3.201 \(\int \frac{\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=212 \[ \frac{(19 A+163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{5 (3 A+19 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}-\frac{(21 A+197 C) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(A+17 C) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

((19*A + 163*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((
A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((A + 17*C)*Sec[c + d*x]^2*Tan[c + d*x]
)/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((21*A + 197*C)*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + (5
*(3*A + 19*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(48*a^3*d)

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Rubi [A]  time = 0.659367, antiderivative size = 212, 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 = {4085, 4019, 4010, 4001, 3795, 203} \[ \frac{(19 A+163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{5 (3 A+19 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}-\frac{(21 A+197 C) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(A+17 C) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((19*A + 163*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((
A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((A + 17*C)*Sec[c + d*x]^2*Tan[c + d*x]
)/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((21*A + 197*C)*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + (5
*(3*A + 19*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(48*a^3*d)

Rule 4085

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*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + 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] && LtQ[m, -2^(-1)] && GtQ[n, 0]

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 )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{\sec ^3(c+d x) \left (-a (A-3 C)-\frac{1}{2} a (3 A+11 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(A+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^2(c+d x) \left (a^2 (A+17 C)-\frac{5}{4} a^2 (3 A+19 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(A+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{5 (3 A+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{\int \frac{\sec (c+d x) \left (-\frac{5}{8} a^3 (3 A+19 C)+\frac{1}{4} a^3 (21 A+197 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(A+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(21 A+197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{5 (3 A+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{(19 A+163 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(A+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(21 A+197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{5 (3 A+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{(19 A+163 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 )}{16 a^2 d}\\ &=\frac{(19 A+163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(A+17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(21 A+197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{5 (3 A+19 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}\\ \end{align*}

Mathematica [A]  time = 2.33556, size = 196, normalized size = 0.92 \[ \frac{\tan (c+d x) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-(81 A+1537 C) \cos (c+d x)-2 (39 A+503 C) \cos (2 (c+d x))-\frac{6 \sqrt{2} (19 A+163 C) \cos ^2(c+d x) (\cos (c+d x)+1)^2 \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )}{\cos (c+d x)-1}-27 A \cos (3 (c+d x))-78 A-299 C \cos (3 (c+d x))-878 C\right )}{96 d (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((-78*A - 878*C - (81*A + 1537*C)*Cos[c + d*x] - 2*(39*A + 503*C)*Cos[2*(c + d*x)] - 27*A*Cos[3*(c + d*x)] - 2
99*C*Cos[3*(c + d*x)] - (6*Sqrt[2]*(19*A + 163*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[c + d*x]^2*(1 +
Cos[c + d*x])^2*Sqrt[-1 + Sec[c + d*x]])/(-1 + Cos[c + d*x]))*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*Tan[c + d*x]
)/(96*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.365, size = 786, normalized size = 3.7 \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))^(5/2),x)

[Out]

-1/192/d/a^3*(-1+cos(d*x+c))^2*(57*A*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)*cos(d*x+c)^3+489*C*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)*cos(d*x+c)^3+171*A
*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(3/2)*sin(d*x+c)*cos(d*x+c)^2+1467*C*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)*cos(d*x+c)^2+171*A*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)+1467*
C*sin(d*x+c)*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))*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(3/2)*cos(d*x+c)+57*A*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+489*C*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)-108*A*cos(d*x+c)^4-1196*C*cos(d*
x+c)^4-48*A*cos(d*x+c)^3-816*C*cos(d*x+c)^3+156*A*cos(d*x+c)^2+1372*C*cos(d*x+c)^2+768*C*cos(d*x+c)-128*C)*(a*
(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.619588, size = 1447, normalized size = 6.83 \begin{align*} \text{result too large to display} \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))^(5/2),x, algorithm="fricas")

[Out]

[-1/192*(3*sqrt(2)*((19*A + 163*C)*cos(d*x + c)^4 + 3*(19*A + 163*C)*cos(d*x + c)^3 + 3*(19*A + 163*C)*cos(d*x
+ c)^2 + (19*A + 163*C)*cos(d*x + c))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
)*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)
) + 4*((27*A + 299*C)*cos(d*x + c)^3 + (39*A + 503*C)*cos(d*x + c)^2 + 160*C*cos(d*x + c) - 32*C)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c
)^2 + a^3*d*cos(d*x + c)), -1/96*(3*sqrt(2)*((19*A + 163*C)*cos(d*x + c)^4 + 3*(19*A + 163*C)*cos(d*x + c)^3 +
3*(19*A + 163*C)*cos(d*x + c)^2 + (19*A + 163*C)*cos(d*x + c))*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))) + 2*((27*A + 299*C)*cos(d*x + c)^3 + (39*A + 503*C)*cos(
d*x + c)^2 + 160*C*cos(d*x + c) - 32*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x +
c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))]

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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 )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, 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))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]  time = 9.61947, size = 419, normalized size = 1.98 \begin{align*} -\frac{\frac{{\left ({\left (3 \,{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} + C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{\sqrt{2}{\left (7 \, A a^{5} + 23 \, C a^{5}\right )}}{a^{6} \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{4 \, \sqrt{2}{\left (15 \, A a^{5} + 167 \, C a^{5}\right )}}{a^{6} \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{3 \, \sqrt{2}{\left (11 \, A a^{5} + 155 \, C a^{5}\right )}}{a^{6} \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 )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{3 \, \sqrt{2}{\left (19 \, A + 163 \, 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} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{96 \, 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))^(5/2),x, algorithm="giac")

[Out]

-1/96*(((3*(2*sqrt(2)*(A*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + sqrt(2)*(
7*A*a^5 + 23*C*a^5)/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 - 4*sqrt(2)*(15*A*a^5 + 167*
C*a^5)/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 + 3*sqrt(2)*(11*A*a^5 + 155*C*a^5)/(a^6*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a)) - 3*sqrt(2)*(19*A + 163*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)))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d