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

Optimal. Leaf size=154 \[ \frac{\left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{2 a b (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a A b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.42167, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3048, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{\left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{2 a b (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a A b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a*b*(2*A + 3*C)*Tan[c + d*x])/(3*d) + (
(2*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*A*b*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + (A*
(a + b*Cos[c + d*x])^2*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 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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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]

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))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x)) \left (2 A b+a (3 A+4 C) \cos (c+d x)+b (A+4 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{12} \int \left (-3 \left (2 A b^2+a^2 (3 A+4 C)\right )-8 a b (2 A+3 C) \cos (c+d x)-3 b^2 (A+4 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-16 a b (2 A+3 C)-3 \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{3} (2 a b (2 A+3 C)) \int \sec ^2(c+d x) \, dx-\frac{1}{8} \left (-4 b^2 (A+2 C)-a^2 (3 A+4 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{(2 a b (2 A+3 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{\left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{2 a b (2 A+3 C) \tan (c+d x)}{3 d}+\frac{\left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.618409, size = 107, normalized size = 0.69 \[ \frac{3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sec (c+d x)+6 a^2 A \sec ^3(c+d x)+16 a b \left (A \tan ^2(c+d x)+3 (A+C)\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(4*A*b^2 + a^2*(3*A + 4*C))*Sec
[c + d*x] + 6*a^2*A*Sec[c + d*x]^3 + 16*a*b*(3*(A + C) + A*Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.059, size = 229, normalized size = 1.5 \begin{align*}{\frac{A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,aAb\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,aAb \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+2\,{\frac{abC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}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))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

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

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Maxima [A]  time = 1.0178, size = 313, normalized size = 2.03 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A a^{2}{\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 )} - 12 \, C a^{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 )} - 12 \, 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 b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, C a b \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b - 3*A*a^2*(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)) - 12*C*a^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*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*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) -
1)) + 96*C*a*b*tan(d*x + c))/d

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Fricas [A]  time = 1.66438, size = 424, normalized size = 2.75 \begin{align*} \frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \,{\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \,{\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (2 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 16 \, A a b \cos \left (d x + c\right ) + 6 \, A a^{2} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, 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))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*((3*A + 4*C)*a^2 + 4*(A + 2*C)*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*((3*A + 4*C)*a^2 + 4*(A +
2*C)*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(2*A + 3*C)*a*b*cos(d*x + c)^3 + 16*A*a*b*cos(d*x + c
) + 6*A*a^2 + 3*((3*A + 4*C)*a^2 + 4*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))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.4491, size = 575, normalized size = 3.73 \begin{align*} \frac{3 \,{\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 80 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 144 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 144 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a^2 + 4*C*a^2 + 4*A*b^2 + 8*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^2 + 4*C*a^2 + 4*
A*b^2 + 8*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^2*tan(1/2*d*
x + 1/2*c)^7 - 48*A*a*b*tan(1/2*d*x + 1/2*c)^7 - 48*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan(1/2*d*x + 1/2*
c)^7 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 14
4*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*t
an(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 144*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*
d*x + 1/2*c)^3 + 15*A*a^2*tan(1/2*d*x + 1/2*c) + 12*C*a^2*tan(1/2*d*x + 1/2*c) + 48*A*a*b*tan(1/2*d*x + 1/2*c)
+ 48*C*a*b*tan(1/2*d*x + 1/2*c) + 12*A*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d