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

Optimal. Leaf size=204 \[ -\frac{b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac{a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac{b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac{(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.645678, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3047, 3033, 3023, 2735, 3770} \[ -\frac{b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac{a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac{b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac{(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

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

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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x))^2 \left (3 A b+2 a B+(2 b B+a (A+2 C)) \cos (c+d x)-2 b (A-C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x)) \left (6 A b^2+6 a b B+a^2 (A+2 C)-b (a A-2 b B-4 a C) \cos (c+d x)-2 b (4 A b+2 a B-b C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{4} \int \left (2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \cos (c+d x)-2 b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{4} \int \left (2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) x-\frac{b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) x+\frac{a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac{b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 3.20042, size = 318, normalized size = 1.56 \[ \frac{2 b (c+d x) \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-2 a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 a^2 (a B+3 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a^2 (a B+3 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{a^3 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^3 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+4 b^2 (3 a C+b B) \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*(c + d*x) - 2*a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] + 2*a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] +
(a^3*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]) - (a^3*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[(c + d*x)/2])/(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*b^2*(b*B + 3*a*C)*Sin[c + d*x] + b^3*C*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.076, size = 267, normalized size = 1.3 \begin{align*} A{b}^{3}x+{\frac{A{b}^{3}c}{d}}+{\frac{{b}^{3}B\sin \left ( dx+c \right ) }{d}}+{\frac{C{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}Cx}{2}}+{\frac{C{b}^{3}c}{2\,d}}+3\,{\frac{aA{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,a{b}^{2}Bx+3\,{\frac{Ba{b}^{2}c}{d}}+3\,{\frac{Ca{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}b\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{a}^{2}bCx+3\,{\frac{C{a}^{2}bc}{d}}+{\frac{A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00006, size = 328, normalized size = 1.61 \begin{align*} \frac{12 \,{\left (d x + c\right )} C a^{2} b + 12 \,{\left (d x + c\right )} B a b^{2} + 4 \,{\left (d x + c\right )} A b^{3} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} - A 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 )} + 2 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A 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 )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 4 \, B b^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, A a^{2} b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*(d*x + c)*C*a^2*b + 12*(d*x + c)*B*a*b^2 + 4*(d*x + c)*A*b^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^3
- A*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)) + 2*C*a^3*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*A*a*
b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*b^2*sin(d*x + c) + 4*B*b^3*sin(d*x + c) + 4*B*a^3
*tan(d*x + c) + 12*A*a^2*b*tan(d*x + c))/d

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Fricas [A]  time = 1.92889, size = 500, normalized size = 2.45 \begin{align*} \frac{2 \,{\left (6 \, C a^{2} b + 6 \, B a b^{2} +{\left (2 \, A + C\right )} b^{3}\right )} d x \cos \left (d x + c\right )^{2} +{\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C b^{3} \cos \left (d x + c\right )^{3} + A a^{3} + 2 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.30229, size = 726, normalized size = 3.56 \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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/2*((6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*(d*x + c) + (A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(t
an(1/2*d*x + 1/2*c) + 1)) - (A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
A*a^3*tan(1/2*d*x + 1/2*c)^7 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*t
an(1/2*d*x + 1/2*c)^7 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^3*tan(1/2*d*x +
1/2*c)^5 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^
5 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3
*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*b^3*tan(1/
2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 + A*a^3*tan(1/2*d*x + 1/2*c) + 2*B*a^3*tan(1/2*d*x + 1/2*c)
+ 6*A*a^2*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b^2*tan(1/2*d*x + 1/2*c) + 2*B*b^3*tan(1/2*d*x + 1/2*c) + C*b^3*tan(1
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d