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

Optimal. Leaf size=155 \[ \frac{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{11 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{4 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}} \]

[Out]

(11*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*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) + (5*A*Sin[c + d*x])/(4*d*Sqrt[a - a*Sec[c + d*x]
]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a - a*Sec[c + d*x]])

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Rubi [A]  time = 0.361976, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{11 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{4 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*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) + (5*A*Sin[c + d*x])/(4*d*Sqrt[a - a*Sec[c + d*x]
]) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[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 ^2(c+d x) (A+A \sec (c+d x))}{\sqrt{a-a \sec (c+d x)}} \, dx &=\frac{A \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{\cos (c+d x) \left (-\frac{5 a A}{2}-\frac{3}{2} a A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a}\\ &=\frac{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}}+\frac{\int \frac{\frac{11 a^2 A}{4}+\frac{5}{4} a^2 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 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{(11 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{8 a}\\ &=\frac{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}}+\frac{(11 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 )}{4 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{11 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{4 \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{5 A \sin (c+d x)}{4 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.62897, size = 297, normalized size = 1.92 \[ \frac{A e^{-\frac{1}{2} i (c+d x)} \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (11 e^{-\frac{1}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+e^{-\frac{1}{2} i (c+d x)} \left (6 e^{-i (c+d x)}+7 e^{i (c+d x)}+e^{-2 i (c+d x)}+6 e^{2 i (c+d x)}+e^{3 i (c+d x)}-16 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+11 \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+7\right )\right )}{8 d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*((11*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])/E^((I/2)*(c + d*x)) + (7 + 6/E^(I*(c + d*x)) +
7*E^(I*(c + d*x)) + E^((-2*I)*(c + d*x)) + 6*E^((2*I)*(c + d*x)) + E^((3*I)*(c + d*x)) - 16*Sqrt[2]*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))])] + 11*Sqrt[1 + E^((
2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])/E^((I/2)*(c + d*x)))*Sec[c + d*x]*(Cos[(c + d*x)/2] +
I*Sin[(c + d*x)/2])*Sin[(c + d*x)/2])/(8*d*E^((I/2)*(c + d*x))*Sqrt[a - a*Sec[c + d*x]])

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Maple [B]  time = 0.345, size = 367, normalized size = 2.4 \begin{align*}{\frac{A\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{24\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( 6\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-16\,\cos \left ( dx+c \right ) \sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}+27\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-16\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}+4\,\cos \left ( dx+c \right ) \sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+48\,\cos \left ( dx+c \right ) \sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +66\,\cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) +15\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+48\,\sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +66\,\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \right ){\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}{\frac{1}{\sqrt{{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*A/d*2^(1/2)*(-1+cos(d*x+c))^2*(6*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-16*cos(d*x+c)*
2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+27*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-16*2
^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+4*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+48*cos(d
*x+c)*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+66*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2))+15*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+48*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2))+66*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)^3

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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 )^{2}}{\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)^2*(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)^2/sqrt(-a*sec(d*x + c) + a), x)

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Fricas [A]  time = 0.538782, size = 1188, normalized size = 7.66 \begin{align*} \left [\frac{8 \, \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 ) - 11 \, 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 (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, a d \sin \left (d x + c\right )}, \frac{8 \, \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 ) - 11 \, 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 (2 \, A \cos \left (d x + c\right )^{3} + 7 \, A \cos \left (d x + c\right )^{2} + 5 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, 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)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/8*(8*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) - 11*
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*cos(
d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 2*(2*A*cos(d*x + c)^3 + 7*A*cos(d*x + c)^2 + 5*A*cos(
d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/(a*d*sin(d*x + c)), 1/4*(8*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) - 11*A*sqrt(a)*arct
an(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) - (2*A*cos(d*x +
c)^3 + 7*A*cos(d*x + c)^2 + 5*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]  time = 0., size = 0, normalized size = 0. \begin{align*} A \left (\int \frac{\cos ^{2}{\left (c + d x \right )}}{\sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx + \int \frac{\cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(Integral(cos(c + d*x)**2/sqrt(-a*sec(c + d*x) + a), x) + Integral(cos(c + d*x)**2*sec(c + d*x)/sqrt(-a*sec(
c + d*x) + a), x))

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Giac [C]  time = 2.01873, size = 379, normalized size = 2.45 \begin{align*} -\frac{A a{\left (\frac{8 \, \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{11 \, \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 (3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} + 10 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{2} 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 (8 i \, \sqrt{2} A \sqrt{-a} \arctan \left (-i\right ) - 11 i \, A \sqrt{-a} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right ) - 7 \, \sqrt{2} A \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{a}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(A*a*(8*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))) - 11*arctan(1/2*sqrt(2)*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))) - sqrt(2)*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2) + 10
*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^2*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn
(tan(1/2*d*x + 1/2*c)))) + (8*I*sqrt(2)*A*sqrt(-a)*arctan(-I) - 11*I*A*sqrt(-a)*arctan(-1/2*I*sqrt(2)) - 7*sqr
t(2)*A*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c))/a)/d