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

Optimal. Leaf size=200 \[ \frac{b \left (34 a^2 A b+19 a^3 B+16 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d}+\frac{b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right )+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b (7 a B+4 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{b B \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.547011, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2990, 3049, 3033, 3023, 2735, 3770} \[ \frac{b \left (34 a^2 A b+19 a^3 B+16 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d}+\frac{b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right )+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b (7 a B+4 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{b B \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2990

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

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

Mathematica [A]  time = 0.580132, size = 210, normalized size = 1.05 \[ \frac{12 (c+d x) \left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right )+24 b \left (24 a^2 A b+16 a^3 B+12 a b^2 B+3 A b^3\right ) \sin (c+d x)+24 b^2 \left (6 a^2 B+4 a A b+b^2 B\right ) \sin (2 (c+d x))-96 a^4 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 a^4 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 b^3 (4 a B+A b) \sin (3 (c+d x))+3 b^4 B \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 319, normalized size = 1.6 \begin{align*}{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{4}Bx+{\frac{B{a}^{4}c}{d}}+4\,A{a}^{3}bx+4\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{B{a}^{3}b\sin \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{2}{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+3\,B{a}^{2}{b}^{2}x+3\,{\frac{B{a}^{2}{b}^{2}c}{d}}+2\,{\frac{Aa{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,Aa{b}^{3}x+2\,{\frac{Aa{b}^{3}c}{d}}+{\frac{4\,B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}a{b}^{3}}{3\,d}}+{\frac{8\,Ba{b}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{4}}{3\,d}}+{\frac{2\,A{b}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,B{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{4}Bx}{8}}+{\frac{3\,B{b}^{4}c}{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12496, size = 281, normalized size = 1.4 \begin{align*} \frac{96 \,{\left (d x + c\right )} B a^{4} + 384 \,{\left (d x + c\right )} A a^{3} b + 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b^{2} + 96 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} - 128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{3} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{4} + 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 )} B b^{4} + 96 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 384 \, B a^{3} b \sin \left (d x + c\right ) + 576 \, A a^{2} 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))^4*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54827, size = 447, normalized size = 2.23 \begin{align*} \frac{12 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} d x +{\left (6 \, B b^{4} \cos \left (d x + c\right )^{3} + 96 \, B a^{3} b + 144 \, A a^{2} b^{2} + 64 \, B a b^{3} + 16 \, A b^{4} + 8 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\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))^4*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.53986, size = 814, normalized size = 4.07 \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))^4*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="giac")

[Out]

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