3.497 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=165 \[ \frac{a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (11 A+14 B+24 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a (A+2 B) \sin (c+d x) \cos (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

(a^(3/2)*(11*A + 14*B + 24*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^2*(19*A + 30
*B + 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.481126, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4086, 4017, 4015, 3774, 203} \[ \frac{a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (11 A+14 B+24 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a (A+2 B) \sin (c+d x) \cos (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(11*A + 14*B + 24*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^2*(19*A + 30
*B + 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*
Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

Rule 4086

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 4015

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

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{3}{2} a (A+2 B)+\frac{1}{2} a (A+6 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{a (A+2 B) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (19 A+30 B+24 C)+\frac{1}{4} a^2 (7 A+6 B+24 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac{a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+2 B) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{1}{16} (a (11 A+14 B+24 C)) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+2 B) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac{\left (a^2 (11 A+14 B+24 C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}\\ &=\frac{a^{3/2} (11 A+14 B+24 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}+\frac{a^2 (19 A+30 B+24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+2 B) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.67123, size = 124, normalized size = 0.75 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (\cos (c+d x) \sqrt{\sec (c+d x)-1} (2 (11 A+6 B) \cos (c+d x)+4 A \cos (2 (c+d x))+37 A+42 B+24 C)+(33 A+42 B+72 C) \tan ^{-1}\left (\sqrt{\sec (c+d x)-1}\right )\right )}{24 d \sqrt{\sec (c+d x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((33*A + 42*B + 72*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]] + Cos[c + d*x]*(37*A + 42*B + 24*C + 2*(11*A + 6*B)*C
os[c + d*x] + 4*A*Cos[2*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(24*
d*Sqrt[-1 + Sec[c + d*x]])

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Maple [B]  time = 0.297, size = 833, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192/d*a*(33*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+42*B*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+72*C*cos(d
*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+66*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arct
anh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+84*B*cos(d*x+c)*2^(1/2)*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin
(d*x+c)+144*C*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+33*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*arctanh(1
/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+42*B*2^(1/2)*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+
c)+72*C*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+64*A*cos(d*x+c)^6+112*A*cos(d*x+c)^5+96*B*cos(d*x+c)^5+88*A*cos(d*x+c)^4+240
*B*cos(d*x+c)^4+192*C*cos(d*x+c)^4-264*A*cos(d*x+c)^3-336*B*cos(d*x+c)^3-192*C*cos(d*x+c)^3)*(a*(cos(d*x+c)+1)
/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)

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Maxima [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)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.916013, size = 1003, normalized size = 6.08 \begin{align*} \left [\frac{3 \,{\left ({\left (11 \, A + 14 \, B + 24 \, C\right )} a \cos \left (d x + c\right ) +{\left (11 \, A + 14 \, B + 24 \, C\right )} a\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (8 \, A a \cos \left (d x + c\right )^{3} + 2 \,{\left (11 \, A + 6 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (11 \, A + 14 \, B + 8 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{3 \,{\left ({\left (11 \, A + 14 \, B + 24 \, C\right )} a \cos \left (d x + c\right ) +{\left (11 \, A + 14 \, B + 24 \, C\right )} a\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left (8 \, A a \cos \left (d x + c\right )^{3} + 2 \,{\left (11 \, A + 6 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (11 \, A + 14 \, B + 8 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*((11*A + 14*B + 24*C)*a*cos(d*x + c) + (11*A + 14*B + 24*C)*a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*s
qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c)
+ 1)) + 2*(8*A*a*cos(d*x + c)^3 + 2*(11*A + 6*B)*a*cos(d*x + c)^2 + 3*(11*A + 14*B + 8*C)*a*cos(d*x + c))*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((11*A + 14*B + 24*C)*a*cos(d
*x + c) + (11*A + 14*B + 24*C)*a)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)
*sin(d*x + c))) - (8*A*a*cos(d*x + c)^3 + 2*(11*A + 6*B)*a*cos(d*x + c)^2 + 3*(11*A + 14*B + 8*C)*a*cos(d*x +
c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + 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)**3*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 7.28482, size = 1612, normalized size = 9.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(11*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 24*C*sqrt(-a)*a*sgn(cos(d*x
+ c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) -
3*(11*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 24*C*sqrt(-a)*a*sgn(cos(d*x + c)))
*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqr
t(2)*(33*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^2*sgn(cos(d*x +
c)) + 42*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a)*a^2*sgn(cos(d*x
+ c)) + 72*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a)*a^2*sgn(cos(d*x
+ c)) - 303*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^3*sgn(cos(d*
x + c)) - 822*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^3*sgn(cos(d
*x + c)) - 888*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a^3*sgn(cos(
d*x + c)) + 2394*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^4*sgn(co
s(d*x + c)) + 3780*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^4*sgn(
cos(d*x + c)) + 3024*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^4*sg
n(cos(d*x + c)) - 1806*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^5*
sgn(cos(d*x + c)) - 2508*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^
5*sgn(cos(d*x + c)) - 1776*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*
a^5*sgn(cos(d*x + c)) + 309*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)
*a^6*sgn(cos(d*x + c)) + 498*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a
)*a^6*sgn(cos(d*x + c)) + 360*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt(-
a)*a^6*sgn(cos(d*x + c)) - 19*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 30*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 24*C*sq
rt(-a)*a^7*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sq
rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d