3.117 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=200 \[ \frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{(7 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{8 d}+\frac{1}{8} a^4 x (35 A+52 C)+\frac{4 a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d} \]

[Out]

(a^4*(35*A + 52*C)*x)/8 + (4*a^4*C*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 4*C)*Sin[c + d*x])/(8*d) + (a*A*Co
s[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*Sin[c + d*x
])/(4*d) + ((7*A + 4*C)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(8*d) - ((35*A - 12*C)*(a^4 + a^
4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

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Rubi [A]  time = 0.573874, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4087, 4017, 4018, 3996, 3770} \[ \frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{(7 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{8 d}+\frac{1}{8} a^4 x (35 A+52 C)+\frac{4 a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(35*A + 52*C)*x)/8 + (4*a^4*C*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 4*C)*Sin[c + d*x])/(8*d) + (a*A*Co
s[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*Sin[c + d*x
])/(4*d) + ((7*A + 4*C)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(8*d) - ((35*A - 12*C)*(a^4 + a^
4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

Rule 4087

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4017

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

Rule 4018

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

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -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+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (4 a A-a (A-4 C) \sec (c+d x)) \, dx}{4 a}\\ &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (3 a^2 (7 A+4 C)-a^2 (7 A-12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{(7 A+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^3 (35 A+36 C)-a^3 (35 A-12 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{(7 A+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^4 (7 A+4 C)+96 a^4 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{(7 A+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac{\int \left (-3 a^5 (35 A+52 C)-96 a^5 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{1}{8} a^4 (35 A+52 C) x+\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{(7 A+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (4 a^4 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^4 (35 A+52 C) x+\frac{4 a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac{(7 A+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac{(35 A-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end{align*}

Mathematica [A]  time = 2.28263, size = 375, normalized size = 1.88 \[ \frac{a^4 \cos ^2(c+d x) (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{96 (7 A+4 C) \sin (c) \cos (d x)}{d}+\frac{24 (7 A+C) \sin (2 c) \cos (2 d x)}{d}+\frac{96 (7 A+4 C) \cos (c) \sin (d x)}{d}+\frac{24 (7 A+C) \cos (2 c) \sin (2 d x)}{d}+\frac{32 A \sin (3 c) \cos (3 d x)}{d}+\frac{3 A \sin (4 c) \cos (4 d x)}{d}+\frac{32 A \cos (3 c) \sin (3 d x)}{d}+\frac{3 A \cos (4 c) \sin (4 d x)}{d}+12 x (35 A+52 C)+\frac{96 C \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{96 C \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{384 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{384 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}\right )}{768 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(A + C*Sec[c + d*x]^2)*(12*(35*A + 52*C)*x - (384*
C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (384*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (96*(7*A
+ 4*C)*Cos[d*x]*Sin[c])/d + (24*(7*A + C)*Cos[2*d*x]*Sin[2*c])/d + (32*A*Cos[3*d*x]*Sin[3*c])/d + (3*A*Cos[4*d
*x]*Sin[4*c])/d + (96*(7*A + 4*C)*Cos[c]*Sin[d*x])/d + (24*(7*A + C)*Cos[2*c]*Sin[2*d*x])/d + (32*A*Cos[3*c]*S
in[3*d*x])/d + (3*A*Cos[4*c]*Sin[4*d*x])/d + (96*C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] -
Sin[(c + d*x)/2])) + (96*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(76
8*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.095, size = 191, normalized size = 1. \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{27\,A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{35\,{a}^{4}Ax}{8}}+{\frac{35\,A{a}^{4}c}{8\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}Cx}{2}}+{\frac{13\,{a}^{4}Cc}{2\,d}}+{\frac{4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+4\,{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/4/d*A*a^4*sin(d*x+c)*cos(d*x+c)^3+27/8/d*A*a^4*cos(d*x+c)*sin(d*x+c)+35/8*a^4*A*x+35/8/d*A*a^4*c+1/2/d*a^4*C
*sin(d*x+c)*cos(d*x+c)+13/2*a^4*C*x+13/2/d*C*a^4*c+4/3/d*A*cos(d*x+c)^2*sin(d*x+c)*a^4+20/3/d*A*a^4*sin(d*x+c)
+4/d*a^4*C*sin(d*x+c)+4/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^4*C*tan(d*x+c)

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Maxima [A]  time = 0.949653, size = 262, normalized size = 1.31 \begin{align*} -\frac{128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{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 )} A a^{4} - 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \,{\left (d x + c\right )} A a^{4} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 576 \,{\left (d x + c\right )} C a^{4} - 192 \, C a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 384 \, C a^{4} \sin \left (d x + c\right ) - 96 \, C a^{4} \tan \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+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 96*(d*x + c)*A*a^4 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))
*C*a^4 - 576*(d*x + c)*C*a^4 - 192*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 384*A*a^4*sin(d*x +
c) - 384*C*a^4*sin(d*x + c) - 96*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.548459, size = 408, normalized size = 2.04 \begin{align*} \frac{3 \,{\left (35 \, A + 52 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 48 \, C a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, C a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \, A a^{4} \cos \left (d x + c\right )^{3} + 3 \,{\left (27 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \,{\left (5 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(35*A + 52*C)*a^4*d*x*cos(d*x + c) + 48*C*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 48*C*a^4*cos(d*x +
c)*log(-sin(d*x + c) + 1) + (6*A*a^4*cos(d*x + c)^4 + 32*A*a^4*cos(d*x + c)^3 + 3*(27*A + 4*C)*a^4*cos(d*x + c
)^2 + 32*(5*A + 3*C)*a^4*cos(d*x + c) + 24*C*a^4)*sin(d*x + c))/(d*cos(d*x + c))

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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+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.27484, size = 329, normalized size = 1.64 \begin{align*} \frac{96 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{48 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + 3 \,{\left (35 \, A a^{4} + 52 \, C a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 276 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 108 \, C a^{4} \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(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(96*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 96*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 48*C*a^4*tan
(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(35*A*a^4 + 52*C*a^4)*(d*x + c) + 2*(105*A*a^4*tan(1/2*d*x
+ 1/2*c)^7 + 84*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 276*C*a^4*tan(1/2*d*x + 1/2*
c)^5 + 511*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 300*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 279*A*a^4*tan(1/2*d*x + 1/2*c) +
108*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d