3/4*a*b*C*sec(d*x+c)*tan(d*x+c)/d+3/
4/d*a*b*C*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*A*b^2*tan(d*x+c)+1/3/d*A*b^2*tan(d*x+c)*sec(d*x+c)^2+8/15*b^2*C*tan(
d*x+c)/d+1/5/d*b^2*C*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^2*C*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.0013, size = 292, normalized size = 1.29 \begin{align*} \frac{40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 8 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 15 \, C a b{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b{\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 )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 + 8*(3*tan(d*x
+ c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^2 - 15*C*a*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x
+ c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*A*a*b*(2*sin(d*x + c
)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*A*a^2*tan(d*x + c))/d

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Fricas [A]  time = 0.539279, size = 455, normalized size = 2.01 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \,{\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \,{\left (3 \, A + 2 \, C\right )} a^{2} + 2 \,{\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right ) + 12 \, C b^{2} + 4 \,{\left (5 \, C a^{2} +{\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*A + 3*C)*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a*b*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(15*(4*A + 3*C)*a*b*cos(d*x + c)^3 + 4*(5*(3*A + 2*C)*a^2 + 2*(5*A + 4*C)*b^2)*cos(d*x + c)^4
+ 30*C*a*b*cos(d*x + c) + 12*C*b^2 + 4*(5*C*a^2 + (5*A + 4*C)*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**2*sec(c + d*x)**2, x)

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Giac [B]  time = 1.20978, size = 718, normalized size = 3.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(4*A*a*b + 3*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a*b + 3*C*a*b)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b*tan(1/2*d*x +
1/2*c)^9 - 75*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b^2*tan(1/2*d*x + 1/2*c)^9
- 240*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 30
*C*a*b*tan(1/2*d*x + 1/2*c)^7 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 80*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2
*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 232*C*b^2*tan(
1/2*d*x + 1/2*c)^5 - 240*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 120*A*a*b*tan(1/2*d
*x + 1/2*c)^3 - 30*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 80*C*b^2*tan(1/2*d*x + 1/
2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b*tan(1/2*d*x + 1/2*c) + 75*C*
a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c
)^2 - 1)^5)/d