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

Optimal. Leaf size=192 \[ \frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}} \]

[Out]

(23*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*Sqrt[a]*d) - (2*Sqrt[2]*A*ArcTan[(Sqrt[a]*Ta
n[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d) + (9*A*Sin[c + d*x])/(8*d*Sqrt[a - a*Sec[c + d*x]
]) + (7*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*Sq
rt[a - a*Sec[c + d*x]])

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Rubi [A]  time = 0.524277, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {4022, 3920, 3774, 203, 3795} \[ \frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*Sqrt[a]*d) - (2*Sqrt[2]*A*ArcTan[(Sqrt[a]*Ta
n[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d) + (9*A*Sin[c + d*x])/(8*d*Sqrt[a - a*Sec[c + d*x]
]) + (7*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*Sq
rt[a - a*Sec[c + d*x]])

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) (A+A \sec (c+d x))}{\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{7 a A}{2}-\frac{5}{2} a A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{7 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{27 a^2 A}{4}+\frac{21}{4} a^2 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 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{69 a^3 A}{8}-\frac{27}{8} a^3 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 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)}}+(2 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx+\frac{(23 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{16 a}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 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{(23 A) \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{(4 A) \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{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 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 [C]  time = 1.88793, size = 330, normalized size = 1.72 \[ \frac{A e^{-4 i (c+d x)} \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (9 e^{i (c+d x)}+40 e^{2 i (c+d x)}+47 e^{3 i (c+d x)}+47 e^{4 i (c+d x)}+40 e^{5 i (c+d x)}+9 e^{6 i (c+d x)}+2 e^{7 i (c+d x)}+69 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-96 \sqrt{2} e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+69 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+2\right )}{48 d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(2 + 9*E^(I*(c + d*x)) + 40*E^((2*I)*(c + d*x)) + 47*E^((3*I)*(c + d*x)) + 47*E^((4*I)*(c + d*x)) + 40*E^((
5*I)*(c + d*x)) + 9*E^((6*I)*(c + d*x)) + 2*E^((7*I)*(c + d*x)) + 69*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcSinh[E^(I*(c + d*x))] - 96*Sqrt[2]*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 +
E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 69*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x)
)]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c + d*x)/
2])/(48*d*E^((4*I)*(c + d*x))*Sqrt[a - a*Sec[c + d*x]])

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Maple [B]  time = 0.404, size = 625, normalized size = 3.3 \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))/(a-a*sec(d*x+c))^(1/2),x)

[Out]

-1/240*A/d*2^(1/2)*(-1+cos(d*x+c))^3*(96*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*2^(1/2)+192*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)+40*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*2^(1/2)+
96*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)-160*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*2^(1/2)+
190*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*2^(1/2)-320*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(3/2)+465*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-160*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(3/2)-49*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+480*arctan(1/(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2))*cos(d*x+c)^2*2^(1/2)+155*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+960*cos(d*x+c)
*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+690*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2))*cos(d*x+c)^2+135*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+480*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2))+1380*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+690*arctan(1/2*2^(1
/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))
^(1/2)/sin(d*x+c)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (A \sec \left (d x + c\right ) + 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+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((A*sec(d*x + c) + A)*cos(d*x + c)^3/sqrt(-a*sec(d*x + c) + a), x)

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Fricas [A]  time = 0.548307, size = 1258, normalized size = 6.55 \begin{align*} \left [\frac{48 \, \sqrt{2} A a \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} -{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 69 \, A \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \,{\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{48 \, a d \sin \left (d x + c\right )}, \frac{48 \, \sqrt{2} A \sqrt{a} \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 ) \sin \left (d x + c\right ) - 69 \, A \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 ) \sin \left (d x + c\right ) -{\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{24 \, a d \sin \left (d x + c\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))/(a-a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/48*(48*sqrt(2)*A*a*sqrt(-1/a)*log(-(2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos
(d*x + c))*sqrt(-1/a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) - 6
9*A*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) - (2*a*co
s(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 2*(8*A*cos(d*x + c)^4 + 22*A*cos(d*x + c)^3 + 41*A*
cos(d*x + c)^2 + 27*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c)), 1/24*(48*sqrt
(2)*A*sqrt(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)))*sin(
d*x + c) - 69*A*sqrt(a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*si
n(d*x + c) - (8*A*cos(d*x + c)^4 + 22*A*cos(d*x + c)^3 + 41*A*cos(d*x + c)^2 + 27*A*cos(d*x + c))*sqrt((a*cos(
d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c))]

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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))/(a-a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [C]  time = 2.21682, size = 412, normalized size = 2.15 \begin{align*} -\frac{A a{\left (\frac{48 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{69 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2}{\left (21 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{5}{2}} + 80 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} a + 108 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{3} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}\right )} + \frac{{\left (48 i \, \sqrt{2} A \sqrt{-a} \arctan \left (-i\right ) - 69 i \, A \sqrt{-a} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right ) - 49 \, \sqrt{2} A \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{a}}{24 \, d} \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))/(a-a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/24*(A*a*(48*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2
- 1)*sgn(tan(1/2*d*x + 1/2*c))) - 69*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(21*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2) +
80*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*a + 108*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2)/((a*tan(1/2*d*x + 1/2
*c)^2 + a)^3*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c)))) + (48*I*sqrt(2)*A*sqrt(-a)*arctan(-
I) - 69*I*A*sqrt(-a)*arctan(-1/2*I*sqrt(2)) - 49*sqrt(2)*A*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c))/a)/d