Integrand size = 42, antiderivative size = 144 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {4 (7 B+6 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d} \]
2/35*(7*B+6*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*a*(7*B+6*C)*tan( d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*a*C*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec (d*x+c))^(1/2)-4/105*(7*B+6*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a \left (8 (7 B+6 C)+4 (7 B+6 C) \sec (c+d x)+3 (7 B+6 C) \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a*(8*(7*B + 6*C) + 4*(7*B + 6*C)*Sec[c + d*x] + 3*(7*B + 6*C)*Sec[c + d *x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x] )])
Time = 0.90 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.262, Rules used = {3042, 4560, 3042, 4504, 3042, 4287, 27, 3042, 4489, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {1}{7} (7 B+6 C) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\) |
(2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + ((7*B + 6*C)*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Ta n[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]] *Tan[c + d*x])/(3*d))/(5*a)))/7
3.4.58.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
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] + Simp[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)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(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)]
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*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(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]
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] :> Simp[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]
Time = 0.71 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {2 \left (56 B \cos \left (d x +c \right )^{3}+48 C \cos \left (d x +c \right )^{3}+28 B \cos \left (d x +c \right )^{2}+24 C \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )+18 C \cos \left (d x +c \right )+15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(108\) |
| parts | \(\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}\) | \(133\) |
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
2/105/d*(56*B*cos(d*x+c)^3+48*C*cos(d*x+c)^3+28*B*cos(d*x+c)^2+24*C*cos(d* x+c)^2+21*B*cos(d*x+c)+18*C*cos(d*x+c)+15*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos (d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^2
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (8 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2 ),x, algorithm="fricas")
2/105*(8*(7*B + 6*C)*cos(d*x + c)^3 + 4*(7*B + 6*C)*cos(d*x + c)^2 + 3*(7* B + 6*C)*cos(d*x + c) + 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin( d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2 ),x, algorithm="maxima")
-8/105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(1/4)*(7*(5*B*sin(4*d*x + 4*c) + (7*B + 6*C)*sin(2*d*x + 2*c))*cos(7/2*ar ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (35*B*cos(4*d*x + 4*c) + 7*(7*B + 6*C)*cos(2*d*x + 2*c) + 14*B + 12*C)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - 105*(((B + 2*C)*d*cos(2*d*x + 2*c) ^4 + (B + 2*C)*d*sin(2*d*x + 2*c)^4 + 4*(B + 2*C)*d*cos(2*d*x + 2*c)^3 + 6 *(B + 2*C)*d*cos(2*d*x + 2*c)^2 + 4*(B + 2*C)*d*cos(2*d*x + 2*c) + 2*((B + 2*C)*d*cos(2*d*x + 2*c)^2 + 2*(B + 2*C)*d*cos(2*d*x + 2*c) + (B + 2*C)*d) *sin(2*d*x + 2*c)^2 + (B + 2*C)*d)*integrate((((cos(10*d*x + 10*c)*cos(2*d *x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2 *d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + s in(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) ) + (cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d* x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2* d*x + 2*c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*s...
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2 ),x, algorithm="giac")
Time = 20.21 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.83 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-\frac {B\,8{}\mathrm {i}}{5\,d}+\frac {C\,16{}\mathrm {i}}{5\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {C\,16{}\mathrm {i}}{35\,d}+\frac {\left (56\,B+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {B\,8{}\mathrm {i}}{7\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,8{}\mathrm {i}}{7\,d}-\frac {\left (8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )-\frac {\left (8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {B\,8{}\mathrm {i}}{3\,d}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (56\,B+48\,C\right )\,1{}\mathrm {i}}{105\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (112\,B+96\,C\right )\,1{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((C*16i)/(5 *d) - (B*8i)/(5*d) + exp(c*1i + d*x*1i)*((C*16i)/(35*d) + ((56*B + 112*C)* 1i)/(35*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)*((B*8i)/(7*d) + exp(c*1i + d*x*1i)*((B*8i)/(7*d) - ((8*B + 16*C)*1i)/(7*d)) - ((8*B + 16* C)*1i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((( B*8i)/(3*d) - (exp(c*1i + d*x*1i)*(56*B + 48*C)*1i)/(105*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1 )*(exp(c*2i + d*x*2i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x *1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(112*B + 96*C)*1i)/(105*d*(exp(c*1i + d*x*1i) + 1))