3.546 \(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=182 \[ \frac{b \left (a^2 (4 A+6 C)+A b^2\right ) \tan (c+d x)}{2 d}+\frac{a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^3 C x \]

[Out]

b^3*C*x + (a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (b*(A*b^2 + a^2*(4*A + 6*C))*
Tan[c + d*x])/(2*d) + (a*(2*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (A*b*(a + b*Cos[c + d*
x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(4*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.599905, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3047, 3031, 3021, 2735, 3770} \[ \frac{b \left (a^2 (4 A+6 C)+A b^2\right ) \tan (c+d x)}{2 d}+\frac{a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^3 C x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

b^3*C*x + (a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (b*(A*b^2 + a^2*(4*A + 6*C))*
Tan[c + d*x])/(2*d) + (a*(2*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (A*b*(a + b*Cos[c + d*
x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(4*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3047

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

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (3 A b+a (3 A+4 C) \cos (c+d x)+4 b C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (3 \left (2 A b^2+a^2 (3 A+4 C)\right )+3 a b (5 A+8 C) \cos (c+d x)+12 b^2 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)-24 b^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )-24 b^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 C x+\frac{b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=b^3 C x+\frac{a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.942651, size = 127, normalized size = 0.7 \[ \frac{a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (a \left (a^2 (3 A+4 C)+12 A b^2\right ) \sec (c+d x)+8 b \left (3 a^2 (A+C)+A b^2\right )+2 a^3 A \sec ^3(c+d x)\right )+8 a^2 A b \tan ^3(c+d x)+8 b^3 C d x}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(8*b^3*C*d*x + a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + (8*b*(A*b^2 + 3*a^2*(A + C)) + a
*(12*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x] + 2*a^3*A*Sec[c + d*x]^3)*Tan[c + d*x] + 8*a^2*A*b*Tan[c + d*x]^3)/
(8*d)

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Maple [A]  time = 0.061, size = 267, normalized size = 1.5 \begin{align*}{\frac{A{b}^{3}\tan \left ( dx+c \right ) }{d}}+{b}^{3}Cx+{\frac{C{b}^{3}c}{d}}+{\frac{3\,aA{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,aA{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{A{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+3\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

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

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Maxima [A]  time = 1.02143, size = 352, normalized size = 1.93 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b + 16 \,{\left (d x + c\right )} C b^{3} - A a^{3}{\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 )} - 4 \, C a^{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 )} - 12 \, A a b^{2}{\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 b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a^{2} b \tan \left (d x + c\right ) + 16 \, A b^{3} \tan \left (d x + c\right )}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2*b + 16*(d*x + c)*C*b^3 - A*a^3*(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)) - 4*C*a^
3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*A*a*b^2*(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*b^2*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 48*C*a^2*b*tan(d*x + c) + 16*A*b^3*tan(d*x + c))/d

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Fricas [A]  time = 1.55796, size = 485, normalized size = 2.66 \begin{align*} \frac{16 \, C b^{3} d x \cos \left (d x + c\right )^{4} +{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \,{\left (A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \,{\left (A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \, A a^{2} b \cos \left (d x + c\right ) + 2 \, A a^{3} + 8 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{2} b + A b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/16*(16*C*b^3*d*x*cos(d*x + c)^4 + ((3*A + 4*C)*a^3 + 12*(A + 2*C)*a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1
) - ((3*A + 4*C)*a^3 + 12*(A + 2*C)*a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*A*a^2*b*cos(d*x + c) +
2*A*a^3 + 8*((2*A + 3*C)*a^2*b + A*b^3)*cos(d*x + c)^3 + ((3*A + 4*C)*a^3 + 12*A*a*b^2)*cos(d*x + c)^2)*sin(d
*x + c))/(d*cos(d*x + c)^4)

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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((a+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.37252, size = 710, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)*C*b^3 + (3*A*a^3 + 4*C*a^3 + 12*A*a*b^2 + 24*C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (3
*A*a^3 + 4*C*a^3 + 12*A*a*b^2 + 24*C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^3*tan(1/2*d*x + 1/2*
c)^7 + 4*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7
+ 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 8*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 4*C*a^
3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*a*b^2*
tan(1/2*d*x + 1/2*c)^5 + 24*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^3*tan(1/2*d*
x + 1/2*c)^3 - 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x
+ 1/2*c)^3 - 24*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^3*tan(1/2*d*x + 1/2*c) + 4*C*a^3*tan(1/2*d*x + 1/2*c) + 2
4*A*a^2*b*tan(1/2*d*x + 1/2*c) + 24*C*a^2*b*tan(1/2*d*x + 1/2*c) + 12*A*a*b^2*tan(1/2*d*x + 1/2*c) + 8*A*b^3*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d