3.392 \(\int \frac{\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)
Optimal. Leaf size=206 \[ \frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}-\frac{(9 B-14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{(B-6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a \sec (c+d x)+a}} \]
[Out]
-((9*B - 14*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(B - C)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + ((7*B - 2*C)*Sin[c + d*x])/(8*d*Sq
rt[a + a*Sec[c + d*x]]) - ((B - 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (B*Cos[c + d
*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.645478, antiderivative size = 206, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4072, 4022, 3920, 3774, 203, 3795} \[ \frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}-\frac{(9 B-14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{(B-6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
-((9*B - 14*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(B - C)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + ((7*B - 2*C)*Sin[c + d*x])/(8*d*Sq
rt[a + a*Sec[c + d*x]]) - ((B - 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (B*Cos[c + d
*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])
Rule 4072
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 4022
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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 - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
Rule 3920
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 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])
Rule 3795
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \sec (c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos ^2(c+d x) \left (-\frac{1}{2} a (B-6 C)+\frac{5}{2} a B \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (\frac{3}{4} a^2 (7 B-2 C)-\frac{3}{4} a^2 (B-6 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-\frac{3}{8} a^3 (9 B-14 C)+\frac{3}{8} a^3 (7 B-2 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{(9 B-14 C) \int \sqrt{a+a \sec (c+d x)} \, dx}{16 a}+(B-C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{(9 B-14 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 )}{8 d}-\frac{(2 (B-C)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{(9 B-14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{(7 B-2 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.762773, size = 150, normalized size = 0.73 \[ \frac{\tan (c+d x) \left (\cos (c+d x) \sqrt{1-\sec (c+d x)} \left (-2 (B-6 C) \cos (c+d x)+8 B \cos ^2(c+d x)+21 B-6 C\right )+(42 C-27 B) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )+24 \sqrt{2} (B-C) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{24 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(((-27*B + 42*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 24*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]
+ Cos[c + d*x]*(21*B - 6*C - 2*(B - 6*C)*Cos[c + d*x] + 8*B*Cos[c + d*x]^2)*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*
x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
________________________________________________________________________________________
Maple [B] time = 0.349, size = 1067, normalized size = 5.2 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)
[Out]
-1/192/d/a*(-27*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)+42*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)-48*B*cos(
d*x+c)^2*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(5/2)*sin(d*x+c)-54*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)+48*C*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+84*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)*s
in(d*x+c)/cos(d*x+c))*sin(d*x+c)-96*B*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c
)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-27*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)+96*C*cos(d*x+c
)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(5/2)*sin(d*x+c)+42*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)-48*B*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(
d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+48*C*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+64*B*cos(d*x+c)^6-8
0*B*cos(d*x+c)^5+96*C*cos(d*x+c)^5+184*B*cos(d*x+c)^4-144*C*cos(d*x+c)^4-168*B*cos(d*x+c)^3+48*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)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{4}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^4/sqrt(a*sec(d*x + c) + a), x)
________________________________________________________________________________________
Fricas [A] time = 5.97143, size = 1426, normalized size = 6.92 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[-1/48*(24*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/co
s(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*
cos(d*x + c) + 1)) - 3*((9*B - 14*C)*cos(d*x + c) + 9*B - 14*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*(8*B*cos(d*x + c)^3 - 2*(B - 6*C)*cos(d*x + c)^2 + 3*(7*B - 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), 1/24*(3*((9*B - 14*C)*cos(d*x + c) + 9*B - 14*C)*sqrt(a)*ar
ctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (8*B*cos(d*x + c)^3 - 2*(B
- 6*C)*cos(d*x + c)^2 + 3*(7*B - 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 24
*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x +
c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]
________________________________________________________________________________________
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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 11.7257, size = 1142, normalized size = 5.54 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
1/48*(24*sqrt(2)*(B - C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a
)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 3*(9*B - 14*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)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 3*(9*B - 14*C)*log(ab
s((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)))/(sqrt(-a)*sgn(
tan(1/2*d*x + 1/2*c)^2 - 1)) + 4*sqrt(2)*(165*(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) - 102*(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)
- 1323*(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 + 954*(sqrt(-a)*ta
n(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a + 3906*(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^2 - 2268*(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^2 - 2118*(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^3 + 1044*(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^3 + 393*(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^4 - 222*(s
qrt(-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^4 - 31*B*sqrt(-a)*a^5 + 18*
C*sqrt(-a)*a^5)/(((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)^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d