\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [365]

Optimal result

Integrand size = 42, antiderivative size = 234 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d} \]

[Out]

4/1155*(187*B+168*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+4/495*a^2*(187*B+168*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))
^(1/2)+2/693*a^2*(187*B+168*C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a^2*(11*B+12*C)*sec(d*x+c
)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-8/3465*a*(187*B+168*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/11*a*C*se
c(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4157, 4103, 4101, 3888, 3885, 4086, 3877} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 (11 B+12 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (187 B+168 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {4 (187 B+168 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac {8 a (187 B+168 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d} \]

[In]

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

[Out]

(4*a^2*(187*B + 168*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(187*B + 168*C)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(11*B + 12*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[
a + a*Sec[c + d*x]]) - (8*a*(187*B + 168*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*C*Sec[c + d
*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(187*B + 168*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(1155*d)

Rule 3877

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(Cot[e + f*x]/(
f*Sqrt[a + b*Csc[e + f*x]])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3885

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3888

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*
Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[2*a*d*((n - 1)/(b*(2
*n - 1))), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 4086

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 4101

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

Rule 4103

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

Rule 4157

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

Rubi steps \begin{align*} \text {integral}& = \int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} (B+C \sec (c+d x)) \, dx \\ & = \frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {2}{11} \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (11 B+8 C)+\frac {1}{2} a (11 B+12 C) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{99} (a (187 B+168 C)) \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {1}{231} (2 a (187 B+168 C)) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {(4 (187 B+168 C)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{1155} \\ & = \frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {1}{495} (2 a (187 B+168 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 a (187 B+168 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.77 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.48 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 \left (2992 B+2688 C+8 (187 B+168 C) \sec (c+d x)+6 (187 B+168 C) \sec ^2(c+d x)+(935 B+840 C) \sec ^3(c+d x)+35 (11 B+21 C) \sec ^4(c+d x)+315 C \sec ^5(c+d x)\right ) \tan (c+d x)}{3465 d \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*a^2*(2992*B + 2688*C + 8*(187*B + 168*C)*Sec[c + d*x] + 6*(187*B + 168*C)*Sec[c + d*x]^2 + (935*B + 840*C)*
Sec[c + d*x]^3 + 35*(11*B + 21*C)*Sec[c + d*x]^4 + 315*C*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 + Sec
[c + d*x])])

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 a \left (2992 B \cos \left (d x +c \right )^{5}+2688 C \cos \left (d x +c \right )^{5}+1496 B \cos \left (d x +c \right )^{4}+1344 C \cos \left (d x +c \right )^{4}+1122 B \cos \left (d x +c \right )^{3}+1008 C \cos \left (d x +c \right )^{3}+935 B \cos \left (d x +c \right )^{2}+840 C \cos \left (d x +c \right )^{2}+385 B \cos \left (d x +c \right )+735 C \cos \left (d x +c \right )+315 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\)	\(153\)
parts	\(\frac {2 B a \left (272 \cos \left (d x +c \right )^{4}+136 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \left (128 \cos \left (d x +c \right )^{5}+64 \cos \left (d x +c \right )^{4}+48 \cos \left (d x +c \right )^{3}+40 \cos \left (d x +c \right )^{2}+35 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{165 d \left (\cos \left (d x +c \right )+1\right )}\)	\(178\)

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

2/3465*a/d*(2992*B*cos(d*x+c)^5+2688*C*cos(d*x+c)^5+1496*B*cos(d*x+c)^4+1344*C*cos(d*x+c)^4+1122*B*cos(d*x+c)^
3+1008*C*cos(d*x+c)^3+935*B*cos(d*x+c)^2+840*C*cos(d*x+c)^2+385*B*cos(d*x+c)+735*C*cos(d*x+c)+315*C)*(a*(1+sec
(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.62 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (16 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{5} + 8 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{4} + 6 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 21 \, C\right )} a \cos \left (d x + c\right ) + 315 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[In]

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

[Out]

2/3465*(16*(187*B + 168*C)*a*cos(d*x + c)^5 + 8*(187*B + 168*C)*a*cos(d*x + c)^4 + 6*(187*B + 168*C)*a*cos(d*x
+ c)^3 + 5*(187*B + 168*C)*a*cos(d*x + c)^2 + 35*(11*B + 21*C)*a*cos(d*x + c) + 315*C*a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x)**4, x)

Maxima [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 27.47 (sec) , antiderivative size = 720, normalized size of antiderivative = 3.08 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2))/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(11*B - 42*C)*16i)/(115
5*d) + (B*a*16i)/(5*d)) + (a*(3*B + 2*C)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) -
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*a*256i)/(33*d) - (B*a*1
6i)/(9*d) + (a*(B + 2*C)*16i)/(3*d)) - (a*(3*B + 2*C)*16i)/(9*d) + (C*a*64i)/(9*d) + (a*(B + 4*C)*16i)/(9*d)))
/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/
2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(3*B + 2*C)*16i)/(11*d) - (a*(2*B + 3*C)*32i)/(11*d) + (B*a*16i)/(11*d)) - (
a*(2*B + 3*C)*32i)/(11*d) + (a*(3*B + 2*C)*16i)/(11*d) + (B*a*16i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2
i + d*x*2i) + 1)^5) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a*(3*B + 2*C)*16i)/(7*d
) - exp(c*1i + d*x*1i)*((a*(11*B + 39*C)*32i)/(693*d) - (B*a*16i)/(7*d) + (a*(B + 3*C)*32i)/(7*d)) + (a*(B - C
)*32i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1
i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(187*B + 168*C)*32i)/(3465*d*(exp(c*1i + d*x*1i) + 1)) - (a*exp(c
*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(187*B + 168*C)*16i)/(3465*d*(exp(
c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))