3.25 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=169 \[ \frac{5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac{a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+a^3 C x+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

a^3*C*x + (a^3*(15*A + 28*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(3*A + 4*C)*Tan[c + d*x])/(8*d) + ((5*A + 4
*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^2*T
an[c + d*x])/(4*a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.496396, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3021, 2735, 3770} \[ \frac{5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac{a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(5 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{8 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{4 a d}+a^3 C x+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

a^3*C*x + (a^3*(15*A + 28*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(3*A + 4*C)*Tan[c + d*x])/(8*d) + ((5*A + 4
*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (A*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^2*T
an[c + d*x])/(4*a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3044

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

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(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] && GtQ[m, 1/2] && LtQ[n, -1] && 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 3021

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^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[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 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+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^3 (3 a A+4 a C \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^2 \left (3 a^2 (5 A+4 C)+12 a^2 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac{(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x)) \left (15 a^3 (3 A+4 C)+24 a^3 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (15 a^4 (3 A+4 C)+\left (24 a^4 C+15 a^4 (3 A+4 C)\right ) \cos (c+d x)+24 a^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac{(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (3 a^4 (15 A+28 C)+24 a^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=a^3 C x+\frac{5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac{(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (a^3 (15 A+28 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 C x+\frac{a^3 (15 A+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^3 (3 A+4 C) \tan (c+d x)}{8 d}+\frac{(5 A+4 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{4 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.42177, size = 334, normalized size = 1.98 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (\sec (c) (23 A \sin (2 c+d x)+88 A \sin (c+2 d x)-8 A \sin (3 c+2 d x)+15 A \sin (2 c+3 d x)+15 A \sin (4 c+3 d x)+24 A \sin (3 c+4 d x)-72 A \sin (c)+23 A \sin (d x)+4 C \sin (2 c+d x)+72 C \sin (c+2 d x)-24 C \sin (3 c+2 d x)+4 C \sin (2 c+3 d x)+4 C \sin (4 c+3 d x)+24 C \sin (3 c+4 d x)+24 C d x \cos (c)+16 C d x \cos (c+2 d x)+16 C d x \cos (3 c+2 d x)+4 C d x \cos (3 c+4 d x)+4 C d x \cos (5 c+4 d x)-72 C \sin (c)+4 C \sin (d x))-8 (15 A+28 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{512 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(-8*(15*A + 28*C)*Cos[c + d*x]^4*(Log[Cos[(c + d*x
)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(24*C*d*x*Cos[c] + 16*C*d*x*Cos[
c + 2*d*x] + 16*C*d*x*Cos[3*c + 2*d*x] + 4*C*d*x*Cos[3*c + 4*d*x] + 4*C*d*x*Cos[5*c + 4*d*x] - 72*A*Sin[c] - 7
2*C*Sin[c] + 23*A*Sin[d*x] + 4*C*Sin[d*x] + 23*A*Sin[2*c + d*x] + 4*C*Sin[2*c + d*x] + 88*A*Sin[c + 2*d*x] + 7
2*C*Sin[c + 2*d*x] - 8*A*Sin[3*c + 2*d*x] - 24*C*Sin[3*c + 2*d*x] + 15*A*Sin[2*c + 3*d*x] + 4*C*Sin[2*c + 3*d*
x] + 15*A*Sin[4*c + 3*d*x] + 4*C*Sin[4*c + 3*d*x] + 24*A*Sin[3*c + 4*d*x] + 24*C*Sin[3*c + 4*d*x])))/(512*d)

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Maple [A]  time = 0.078, size = 180, normalized size = 1.1 \begin{align*} 3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{a}^{3}Cx+{\frac{C{a}^{3}c}{d}}+{\frac{15\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{7\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

3/d*A*a^3*tan(d*x+c)+a^3*C*x+1/d*a^3*C*c+15/8/d*A*a^3*sec(d*x+c)*tan(d*x+c)+15/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x
+c))+7/2/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+3/d*a^3*C*tan(d*x+c)+1/4/d*A*a^3*
tan(d*x+c)*sec(d*x+c)^3+1/2/d*a^3*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 1.01213, size = 347, normalized size = 2.05 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \,{\left (d x + c\right )} C a^{3} - A a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3}{\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 )} - 4 \, C a^{3}{\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 )} + 24 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(d*x + c)*C*a^3 - A*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x
+ c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*A*a^3
*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*C*a^3*(2*sin(d*x +
c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^3*(log(sin(d*x + c) + 1) - l
og(sin(d*x + c) - 1)) + 16*A*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d

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Fricas [A]  time = 1.46781, size = 386, normalized size = 2.28 \begin{align*} \frac{16 \, C a^{3} d x \cos \left (d x + c\right )^{4} +{\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (15 \, A + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \,{\left (A + C\right )} a^{3} \cos \left (d x + c\right )^{3} +{\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, A a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/16*(16*C*a^3*d*x*cos(d*x + c)^4 + (15*A + 28*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (15*A + 28*C)*a^3
*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*(A + C)*a^3*cos(d*x + c)^3 + (15*A + 4*C)*a^3*cos(d*x + c)^2 +
8*A*a^3*cos(d*x + c) + 2*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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+a*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.2644, size = 300, normalized size = 1.78 \begin{align*} \frac{8 \,{\left (d x + c\right )} C a^{3} +{\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (15 \, A a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 55 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 68 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 49 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 28 \, C a^{3} \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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)*C*a^3 + (15*A*a^3 + 28*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (15*A*a^3 + 28*C*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 20*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 55*A*a
^3*tan(1/2*d*x + 1/2*c)^5 - 68*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 76*C*a^3*tan(1
/2*d*x + 1/2*c)^3 - 49*A*a^3*tan(1/2*d*x + 1/2*c) - 28*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1
)^4)/d