3.308 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=216 \[ \frac{a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac{a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{b^4 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

((3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*x)/8 + (b^4*B*ArcTanh[Sin[c + d*x]])/d + (a*(16*
a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^2*B)*Sin[c + d*x])/(6*d) + (a^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Cos[c +
d*x]*Sin[c + d*x])/(24*d) + (a*(7*A*b + 4*a*B)*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(12*d) + (a
*A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.609493, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4025, 4094, 4074, 4047, 8, 4045, 3770} \[ \frac{a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac{a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{b^4 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

((3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*x)/8 + (b^4*B*ArcTanh[Sin[c + d*x]])/d + (a*(16*
a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^2*B)*Sin[c + d*x])/(6*d) + (a^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Cos[c +
d*x]*Sin[c + d*x])/(24*d) + (a*(7*A*b + 4*a*B)*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(12*d) + (a
*A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 4025

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

Rule 4094

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

Rule 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 8

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

Rule 4045

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

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 \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (-a (7 A b+4 a B)-\left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x)-4 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{12} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (-a \left (9 a^2 A+26 A b^2+32 a b B\right )-\left (23 a^2 A b+12 A b^3+8 a^3 B+36 a b^2 B\right ) \sec (c+d x)-12 b^3 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{24} \int \cos (c+d x) \left (4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )+3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \sec (c+d x)+24 b^4 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{24} \int \cos (c+d x) \left (4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )+24 b^4 B \sec ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) x+\frac{a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\left (b^4 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) x+\frac{b^4 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac{a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.601376, size = 210, normalized size = 0.97 \[ \frac{12 (c+d x) \left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right )+24 a^2 \left (a^2 A+4 a b B+6 A b^2\right ) \sin (2 (c+d x))+24 a \left (12 a^2 A b+3 a^3 B+24 a b^2 B+16 A b^3\right ) \sin (c+d x)+8 a^3 (a B+4 A b) \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))-96 b^4 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 b^4 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(12*(3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*(c + d*x) - 96*b^4*B*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 96*b^4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 24*a*(12*a^2*A*b + 16*A*b^3 + 3*a^3*B +
24*a*b^2*B)*Sin[c + d*x] + 24*a^2*(a^2*A + 6*A*b^2 + 4*a*b*B)*Sin[2*(c + d*x)] + 8*a^3*(4*A*b + a*B)*Sin[3*(c
+ d*x)] + 3*a^4*A*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.071, size = 319, normalized size = 1.5 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{4}Ax}{8}}+{\frac{3\,A{a}^{4}c}{8\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{2\,B{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}b}{3\,d}}+{\frac{8\,A{a}^{3}b\sin \left ( dx+c \right ) }{3\,d}}+2\,{\frac{B{a}^{3}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+2\,B{a}^{3}bx+2\,{\frac{B{a}^{3}bc}{d}}+3\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+3\,A{a}^{2}{b}^{2}x+3\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{B{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\sin \left ( dx+c \right ) }{d}}+4\,Ba{b}^{3}x+4\,{\frac{Ba{b}^{3}c}{d}}+A{b}^{4}x+{\frac{A{b}^{4}c}{d}}+{\frac{B{b}^{4}\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(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/4/d*A*a^4*sin(d*x+c)*cos(d*x+c)^3+3/8/d*A*a^4*sin(d*x+c)*cos(d*x+c)+3/8*a^4*A*x+3/8/d*A*a^4*c+1/3/d*B*sin(d*
x+c)*cos(d*x+c)^2*a^4+2/3/d*B*a^4*sin(d*x+c)+4/3/d*A*sin(d*x+c)*cos(d*x+c)^2*a^3*b+8/3/d*A*a^3*b*sin(d*x+c)+2/
d*B*a^3*b*sin(d*x+c)*cos(d*x+c)+2*B*a^3*b*x+2/d*B*a^3*b*c+3/d*A*a^2*b^2*sin(d*x+c)*cos(d*x+c)+3*A*a^2*b^2*x+3/
d*A*a^2*b^2*c+6/d*B*a^2*b^2*sin(d*x+c)+4/d*A*a*b^3*sin(d*x+c)+4*B*a*b^3*x+4/d*B*a*b^3*c+A*b^4*x+1/d*A*b^4*c+1/
d*B*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.9832, size = 290, normalized size = 1.34 \begin{align*} \frac{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 )} A a^{4} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} b + 96 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b + 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 384 \,{\left (d x + c\right )} B a b^{3} + 96 \,{\left (d x + c\right )} A b^{4} + 48 \, B b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 576 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3*b + 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3*b + 144*(2*d*
x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 + 384*(d*x + c)*B*a*b^3 + 96*(d*x + c)*A*b^4 + 48*B*b^4*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 576*B*a^2*b^2*sin(d*x + c) + 384*A*a*b^3*sin(d*x + c))/d

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Fricas [A]  time = 0.594929, size = 447, normalized size = 2.07 \begin{align*} \frac{12 \, B b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, B b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} d x +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 16 \, B a^{4} + 64 \, A a^{3} b + 144 \, B a^{2} b^{2} + 96 \, A a b^{3} + 8 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\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(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(12*B*b^4*log(sin(d*x + c) + 1) - 12*B*b^4*log(-sin(d*x + c) + 1) + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^
2 + 32*B*a*b^3 + 8*A*b^4)*d*x + (6*A*a^4*cos(d*x + c)^3 + 16*B*a^4 + 64*A*a^3*b + 144*B*a^2*b^2 + 96*A*a*b^3 +
8*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2)*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(cos(d*x+c)**4*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.30325, size = 814, normalized size = 3.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(24*B*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(3*A*a^4 +
16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*(d*x + c) - 2*(15*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^4*t
an(1/2*d*x + 1/2*c)^7 - 96*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 72*A*a^2*b^2*t
an(1/2*d*x + 1/2*c)^7 - 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 96*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^4*tan
(1/2*d*x + 1/2*c)^5 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 48*B*a^3*b*tan(1/
2*d*x + 1/2*c)^5 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*A*a*b^3*ta
n(1/2*d*x + 1/2*c)^5 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 160*A*a^3*b*tan(1/2*
d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*a^2*b^2*tan(1
/2*d*x + 1/2*c)^3 - 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^4*tan(1/2*d*x + 1/2*c) - 24*B*a^4*tan(1/2*d*x
+ 1/2*c) - 96*A*a^3*b*tan(1/2*d*x + 1/2*c) - 48*B*a^3*b*tan(1/2*d*x + 1/2*c) - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*
c) - 144*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 96*A*a*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d