3.179 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=192 \[ \frac{a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (3 A-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d}+\frac{5 a^{5/2} (5 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}+\frac{5 a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d} \]
[Out]
(5*a^(5/2)*(5*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(49*A - 24*C)*Sin
[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(3*A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + (
5*a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5
/2)*Sin[c + d*x])/(3*d)
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Rubi [A] time = 0.619128, antiderivative size = 192, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{4087, 4017, 4018, 4015, 3774, 203} \[ \frac{a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (3 A-8 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{4 d}+\frac{5 a^{5/2} (5 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d}+\frac{5 a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(5*a^(5/2)*(5*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(49*A - 24*C)*Sin
[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(3*A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + (
5*a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(5
/2)*Sin[c + d*x])/(3*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 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 ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}-\frac{1}{2} a (A-6 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{5 a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (31 A+24 C)-\frac{3}{4} a^2 (3 A-8 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=-\frac{a^2 (3 A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{5 a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (49 A-24 C)+\frac{1}{8} a^3 (13 A+72 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac{a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (3 A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{5 a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{16} \left (5 a^2 (5 A+8 C)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (3 A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{5 a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}-\frac{\left (5 a^3 (5 A+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 )}{8 d}\\ &=\frac{5 a^{5/2} (5 A+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}+\frac{a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (3 A-8 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{5 a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 1.48149, size = 132, normalized size = 0.69 \[ \frac{a^2 \sin (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{\sec (c+d x)-1} (3 (27 A+8 C) \cos (c+d x)+17 A \cos (2 (c+d x))+2 A \cos (3 (c+d x))+17 A+48 C)+15 (5 A+8 C) \tan ^{-1}\left (\sqrt{\sec (c+d x)-1}\right )\right )}{24 d (\cos (c+d x)+1) \sqrt{\sec (c+d x)-1}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
(a^2*(15*(5*A + 8*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]] + (17*A + 48*C + 3*(27*A + 8*C)*Cos[c + d*x] + 17*A*Cos[2
*(c + d*x)] + 2*A*Cos[3*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(24*d*(1
+ Cos[c + d*x])*Sqrt[-1 + Sec[c + d*x]])
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Maple [B] time = 0.406, 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)^3*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
[Out]
-1/192/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(75*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+120*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+150*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)+240*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)+75*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^(1/2)*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+120*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))^(5/2)*sin(d*x+c)+64*A*cos(d*x+c)^6+208*A*cos(d*x+c)^5+328*A
*cos(d*x+c)^4+192*C*cos(d*x+c)^4-600*A*cos(d*x+c)^3+192*C*cos(d*x+c)^3-384*C*cos(d*x+c)^2)/cos(d*x+c)^2/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)^3*(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 = 0.67021, size = 968, normalized size = 5.04 \begin{align*} \left [\frac{15 \,{\left ({\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (5 \, A + 8 \, 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 (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 34 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (25 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, 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 )}{48 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{15 \,{\left ({\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (5 \, A + 8 \, 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 (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 34 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (25 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, 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 ) + d\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/48*(15*((5*A + 8*C)*a^2*cos(d*x + c) + (5*A + 8*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*(8*
A*a^2*cos(d*x + c)^3 + 34*A*a^2*cos(d*x + c)^2 + 3*(25*A + 8*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(15*((5*A + 8*C)*a^2*cos(d*x + c) + (5*A + 8*
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))) - (8*A*a^2
*cos(d*x + c)^3 + 34*A*a^2*cos(d*x + c)^2 + 3*(25*A + 8*C)*a^2*cos(d*x + c) + 48*C*a^2)*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)**3*(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 = 7.3483, size = 1261, normalized size = 6.57 \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+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/48*(96*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*C*a^3*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*
d*x + 1/2*c)^2 - a) + 15*(5*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sq
rt(-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))) - 15*(5*A*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*t
an(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(75*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 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)*a^3*sgn(cos(d*x + c)) - 1125*(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)) - 888*(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)) + 6174*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 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^5*sgn(cos(d*x + c)) - 4314*(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)) - 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^6*sgn(cos(d*x + c)) + 807*(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)) + 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^7*sgn(cos(d*x + c)) - 49*A*sqrt(-a)*a^8*sgn(cos(d*x
+ c)) - 24*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)^3)/d