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

Optimal. Leaf size=193 \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (35 A+31 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}-\frac{4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}}-\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*(35*A + 3
7*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) - (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(35*d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 31*C)*Sqrt[a + a*Sec
[c + d*x]]*Tan[c + d*x])/(105*a*d)

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Rubi [A]  time = 0.593652, antiderivative size = 193, 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 = {4089, 4021, 4010, 4001, 3795, 203} \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (35 A+31 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}-\frac{4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}}-\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*(35*A + 3
7*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) - (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(35*d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 31*C)*Sqrt[a + a*Sec
[c + d*x]]*Tan[c + d*x])/(105*a*d)

Rule 4089

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(m + n + 1)
), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + a
*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1
)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 4021

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*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] && GtQ[n, 1]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

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
]

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 \frac{\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^3(c+d x) \left (\frac{1}{2} a (7 A+6 C)-\frac{1}{2} a C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{7 a}\\ &=-\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\sec ^2(c+d x) \left (-a^2 C+\frac{1}{4} a^2 (35 A+31 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=-\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac{8 \int \frac{\sec (c+d x) \left (\frac{1}{8} a^3 (35 A+31 C)-\frac{1}{4} a^3 (35 A+37 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(A+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}-\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{\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{4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}\\ \end{align*}

Mathematica [A]  time = 6.11893, size = 173, normalized size = 0.9 \[ \frac{2 \cos ^2(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (A+C \sec ^2(c+d x)\right ) \left (105 \sqrt{2} (A+C) \cot (c+d x) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )+2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) ((35 A+43 C) \cos (2 (c+d x))+35 A+24 C \cos (c+d x)+73 C)\right )}{105 a d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(A + C*Sec[c + d*x]^2)*(105*Sqrt[2]*(A + C)*ArcTan[Sqrt[-1 + Sec[
c + d*x]]/Sqrt[2]]*Cot[c + d*x]*Sqrt[-1 + Sec[c + d*x]] + 2*(35*A + 73*C + 24*C*Cos[c + d*x] + (35*A + 43*C)*C
os[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(105*a*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.366, size = 776, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/840/d/a*(105*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+c
os(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3+105*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)^3+315*A*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*
cos(d*x+c)^2+315*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)^2+315*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)+315*C*(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*c
os(d*x+c)+105*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)+105*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)-560*A*cos(d*x+c)^4-688*C*cos(d*x+c)^4+1120*A*cos(d*x
+c)^3+1184*C*cos(d*x+c)^3-560*A*cos(d*x+c)^2-544*C*cos(d*x+c)^2+288*C*cos(d*x+c)-240*C)*(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(sec(d*x+c)^3*(A+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 = 0.611454, size = 1098, normalized size = 5.69 \begin{align*} \left [\frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} +{\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, -\frac{2 \,{\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} +{\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac{\sqrt{2} \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 )}{\sqrt{a}}}{105 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*c
os(d*x + c) + a)/cos(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)) - 4*((35*A + 43*C)*cos(d*x + c)^3 - (35*A + 31*C)*cos(d*x + c)^2 + 3*C*c
os(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x +
c)^3), -1/105*(2*((35*A + 43*C)*cos(d*x + c)^3 - (35*A + 31*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 15*C)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x +
c)^3)*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)^4 + a*d*cos(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [A]  time = 9.30975, size = 333, normalized size = 1.73 \begin{align*} \frac{\frac{105 \, \sqrt{2}{\left (A + C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{4 \,{\left ({\left (\frac{\sqrt{2}{\left (35 \, A a^{3} + 46 \, C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{14 \, \sqrt{2}{\left (5 \, A a^{3} + 4 \, C a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{35 \, \sqrt{2}{\left (A a^{3} + 2 \, C a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/105*(105*sqrt(2)*(A + 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)))/(sqr
t(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 4*((sqrt(2)*(35*A*a^3 + 46*C*a^3)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*
d*x + 1/2*c)^2 - 1) - 14*sqrt(2)*(5*A*a^3 + 4*C*a^3)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2 +
35*sqrt(2)*(A*a^3 + 2*C*a^3)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^3/((a*tan(1/2*d*x + 1/2*c)
^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d