3.505 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=197 \[ \frac{a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-4 B-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+20 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \]
[Out]
(a^(5/2)*(19*A + 20*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 12*
B - 56*C)*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 4*B - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c
+ d*x])/(2*d) - (a*(3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c +
d*x])^(5/2)*Sin[c + d*x])/(2*d)
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Rubi [A] time = 0.631031, antiderivative size = 197, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used =
{4086, 4018, 4015, 3774, 203} \[ \frac{a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (A-4 B-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{a^{5/2} (19 A+20 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(5/2)*(19*A + 20*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 12*
B - 56*C)*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 4*B - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c
+ d*x])/(2*d) - (a*(3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c +
d*x])^(5/2)*Sin[c + d*x])/(2*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 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 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 ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (5 A+4 B)-\frac{1}{2} a (3 A-4 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (21 A+12 B-8 C)-\frac{3}{4} a^2 (A-4 B-8 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=-\frac{a^2 (A-4 B-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{2 \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (27 A-12 B-56 C)+\frac{1}{8} a^3 (15 A+36 B+40 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac{1}{8} \left (a^2 (19 A+20 B+8 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}-\frac{\left (a^3 (19 A+20 B+8 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 )}{4 d}\\ &=\frac{a^{5/2} (19 A+20 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (A-4 B-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 1.19124, size = 152, normalized size = 0.77 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (6 \sqrt{2} (19 A+20 B+8 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((9 A+48 B+128 C) \cos (c+d x)+3 (11 A+4 B) \cos (2 (c+d x))+3 A \cos (3 (c+d x))+33 A+12 B+16 C)\right )}{48 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(19*A + 20*B + 8*C)*ArcSin[Sqrt[2]*Si
n[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(33*A + 12*B + 16*C + (9*A + 48*B + 128*C)*Cos[c + d*x] + 3*(11*A + 4*B
)*Cos[2*(c + d*x)] + 3*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
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Maple [B] time = 0.388, size = 583, normalized size = 3. \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)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
1/48/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(57*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+60*B*cos(d*x+c)*sin(
d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*s
in(d*x+c)/cos(d*x+c))+24*C*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))+57*A*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+60*B*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(3/2)*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))*2^(
1/2)*sin(d*x+c)+24*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))*2^(1/2)*sin(d*x+c)-24*A*cos(d*x+c)^4-108*A*cos(d*x+c)^3-48*B*cos(d*x+c)^3+132*A*co
s(d*x+c)^2-48*B*cos(d*x+c)^2-256*C*cos(d*x+c)^2+96*B*cos(d*x+c)+224*C*cos(d*x+c)+32*C)/cos(d*x+c)/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)^2*(a+a*sec(d*x+c))^(5/2)*(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 = 0.955547, size = 1111, normalized size = 5.64 \begin{align*} \left [\frac{3 \,{\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\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 )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac{3 \,{\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\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 (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/24*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 20*B + 8*C)*a^2*cos(d*x + c))*sqrt(-a)*log((2*a*cos
(d*x + c)^2 - 2*sqrt(-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*(6*A*a^2*cos(d*x + c)^3 + 3*(11*A + 4*B)*a^2*cos(d*x + c)^2 + 8*(3*B + 8*C)*a^2*cos
(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
, -1/12*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 20*B + 8*C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (6*A*a^2*cos(d*x + c)^3 + 3*(11*A +
4*B)*a^2*cos(d*x + c)^2 + 8*(3*B + 8*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
in(d*x + c))/(d*cos(d*x + c)^2 + 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)**2*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [B] time = 7.08242, size = 1095, normalized size = 5.56 \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)^2*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/24*(3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a^2*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*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 20*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a^2*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)
)) + 16*(3*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 9*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (3*sqrt(2)*B*a^4*sgn(cos(d*x
+ c)) + 7*sqrt(2)*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*
c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) + 12*sqrt(2)*(19*(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^3*sgn(cos(d*x + c)) + 12*(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^3*sgn(cos(d*x + c)) - 171*(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^4*sgn(cos(d*x + c)) - 76*(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^4*sgn(cos(d*x + c)) + 89*(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^5*sgn(cos(d*x + c)) + 36*(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^5*sgn(cos(d*x + c)) - 9*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 4*B*sqrt(-a
)*a^6*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*(sqrt(-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)^2)/d