3.175 \(\int \frac{A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}-\frac{7 A \tan (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \tan (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

[Out]

(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(5/2)*d) - (23*A*ArcTan[(Sqrt[a]*Tan[c + d*x]
)/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) - (A*Tan[c + d*x])/(2*d*(a - a*Sec[c + d*x])^(5/2
)) - (7*A*Tan[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.206017, antiderivative size = 185, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3904, 3887, 471, 527, 522, 203} \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}+\frac{7 A \sin (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{16 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \sin (c+d x) \cos (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{8 a^2 d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(5/2),x]

[Out]

(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(5/2)*d) - (23*A*ArcTan[(Sqrt[a]*Tan[c + d*x]
)/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) + (7*A*Csc[(c + d*x)/2]^2*Sin[c + d*x])/(16*a^2*d
*Sqrt[a - a*Sec[c + d*x]]) - (A*Cos[c + d*x]*Csc[(c + d*x)/2]^4*Sin[c + d*x])/(8*a^2*d*Sqrt[a - a*Sec[c + d*x]
])

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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{A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx &=-\left ((a A) \int \frac{\tan ^2(c+d x)}{(a-a \sec (c+d x))^{7/2}} \, dx\right )\\ &=\frac{(2 A) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a d}\\ &=-\frac{A \cos (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \operatorname{Subst}\left (\int \frac{1-3 a x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{2 a^2 d}\\ &=\frac{7 A \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \cos (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \operatorname{Subst}\left (\int \frac{9 a-7 a^2 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^3 d}\\ &=\frac{7 A \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \cos (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^2 d}+\frac{(23 A) \operatorname{Subst}\left (\int \frac{1}{2+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^2 d}\\ &=\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}+\frac{7 A \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{A \cos (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.73887, size = 387, normalized size = 2.55 \[ A \left (\frac{\sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \left (\frac{11 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{d}-\frac{11 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{d}-\frac{\cot \left (\frac{c}{2}\right ) \csc ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{15 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{15 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}\right )}{(a-a \sec (c+d x))^{5/2}}+\frac{e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{5}{2}}(c+d x) \left (8 \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac{23 \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{2}}+8 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt{2} d (a-a \sec (c+d x))^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(5/2),x]

[Out]

A*((Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(8*ArcSinh[E^(I*(c + d*x))]
- (23*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] + 8*ArcTanh[Sqrt[1 + E^(
(2*I)*(c + d*x))]])*Sec[c + d*x]^(5/2)*Sin[c/2 + (d*x)/2]^5)/(Sqrt[2]*d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d*x
])^(5/2)) + (Sec[c + d*x]^3*((-11*Cos[c/2]*Cos[(d*x)/2])/d + (15*Cot[c/2]*Csc[c/2 + (d*x)/2])/(2*d) - (Cot[c/2
]*Csc[c/2 + (d*x)/2]^3)/d - (15*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/(2*d) + (Csc[c/2]*Csc[c/2 + (d*x)/
2]^4*Sin[(d*x)/2])/d + (11*Sin[c/2]*Sin[(d*x)/2])/d)*Sin[c/2 + (d*x)/2]^5)/(a - a*Sec[c + d*x])^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.257, size = 695, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x)

[Out]

-1/12*A/d*2^(1/2)*(-1+cos(d*x+c))^4*(21*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^3*2^(1/2)+33*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*2^(1/2)+23*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^3*2^(1/2)+
3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)-23*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2
*2^(1/2)-9*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)-5*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)-69*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3*2^(1/2)-23*cos(d*x+c)*2^(1/2)*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(3/2)-96*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3-11*cos(d*x+
c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+69*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)
^2*2^(1/2)+23*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+96*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2))*cos(d*x+c)^2+37*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+69*cos(d*x+c)*2^(1/2)*arctan(
1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+96*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))
-21*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-69*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-96*
arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(a*(-1+cos(d*x+c))/cos(d*x+c))^(5/2)/sin(d*x+c)^7/(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A \sec \left (d x + c\right ) + A}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((A*sec(d*x + c) + A)/(-a*sec(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

Fricas [B]  time = 0.543397, size = 1544, normalized size = 10.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/32*(23*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x
+ c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c)
- 1)*sin(d*x + c)))*sin(d*x + c) + 32*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + 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) - 4*(11*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 - 7*A*cos(d*x + c))*sqrt((a*cos(d*x + c)
- a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c)), 1/16*(23*sqrt(2)*(A*
cos(d*x + c)^2 - 2*A*cos(d*x + c) + 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) - 32*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + 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)))*sin(d*x + c) + 2*(11*A*cos(d*x + c)^3 +
4*A*cos(d*x + c)^2 - 7*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^
3*d*cos(d*x + c) + a^3*d)*sin(d*x + c))]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} A \left (\int \frac{\sec{\left (c + d x \right )}}{a^{2} \sqrt{- a \sec{\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt{- a \sec{\left (c + d x \right )} + a} \sec{\left (c + d x \right )} + a^{2} \sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx + \int \frac{1}{a^{2} \sqrt{- a \sec{\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt{- a \sec{\left (c + d x \right )} + a} \sec{\left (c + d x \right )} + a^{2} \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((A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(5/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 2.03052, size = 296, normalized size = 1.95 \begin{align*} -\frac{A{\left (\frac{23 \, \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{5}{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{32 \, \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{5}{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 (9 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} + 7 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a\right )}}{a^{4} \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 ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/16*A*(23*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
)*sgn(tan(1/2*d*x + 1/2*c))) - 32*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*sgn(
tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(9*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2) + 7*s
qrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a)/(a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))*tan(1/2*d*
x + 1/2*c)^4))/d