3.180 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=200 \[ \frac{a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{5/2} (163 A+304 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (17 A+16 C) \sin (c+d x) \cos (c+d x) \sqrt{a \sec (c+d x)+a}}{32 d}+\frac{5 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]
[Out]
(a^(5/2)*(163*A + 304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(299*A + 432*C
)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(17*A + 16*C)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Si
n[c + d*x])/(32*d) + (5*a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^3
*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d)
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Rubi [A] time = 0.654743, antiderivative size = 200, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4087, 4017, 4015, 3774, 203} \[ \frac{a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{5/2} (163 A+304 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (17 A+16 C) \sin (c+d x) \cos (c+d x) \sqrt{a \sec (c+d x)+a}}{32 d}+\frac{5 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(a^(5/2)*(163*A + 304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(299*A + 432*C
)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(17*A + 16*C)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Si
n[c + d*x])/(32*d) + (5*a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^3
*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d)
Rule 4087
Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (A+8 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{3}{4} a^2 (17 A+16 C)+\frac{1}{4} a^2 (11 A+48 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{a^2 (17 A+16 C) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (299 A+432 C)+\frac{5}{8} a^3 (19 A+48 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (17 A+16 C) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{1}{128} \left (a^2 (163 A+304 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (17 A+16 C) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac{\left (a^3 (163 A+304 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 )}{64 d}\\ &=\frac{a^{5/2} (163 A+304 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{64 d}+\frac{a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (17 A+16 C) \cos (c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.59983, size = 143, normalized size = 0.72 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (3 \sqrt{2} (163 A+304 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}+\left (\sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) ((362 A+96 C) \cos (c+d x)+92 A \cos (2 (c+d x))+12 A \cos (3 (c+d x))+581 A+528 C)\right )}{384 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(163*A + 304*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*S
qrt[Cos[c + d*x]] + (581*A + 528*C + (362*A + 96*C)*Cos[c + d*x] + 92*A*Cos[2*(c + d*x)] + 12*A*Cos[3*(c + d*x
)])*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(384*d)
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Maple [B] time = 0.29, size = 754, normalized size = 3.8 \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)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
[Out]
1/3072/d*a^2*(489*A*sin(d*x+c)*cos(d*x+c)^3*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))^(7/2)*2^(1/2)+912*C*sin(d*x+c)*cos(d*x+c)^3*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))^(7/2)*2^(1/2)+1467*A
*sin(d*x+c)*cos(d*x+c)^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*c
os(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+2736*C*sin(d*x+c)*cos(d*x+c)^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))^(7/2)*2^(1/2)+1467*A*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))^(7/2)*2^(1/2)+2736*C*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))^(7/2)*2^(1/2)+489*A*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))^(7/2)*2^(1/2)*sin(d*x+c)+912*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^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(7/2)*sin(d*x+c)-768*A*cos(d*x+c)^8-2176*A*cos(d*x+c)^7-2272*A*cos(d*x+c)^6-1536*C*cos(d*x+c)^6-2608*A*cos(d*
x+c)^5-6912*C*cos(d*x+c)^5+7824*A*cos(d*x+c)^4+8448*C*cos(d*x+c)^4)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*
x+c)^3/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)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 0.751069, size = 1076, normalized size = 5.38 \begin{align*} \left [\frac{3 \,{\left ({\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (163 \, A + 304 \, C\right )} a^{2}\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 (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 184 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (163 \, A + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (163 \, A + 176 \, C\right )} a^{2} \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 )}{384 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{3 \,{\left ({\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (163 \, A + 304 \, C\right )} a^{2}\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 (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 184 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (163 \, A + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (163 \, A + 176 \, C\right )} a^{2} \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 )}{192 \,{\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)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/384*(3*((163*A + 304*C)*a^2*cos(d*x + c) + (163*A + 304*C)*a^2)*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*(48*A*a^2*cos(d*x + c)^4 + 184*A*a^2*cos(d*x + c)^3 + 2*(163*A + 48*C)*a^2*cos(d*x + c)^2 + 3*(163*A + 17
6*C)*a^2*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), -1/192*(3*(
(163*A + 304*C)*a^2*cos(d*x + c) + (163*A + 304*C)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos(d*x + c)^4 + 184*A*a^2*cos(d*x + c)^3 + 2*(163*A + 48*C)
*a^2*cos(d*x + c)^2 + 3*(163*A + 176*C)*a^2*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)**4*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [B] time = 7.72581, size = 1480, normalized size = 7.4 \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)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/384*(3*(163*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 304*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*(163*A*sqrt(-a)*a^2*sgn(co
s(d*x + c)) + 304*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))) + 4*sqrt(2)*(489*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^14*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 912*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/
2*d*x + 1/2*c)^2 + a))^14*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 10269*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 19152*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-
a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 69885*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 137424*(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^5*sgn(cos(d*x + c)) - 259233*(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^6*sgn(cos(d*x + c)) - 374544*(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^6*sgn(cos(d*x + c)) + 209979*(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^7*sgn(cos(d*x + c)) + 266928*(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^7*sgn(cos(d*x + c)) - 55511*(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^8*sgn(cos(d*x + c)) - 75888*(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^8*sgn(cos(d*x + c)) + 6687*(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^9*sgn(cos(d*x + c)) + 9456*(sq
rt(-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^9*sgn(cos(d*x + c)) - 299*A*
sqrt(-a)*a^10*sgn(cos(d*x + c)) - 432*C*sqrt(-a)*a^10*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr
t(-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)^4)/d