3.464 \(\int \frac{\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=156 \[ -\frac{2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac{(7 A-4 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{x (7 A-4 B+2 C)}{2 a^2}-\frac{(A-B+C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((7*A - 4*B + 2*C)*x)/(2*a^2) - (2*(8*A - 5*B + 2*C)*Sin[c + d*x])/(3*a^2*d) + ((7*A - 4*B + 2*C)*Cos[c + d*x]
*Sin[c + d*x])/(2*a^2*d) - ((8*A - 5*B + 2*C)*Cos[c + d*x]*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A -
B + C)*Cos[c + d*x]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.334872, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4084, 4020, 3787, 2635, 8, 2637} \[ -\frac{2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac{(7 A-4 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{x (7 A-4 B+2 C)}{2 a^2}-\frac{(A-B+C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A - 4*B + 2*C)*x)/(2*a^2) - (2*(8*A - 5*B + 2*C)*Sin[c + d*x])/(3*a^2*d) + ((7*A - 4*B + 2*C)*Cos[c + d*x]
*Sin[c + d*x])/(2*a^2*d) - ((8*A - 5*B + 2*C)*Cos[c + d*x]*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A -
B + C)*Cos[c + d*x]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 4084

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

Rule 4020

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 1.51753, size = 377, normalized size = 2.42 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (36 d x (7 A-4 B+2 C) \cos \left (c+\frac{d x}{2}\right )+36 d x (7 A-4 B+2 C) \cos \left (\frac{d x}{2}\right )+147 A \sin \left (c+\frac{d x}{2}\right )-239 A \sin \left (c+\frac{3 d x}{2}\right )-63 A \sin \left (2 c+\frac{3 d x}{2}\right )-15 A \sin \left (2 c+\frac{5 d x}{2}\right )-15 A \sin \left (3 c+\frac{5 d x}{2}\right )+3 A \sin \left (3 c+\frac{7 d x}{2}\right )+3 A \sin \left (4 c+\frac{7 d x}{2}\right )+84 A d x \cos \left (c+\frac{3 d x}{2}\right )+84 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 A \sin \left (\frac{d x}{2}\right )-120 B \sin \left (c+\frac{d x}{2}\right )+164 B \sin \left (c+\frac{3 d x}{2}\right )+36 B \sin \left (2 c+\frac{3 d x}{2}\right )+12 B \sin \left (2 c+\frac{5 d x}{2}\right )+12 B \sin \left (3 c+\frac{5 d x}{2}\right )-48 B d x \cos \left (c+\frac{3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+264 B \sin \left (\frac{d x}{2}\right )+96 C \sin \left (c+\frac{d x}{2}\right )-80 C \sin \left (c+\frac{3 d x}{2}\right )+24 C d x \cos \left (c+\frac{3 d x}{2}\right )+24 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-144 C \sin \left (\frac{d x}{2}\right )\right )}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(36*(7*A - 4*B + 2*C)*d*x*Cos[(d*x)/2] + 36*(7*A - 4*B + 2*C)*d*x*Cos[c + (d*x)/2
] + 84*A*d*x*Cos[c + (3*d*x)/2] - 48*B*d*x*Cos[c + (3*d*x)/2] + 24*C*d*x*Cos[c + (3*d*x)/2] + 84*A*d*x*Cos[2*c
+ (3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 24*C*d*x*Cos[2*c + (3*d*x)/2] - 381*A*Sin[(d*x)/2] + 264*B*Sin
[(d*x)/2] - 144*C*Sin[(d*x)/2] + 147*A*Sin[c + (d*x)/2] - 120*B*Sin[c + (d*x)/2] + 96*C*Sin[c + (d*x)/2] - 239
*A*Sin[c + (3*d*x)/2] + 164*B*Sin[c + (3*d*x)/2] - 80*C*Sin[c + (3*d*x)/2] - 63*A*Sin[2*c + (3*d*x)/2] + 36*B*
Sin[2*c + (3*d*x)/2] - 15*A*Sin[2*c + (5*d*x)/2] + 12*B*Sin[2*c + (5*d*x)/2] - 15*A*Sin[3*c + (5*d*x)/2] + 12*
B*Sin[3*c + (5*d*x)/2] + 3*A*Sin[3*c + (7*d*x)/2] + 3*A*Sin[4*c + (7*d*x)/2]))/(192*a^2*d)

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Maple [B]  time = 0.106, size = 309, normalized size = 2. \begin{align*}{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-3\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+7\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-4\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-7/2/d/a^2*A
*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-3/2/d/a^2*C*tan(1/2*d*x+1/2*c)-5/d/a^2/(1+tan(1/2*d*x+1/2*c
)^2)^2*tan(1/2*d*x+1/2*c)^3*A+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-3/d/a^2/(1+tan(1/2*d*x
+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+7/d/a^2*A*arctan(tan
(1/2*d*x+1/2*c))-4/d/a^2*B*arctan(tan(1/2*d*x+1/2*c))+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 1.44225, size = 475, normalized size = 3.04 \begin{align*} -\frac{A{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(A*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*((15*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + C*((9*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2))/d

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Fricas [A]  time = 0.49705, size = 383, normalized size = 2.46 \begin{align*} \frac{3 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} d x +{\left (3 \, A \cos \left (d x + c\right )^{3} - 6 \,{\left (A - B\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, A - 28 \, B + 10 \, C\right )} \cos \left (d x + c\right ) - 32 \, A + 20 \, B - 8 \, C\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(7*A - 4*B + 2*C)*d*x*cos(d*x + c)^2 + 6*(7*A - 4*B + 2*C)*d*x*cos(d*x + c) + 3*(7*A - 4*B + 2*C)*d*x +
(3*A*cos(d*x + c)^3 - 6*(A - B)*cos(d*x + c)^2 - (43*A - 28*B + 10*C)*cos(d*x + c) - 32*A + 20*B - 8*C)*sin(d
*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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(cos(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24307, size = 267, normalized size = 1.71 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}{\left (7 \, A - 4 \, B + 2 \, C\right )}}{a^{2}} - \frac{6 \,{\left (5 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B \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 )}^{2} a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(d*x + c)*(7*A - 4*B + 2*C)/a^2 - 6*(5*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + 3*A*tan(
1/2*d*x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c
)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 21*A*a^4*tan(1/2*d*x + 1/2*c) + 15*B*a^4*t
an(1/2*d*x + 1/2*c) - 9*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d