3.67 \(\int \frac{\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=152 \[ \frac{(16 A-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(8 A-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{C x}{a^4}-\frac{(A+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{2 (2 A-5 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[Out]

(C*x)/a^4 - ((8*A - 55*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((16*A - 215*C)*Sin[c + d*x])/(105*
a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + (2*(2*A - 5*C
)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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Rubi [A]  time = 0.444877, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3042, 2977, 2968, 3019, 2735, 2648} \[ \frac{(16 A-215 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(8 A-55 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{C x}{a^4}-\frac{(A+C) \sin (c+d x) \cos ^3(c+d x)}{7 d (a \cos (c+d x)+a)^4}+\frac{2 (2 A-5 C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/a^4 - ((8*A - 55*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((16*A - 215*C)*Sin[c + d*x])/(105*
a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + (2*(2*A - 5*C
)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 3042

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

Rule 2977

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3019

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 - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx &=-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{\int \frac{\cos ^2(c+d x) (a (4 A-3 C)+7 a C \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos (c+d x) \left (4 a^2 (2 A-5 C)+35 a^2 C \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{\int \frac{4 a^2 (2 A-5 C) \cos (c+d x)+35 a^2 C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac{\int \frac{-2 a^3 (8 A-55 C)-105 a^3 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=\frac{C x}{a^4}-\frac{(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(16 A-215 C) \int \frac{1}{a+a \cos (c+d x)} \, dx}{105 a^3}\\ &=\frac{C x}{a^4}-\frac{(8 A-55 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac{(A+C) \cos ^3(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac{2 (2 A-5 C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac{(16 A-215 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.709024, size = 315, normalized size = 2.07 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (-350 A \sin \left (c+\frac{d x}{2}\right )+336 A \sin \left (c+\frac{3 d x}{2}\right )-210 A \sin \left (2 c+\frac{3 d x}{2}\right )+182 A \sin \left (2 c+\frac{5 d x}{2}\right )+26 A \sin \left (3 c+\frac{7 d x}{2}\right )+560 A \sin \left (\frac{d x}{2}\right )+8260 C \sin \left (c+\frac{d x}{2}\right )-7140 C \sin \left (c+\frac{3 d x}{2}\right )+3780 C \sin \left (2 c+\frac{3 d x}{2}\right )-2800 C \sin \left (2 c+\frac{5 d x}{2}\right )+840 C \sin \left (3 c+\frac{5 d x}{2}\right )-520 C \sin \left (3 c+\frac{7 d x}{2}\right )+3675 C d x \cos \left (c+\frac{d x}{2}\right )+2205 C d x \cos \left (c+\frac{3 d x}{2}\right )+2205 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+735 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+735 C d x \cos \left (3 c+\frac{5 d x}{2}\right )+105 C d x \cos \left (3 c+\frac{7 d x}{2}\right )+105 C d x \cos \left (4 c+\frac{7 d x}{2}\right )-9940 C \sin \left (\frac{d x}{2}\right )+3675 C d x \cos \left (\frac{d x}{2}\right )\right )}{13440 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*C*d*x*Cos[(d*x)/2] + 3675*C*d*x*Cos[c + (d*x)/2] + 2205*C*d*x*Cos[c + (3*d*
x)/2] + 2205*C*d*x*Cos[2*c + (3*d*x)/2] + 735*C*d*x*Cos[2*c + (5*d*x)/2] + 735*C*d*x*Cos[3*c + (5*d*x)/2] + 10
5*C*d*x*Cos[3*c + (7*d*x)/2] + 105*C*d*x*Cos[4*c + (7*d*x)/2] + 560*A*Sin[(d*x)/2] - 9940*C*Sin[(d*x)/2] - 350
*A*Sin[c + (d*x)/2] + 8260*C*Sin[c + (d*x)/2] + 336*A*Sin[c + (3*d*x)/2] - 7140*C*Sin[c + (3*d*x)/2] - 210*A*S
in[2*c + (3*d*x)/2] + 3780*C*Sin[2*c + (3*d*x)/2] + 182*A*Sin[2*c + (5*d*x)/2] - 2800*C*Sin[2*c + (5*d*x)/2] +
840*C*Sin[3*c + (5*d*x)/2] + 26*A*Sin[3*c + (7*d*x)/2] - 520*C*Sin[3*c + (7*d*x)/2]))/(13440*a^4*d)

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Maple [A]  time = 0.027, size = 177, normalized size = 1.2 \begin{align*}{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{C}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{11\,C}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{15\,C}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A+1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*C-1/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a^
4*C*tan(1/2*d*x+1/2*c)^5-1/24/d/a^4*tan(1/2*d*x+1/2*c)^3*A+11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3+1/8/d/a^4*A*tan(
1/2*d*x+1/2*c)-15/8/d/a^4*C*tan(1/2*d*x+1/2*c)+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.56561, size = 271, normalized size = 1.78 \begin{align*} -\frac{5 \, C{\left (\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{336 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac{A{\left (\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^4) - A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

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Fricas [A]  time = 1.38198, size = 477, normalized size = 3.14 \begin{align*} \frac{105 \, C d x \cos \left (d x + c\right )^{4} + 420 \, C d x \cos \left (d x + c\right )^{3} + 630 \, C d x \cos \left (d x + c\right )^{2} + 420 \, C d x \cos \left (d x + c\right ) + 105 \, C d x +{\left (13 \,{\left (A - 20 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (13 \, A - 155 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (32 \, A - 535 \, C\right )} \cos \left (d x + c\right ) + 8 \, A - 160 \, C\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*C*d*x*cos(d*x + c)^4 + 420*C*d*x*cos(d*x + c)^3 + 630*C*d*x*cos(d*x + c)^2 + 420*C*d*x*cos(d*x + c)
+ 105*C*d*x + (13*(A - 20*C)*cos(d*x + c)^3 + 4*(13*A - 155*C)*cos(d*x + c)^2 + (32*A - 535*C)*cos(d*x + c) +
8*A - 160*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*
cos(d*x + c) + a^4*d)

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Sympy [A]  time = 24.1228, size = 192, normalized size = 1.26 \begin{align*} \begin{cases} \frac{A \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} - \frac{A \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{40 a^{4} d} - \frac{A \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{4} d} + \frac{A \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{C x}{a^{4}} + \frac{C \tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{4} d} - \frac{C \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} + \frac{11 C \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{4} d} - \frac{15 C \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{4} d} & \text{for}\: d \neq 0 \\\frac{x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40*a**4*d) - A*tan(c/2 + d*x/2)**3/(24*a
**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) + C*x/a**4 + C*tan(c/2 + d*x/2)**7/(56*a**4*d) - C*tan(c/2 + d*x/2)**5/
(8*a**4*d) + 11*C*tan(c/2 + d*x/2)**3/(24*a**4*d) - 15*C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + C*cos
(c)**2)*cos(c)**2/(a*cos(c) + a)**4, True))

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Giac [A]  time = 1.21489, size = 208, normalized size = 1.37 \begin{align*} \frac{\frac{840 \,{\left (d x + c\right )} C}{a^{4}} + \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 21 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(840*(d*x + c)*C/a^4 + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 21*A*a^24*
tan(1/2*d*x + 1/2*c)^5 - 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 - 35*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan
(1/2*d*x + 1/2*c)^3 + 105*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d