3.245 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=209 \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \sin (c+d x)}{2 d}+\frac{a^2 \left (a^2 A+8 a b B+12 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} b^2 x \left (12 a^2 B+8 a A b+b^2 B\right )+\frac{a (2 a B+5 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

[Out]

(b^2*(8*a*A*b + 12*a^2*B + b^2*B)*x)/2 + (a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (b*(
13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Sin[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b^2*B)*Cos[c + d*x
]*Sin[c + d*x])/(2*d) + (a*(5*A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(2*d) + (a*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 = 0.615265, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3047, 3033, 3023, 2735, 3770} \[ -\frac{b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \sin (c+d x)}{2 d}+\frac{a^2 \left (a^2 A+8 a b B+12 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} b^2 x \left (12 a^2 B+8 a A b+b^2 B\right )+\frac{a (2 a B+5 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]

[Out]

(b^2*(8*a*A*b + 12*a^2*B + b^2*B)*x)/2 + (a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (b*(
13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Sin[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b^2*B)*Cos[c + d*x
]*Sin[c + d*x])/(2*d) + (a*(5*A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(2*d) + (a*A*(a + b*Cos[c + d*
x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2989

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

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))^4 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac{a 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 (a (5 A b+2 a B)+\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-2 b (a A-b B) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{a 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 (a \left (a^2 A+12 A b^2+8 a b B\right )-b \left (a^2 A-2 A b^2-6 a b B\right ) \cos (c+d x)-2 b \left (6 a A b+2 a^2 B-b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{a 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^2 \left (a^2 A+12 A b^2+8 a b B\right )+2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \cos (c+d x)-2 b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{a 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^2 \left (a^2 A+12 A b^2+8 a b B\right )+2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) x-\frac{b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 \left (a^2 A+12 A b^2+8 a b B\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) x+\frac{a^2 \left (a^2 A+12 A b^2+8 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sin (c+d x)}{2 d}-\frac{b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a (5 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 2.38423, size = 310, normalized size = 1.48 \[ \frac{2 b^2 (c+d x) \left (12 a^2 B+8 a A b+b^2 B\right )-2 a^2 \left (a^2 A+8 a b B+12 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2 \left (a^2 A+8 a b B+12 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 a^3 (a B+4 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a^3 (a B+4 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{a^4 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^4 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+4 b^3 (4 a B+A b) \sin (c+d x)+b^4 B \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]

[Out]

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

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Maple [A]  time = 0.092, size = 236, normalized size = 1.1 \begin{align*}{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\tan \left ( dx+c \right ) }{d}}+4\,{\frac{B{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,B{a}^{2}{b}^{2}x+6\,{\frac{B{a}^{2}{b}^{2}c}{d}}+4\,Aa{b}^{3}x+4\,{\frac{Aa{b}^{3}c}{d}}+4\,{\frac{Ba{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{4}Bx}{2}}+{\frac{B{b}^{4}c}{2\,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)^3,x)

[Out]

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

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Maxima [A]  time = 1.09024, size = 282, normalized size = 1.35 \begin{align*} \frac{24 \,{\left (d x + c\right )} B a^{2} b^{2} + 16 \,{\left (d x + c\right )} A a b^{3} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{4} - A a^{4}{\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 )} + 8 \, B a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, B a b^{3} \sin \left (d x + c\right ) + 4 \, A b^{4} \sin \left (d x + c\right ) + 4 \, B a^{4} \tan \left (d x + c\right ) + 16 \, A a^{3} 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))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/4*(24*(d*x + c)*B*a^2*b^2 + 16*(d*x + c)*A*a*b^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^4 - A*a^4*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 8*B*a^3*b*(log(sin(d*x + c) + 1)
- log(sin(d*x + c) - 1)) + 12*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 16*B*a*b^3*sin(d*x
+ c) + 4*A*b^4*sin(d*x + c) + 4*B*a^4*tan(d*x + c) + 16*A*a^3*b*tan(d*x + c))/d

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Fricas [A]  time = 1.5289, size = 479, normalized size = 2.29 \begin{align*} \frac{2 \,{\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} d x \cos \left (d x + c\right )^{2} +{\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B b^{4} \cos \left (d x + c\right )^{3} + A a^{4} + 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (B a^{4} + 4 \, A a^{3} 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))^4*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.55699, size = 710, normalized size = 3.4 \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)^3,x, algorithm="giac")

[Out]

1/2*((12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*(d*x + c) + (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1
/2*c) + 1)) - (A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a^4*tan(1/2*d*x + 1
/2*c)^7 - 2*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^7
+ 2*A*b^4*tan(1/2*d*x + 1/2*c)^7 - B*b^4*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 2*B*a^4*ta
n(1/2*d*x + 1/2*c)^5 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 2*A*b^4*tan(1/2*d
*x + 1/2*c)^5 + 3*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c)
^3 + 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 8*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*
B*b^4*tan(1/2*d*x + 1/2*c)^3 + A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b*tan(1/2*d
*x + 1/2*c) + 8*B*a*b^3*tan(1/2*d*x + 1/2*c) + 2*A*b^4*tan(1/2*d*x + 1/2*c) + B*b^4*tan(1/2*d*x + 1/2*c))/(tan
(1/2*d*x + 1/2*c)^4 - 1)^2)/d