3.190 \(\int \frac{\cos ^3(c+d x) (A+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=200 \[ \frac{(7 A+8 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}-\frac{(9 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}-\frac{A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((9*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(A + C)*ArcTan
[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + ((7*A + 8*C)*Sin[c + d*x])/(8*d*Sqr
t[a + a*Sec[c + d*x]]) - (A*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (A*Cos[c + d*x]^2*Sin
[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.56293, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4087, 4022, 3920, 3774, 203, 3795} \[ \frac{(7 A+8 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}-\frac{(9 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 \sqrt{a} d}+\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}-\frac{A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((9*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(A + C)*ArcTan
[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + ((7*A + 8*C)*Sin[c + d*x])/(8*d*Sqr
t[a + a*Sec[c + d*x]]) - (A*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (A*Cos[c + d*x]^2*Sin
[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

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 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 ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{A \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{a A}{2}+\frac{1}{2} a (5 A+6 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{A \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 A+8 C)-\frac{3}{4} a^2 A \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac{(7 A+8 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{A \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 A+8 C)+\frac{3}{8} a^3 (7 A+8 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac{(7 A+8 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+(A+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx-\frac{(9 A+8 C) \int \sqrt{a+a \sec (c+d x)} \, dx}{16 a}\\ &=\frac{(7 A+8 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{(2 (A+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 A+8 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{(9 A+8 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} (A+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 A+8 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}-\frac{A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.555509, size = 145, normalized size = 0.72 \[ \frac{\tan (c+d x) \left (\cos (c+d x) \sqrt{1-\sec (c+d x)} \left (8 A \cos ^2(c+d x)-2 A \cos (c+d x)+21 A+24 C\right )-3 (9 A+8 C) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )+24 \sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{24 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(9*A + 8*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 24*Sqrt[2]*(A + C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]
+ Cos[c + d*x]*(21*A + 24*C - 2*A*Cos[c + d*x] + 8*A*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 = 0.332, size = 1056, 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)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/192/d/a*(27*A*sin(d*x+c)*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))^(5/2)*cos(d*x+c)^2+24*C*sin(d*x+c)*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))^(5/2)*cos(d*x+c)^2+54*A*sin(d*
x+c)*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)/(c
os(d*x+c)+1))^(5/2)*cos(d*x+c)+48*A*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)*cos(d*x+c)^2+48*C*sin(d*x+c)*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))^(5/2)*cos(d*x+c)+4
8*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)*cos(d*x+c)^2+27*A*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^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+96*A*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)*cos(d*x+c)+24*C*arcta
nh(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))^(5/2)*sin(d*x+c)+96*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)*cos(d*x+c)+48*A*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*A*cos(
d*x+c)^6+80*A*cos(d*x+c)^5-184*A*cos(d*x+c)^4-192*C*cos(d*x+c)^4+168*A*cos(d*x+c)^3+192*C*cos(d*x+c)^3)*(a*(co
s(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{3}}{\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)^3*(A+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 + A)*cos(d*x + c)^3/sqrt(a*sec(d*x + c) + a), x)

________________________________________________________________________________________

Fricas [A]  time = 5.90081, size = 1399, normalized size = 7. \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+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/48*(24*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + 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*A + 8*C)*cos(d*x + c) + 9*A + 8*C)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sq
rt((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*cos(d*x + c)^3 - 2*A*cos(d*x + c)^2 + 3*(7*A + 8*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*A + 8*C)*cos(d*x + c) + 9*A + 8*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))) + (8*A*cos(d*x + c)^3 - 2*A*cos(d*x + c)
^2 + 3*(7*A + 8*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 24*sqrt(2)*((A + C)*a*
cos(d*x + c) + (A + 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)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 12.0926, size = 1153, normalized size = 5.76 \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+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/48*(24*sqrt(2)*(A*sqrt(-a) + C*sqrt(-a))*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a))^2)/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 3*(9*A + 8*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*A + 8
*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)))/(sqr
t(-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*A*sqrt(-a) + 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) - 1323*(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 - 888*(sq
rt(-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 + 3906*(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^2 + 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^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*A*sqrt(-a)*a^3 - 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^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*A*sqrt(-a)*a^4
+ 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^4 - 31*A*sqrt(-a)*
a^5 - 24*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