3.490 \(\int \cos ^4(c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=209 \[ \frac{a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{a} (35 A+40 B+48 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a (35 A+40 B+48 C) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{A \sin (c+d x) \cos ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{4 d} \]
[Out]
(Sqrt[a]*(35*A + 40*B + 48*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a*(35*A + 40*
B + 48*C)*Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(35*A + 40*B + 48*C)*Cos[c + d*x]*Sin[c + d*x])/(
96*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (
A*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d)
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Rubi [A] time = 0.45412, antiderivative size = 209, 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, 4015, 3805, 3774, 203} \[ \frac{a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{a} (35 A+40 B+48 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a (35 A+40 B+48 C) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a (A+8 B) \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{A \sin (c+d x) \cos ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Sqrt[a]*(35*A + 40*B + 48*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a*(35*A + 40*
B + 48*C)*Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(35*A + 40*B + 48*C)*Cos[c + d*x]*Sin[c + d*x])/(
96*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (
A*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*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 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 3805
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]
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) \sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{\int \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (A+8 B)+\frac{1}{2} a (5 A+8 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{48} (35 A+40 B+48 C) \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a (35 A+40 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{64} (35 A+40 B+48 C) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a (35 A+40 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{128} (35 A+40 B+48 C) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a (35 A+40 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}-\frac{(a (35 A+40 B+48 C)) \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{\sqrt{a} (35 A+40 B+48 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{64 d}+\frac{a (35 A+40 B+48 C) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a (35 A+40 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a (A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}\\ \end{align*}
Mathematica [C] time = 0.238655, size = 90, normalized size = 0.43 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (A \text{Hypergeometric2F1}\left (\frac{1}{2},5,\frac{3}{2},1-\sec (c+d x)\right )+B \text{Hypergeometric2F1}\left (\frac{1}{2},4,\frac{3}{2},1-\sec (c+d x)\right )+C \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )\right )}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(2*(C*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]] + B*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]] +
A*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/d
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Maple [B] time = 0.398, size = 1105, normalized size = 5.3 \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)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)
[Out]
1/3072/d*(105*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)/co
s(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+120*B*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*2^(1/2)*(-2*c
os(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)+144*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)+315*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*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+360*B*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)+432*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)+315*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)+360*B*s
in(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)+432*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)+105*A*arctanh(1/2*2^(1/2)*(-2*c
os(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)
+120*B*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)+144*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)^8-128*A*cos(d*x+c)^7-1024*B*
cos(d*x+c)^7-224*A*cos(d*x+c)^6-256*B*cos(d*x+c)^6-1536*C*cos(d*x+c)^6-560*A*cos(d*x+c)^5-640*B*cos(d*x+c)^5-7
68*C*cos(d*x+c)^5+1680*A*cos(d*x+c)^4+1920*B*cos(d*x+c)^4+2304*C*cos(d*x+c)^4)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(
1/2)/sin(d*x+c)/cos(d*x+c)^3
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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+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 1.29231, size = 1084, normalized size = 5.19 \begin{align*} \left [\frac{3 \,{\left ({\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right ) + 35 \, A + 40 \, B + 48 \, C\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 \cos \left (d x + c\right )^{4} + 8 \,{\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (35 \, A + 40 \, B + 48 \, C\right )} \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 (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right ) + 35 \, A + 40 \, B + 48 \, C\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 \cos \left (d x + c\right )^{4} + 8 \,{\left (7 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (35 \, A + 40 \, B + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (35 \, A + 40 \, B + 48 \, C\right )} \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+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[1/384*(3*((35*A + 40*B + 48*C)*cos(d*x + c) + 35*A + 40*B + 48*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*(48*A*cos(d*x + c)^4 + 8*(7*A + 8*B)*cos(d*x + c)^3 + 2*(35*A + 40*B + 48*C)*cos(d*x + c)^2 + 3*(35*A + 4
0*B + 48*C)*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*((35*A + 40*B + 48*C)*cos(d*x + c) + 35*A + 40*B + 48*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))) - (48*A*cos(d*x + c)^4 + 8*(7*A + 8*B)*cos(d*x + c)^3 + 2*(35*A + 40*
B + 48*C)*cos(d*x + c)^2 + 3*(35*A + 40*B + 48*C)*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+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [B] time = 7.16694, size = 2049, normalized size = 9.8 \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+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
-1/384*(3*(35*A*sqrt(-a)*sgn(cos(d*x + c)) + 40*B*sqrt(-a)*sgn(cos(d*x + c)) + 48*C*sqrt(-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*(3
5*A*sqrt(-a)*sgn(cos(d*x + c)) + 40*B*sqrt(-a)*sgn(cos(d*x + c)) + 48*C*sqrt(-a)*sgn(cos(d*x + c)))*log(abs((s
qrt(-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)*(279*(
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*sgn(cos(d*x + c)) - 504*(
sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 240*(
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*sgn(cos(d*x + c)) + 285*(
sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 597
6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) -
1968*(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^2*sgn(cos(d*x + c))
- 4605*(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^3*sgn(cos(d*x +
c)) - 31320*(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^3*sgn(cos(d*
x + c)) - 2640*(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^3*sgn(cos
(d*x + c)) + 37281*(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^4*sgn(
cos(d*x + c)) + 90168*(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^4*s
gn(cos(d*x + c)) + 41616*(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^
4*sgn(cos(d*x + c)) - 35643*(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^5*sgn(cos(d*x + c)) - 66024*(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^5*sgn(cos(d*x + c)) - 42288*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sq
rt(-a)*a^5*sgn(cos(d*x + c)) + 9175*(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^6*sgn(cos(d*x + c)) + 16904*(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^6*sgn(cos(d*x + c)) + 12528*(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^6*sgn(cos(d*x + c)) - 1311*(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^7*sgn(cos(d*x + c)) - 1992*(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^7*sgn(cos(d*x + c)) - 1392*(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^7*sgn(cos(d*x + c)) + 43*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 104*B*sqrt(-a)*a^8*sgn(cos(
d*x + c)) + 48*C*sqrt(-a)*a^8*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)^4)/d