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

Optimal. Leaf size=207 \[ \frac{\sin (c+d x) \left (16 a^2 b B+3 a^3 C+6 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{b \sin (c+d x) \cos (c+d x) \left (6 a^2 C+20 a b B+12 A b^2+9 b^2 C\right )}{24 d}+\frac{1}{8} x \left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right )+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{(3 a C+4 b B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

((8*a^3*B + 12*a*b^2*B + 12*a^2*b*(2*A + C) + b^3*(4*A + 3*C))*x)/8 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + ((16*a
^2*b*B + 4*b^3*B + 3*a^3*C + 6*a*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) + (b*(12*A*b^2 + 20*a*b*B + 6*a^2*C + 9*
b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*b*B + 3*a*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (C*(
a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.563557, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3049, 3033, 3023, 2735, 3770} \[ \frac{\sin (c+d x) \left (16 a^2 b B+3 a^3 C+6 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 d}+\frac{b \sin (c+d x) \cos (c+d x) \left (6 a^2 C+20 a b B+12 A b^2+9 b^2 C\right )}{24 d}+\frac{1}{8} x \left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right )+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{(3 a C+4 b B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

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],x]

[Out]

((8*a^3*B + 12*a*b^2*B + 12*a^2*b*(2*A + C) + b^3*(4*A + 3*C))*x)/8 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + ((16*a
^2*b*B + 4*b^3*B + 3*a^3*C + 6*a*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) + (b*(12*A*b^2 + 20*a*b*B + 6*a^2*C + 9*
b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*b*B + 3*a*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (C*(
a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

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 (c+d x) \, dx &=\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (4 a A+(4 A b+4 a B+3 b C) \cos (c+d x)+(4 b B+3 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (12 a^2 A+\left (24 a A b+12 a^2 B+8 b^2 B+15 a b C\right ) \cos (c+d x)+\left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^3 A+3 \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \cos (c+d x)+4 \left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{\left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^3 A+3 \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{8} \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) x+\frac{\left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} \left (8 a^3 B+12 a b^2 B+12 a^2 b (2 A+C)+b^3 (4 A+3 C)\right ) x+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\left (16 a^2 b B+4 b^3 B+3 a^3 C+6 a b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+\frac{b \left (12 A b^2+20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.934392, size = 218, normalized size = 1.05 \[ \frac{12 (c+d x) \left (12 a^2 b (2 A+C)+8 a^3 B+12 a b^2 B+b^3 (4 A+3 C)\right )+24 \sin (c+d x) \left (12 a^2 b B+4 a^3 C+3 a b^2 (4 A+3 C)+3 b^3 B\right )+24 b \sin (2 (c+d x)) \left (3 a^2 C+3 a b B+A b^2+b^2 C\right )-96 a^3 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 a^3 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 b^2 (3 a C+b B) \sin (3 (c+d x))+3 b^3 C \sin (4 (c+d x))}{96 d} \]

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],x]

[Out]

(12*(8*a^3*B + 12*a*b^2*B + 12*a^2*b*(2*A + C) + b^3*(4*A + 3*C))*(c + d*x) - 96*a^3*A*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + 96*a^3*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 24*(12*a^2*b*B + 3*b^3*B + 4*a^3*C + 3
*a*b^2*(4*A + 3*C))*Sin[c + d*x] + 24*b*(A*b^2 + 3*a*b*B + 3*a^2*C + b^2*C)*Sin[2*(c + d*x)] + 8*b^2*(b*B + 3*
a*C)*Sin[3*(c + d*x)] + 3*b^3*C*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.061, size = 362, normalized size = 1.8 \begin{align*}{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+3\,A{a}^{2}bx+{\frac{A{b}^{3}c}{2\,d}}+{\frac{3\,C{b}^{3}c}{8\,d}}+{\frac{3\,{a}^{2}bCx}{2}}+{\frac{2\,{b}^{3}B\sin \left ( dx+c \right ) }{3\,d}}+{\frac{3\,a{b}^{2}Bx}{2}}+{\frac{B{a}^{3}c}{d}}+{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{b}^{3}Cx}{8}}+{\frac{3\,a{b}^{2}Bc}{2\,d}}+{\frac{3\,C{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+2\,{\frac{Ca{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}bCc}{2\,d}}+{\frac{C{b}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+3\,{\frac{aA{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bB\sin \left ( dx+c \right ) }{d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{3\,d}}+{\frac{A\cos \left ( dx+c \right ){b}^{3}\sin \left ( dx+c \right ) }{2\,d}}+3\,{\frac{A{a}^{2}bc}{d}}+{a}^{3}Bx+{\frac{3\,B\cos \left ( dx+c \right ) a{b}^{2}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}+{\frac{3\,C\cos \left ( dx+c \right ){a}^{2}b\sin \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{3}x}{2}} \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),x)

[Out]

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

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Maxima [A]  time = 0.994961, size = 323, normalized size = 1.56 \begin{align*} \frac{96 \,{\left (d x + c\right )} B a^{3} + 288 \,{\left (d x + c\right )} A a^{2} b + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} - 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{2} + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{3} + 3 \,{\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 b^{3} + 96 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, C a^{3} \sin \left (d x + c\right ) + 288 \, B a^{2} b \sin \left (d x + c\right ) + 288 \, A a b^{2} \sin \left (d x + c\right )}{96 \, 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),x, algorithm="maxima")

[Out]

1/96*(96*(d*x + c)*B*a^3 + 288*(d*x + c)*A*a^2*b + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b + 72*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a*b^2 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a*b^2 + 24*(2*d*x + 2*c + sin(2*d*x +
2*c))*A*b^3 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^3 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x +
2*c))*C*b^3 + 96*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 96*C*a^3*sin(d*x + c) + 288*B*a^2*b*sin(d*x + c) +
288*A*a*b^2*sin(d*x + c))/d

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Fricas [A]  time = 1.9642, size = 463, normalized size = 2.24 \begin{align*} \frac{12 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (8 \, B a^{3} + 12 \,{\left (2 \, A + C\right )} a^{2} b + 12 \, B a b^{2} +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} d x +{\left (6 \, C b^{3} \cos \left (d x + c\right )^{3} + 24 \, C a^{3} + 72 \, B a^{2} b + 24 \,{\left (3 \, A + 2 \, C\right )} a b^{2} + 16 \, B b^{3} + 8 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (12 \, C a^{2} b + 12 \, B a b^{2} +{\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, 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),x, algorithm="fricas")

[Out]

1/24*(12*A*a^3*log(sin(d*x + c) + 1) - 12*A*a^3*log(-sin(d*x + c) + 1) + 3*(8*B*a^3 + 12*(2*A + C)*a^2*b + 12*
B*a*b^2 + (4*A + 3*C)*b^3)*d*x + (6*C*b^3*cos(d*x + c)^3 + 24*C*a^3 + 72*B*a^2*b + 24*(3*A + 2*C)*a*b^2 + 16*B
*b^3 + 8*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^2 + 3*(12*C*a^2*b + 12*B*a*b^2 + (4*A + 3*C)*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((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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Giac [B]  time = 1.22459, size = 976, normalized size = 4.71 \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),x, algorithm="giac")

[Out]

1/24*(24*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(8*B*a^3 +
24*A*a^2*b + 12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 3*C*b^3)*(d*x + c) + 2*(24*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 72
*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 36*B
*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 72*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*B*b^3
*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 216*B*a^2*b*tan(
1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 216*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 36*B*a*b^2*tan(1
/2*d*x + 1/2*c)^5 + 120*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 40*B*b^3*tan(1/2*d*
x + 1/2*c)^5 + 9*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 216*B*a^2*b*tan(1/2*d*x + 1/
2*c)^3 + 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 216*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 36*B*a*b^2*tan(1/2*d*x + 1/2
*c)^3 + 120*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 40*B*b^3*tan(1/2*d*x + 1/2*c)^3
- 9*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^3*tan(1/2*d*x + 1/2*c) + 72*B*a^2*b*tan(1/2*d*x + 1/2*c) + 36*C*a^2
*b*tan(1/2*d*x + 1/2*c) + 72*A*a*b^2*tan(1/2*d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 72*C*a*b^2*tan(1
/2*d*x + 1/2*c) + 12*A*b^3*tan(1/2*d*x + 1/2*c) + 24*B*b^3*tan(1/2*d*x + 1/2*c) + 15*C*b^3*tan(1/2*d*x + 1/2*c
))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d