Integrand size = 31, antiderivative size = 208 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {34 c^2 \tan (2 a+2 b x)}{45 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {68 c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{315 b}+\frac {34 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{105 b} \]
34/105*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b+34/45*c^2*tan(2*b*x+2* a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-17/63*c^2*sec(2*b*x+2*a)^3*tan(2*b*x+2*a) /b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/9*c^2*sec(2*b*x+2*a)^4*tan(2*b*x+2*a)/b/( -c+c*sec(2*b*x+2*a))^(1/2)+68/315*c*(-c+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+ 2*a)/b
Time = 1.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.41 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {\cot (a+b x) \left (-84+188 \cot (a+b x) \cot (2 (a+b x))+52 \sec (2 (a+b x))-50 \sec ^2(2 (a+b x))+35 \sec ^3(2 (a+b x))\right ) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{315 b} \]
(Cot[a + b*x]*(-84 + 188*Cot[a + b*x]*Cot[2*(a + b*x)] + 52*Sec[2*(a + b*x )] - 50*Sec[2*(a + b*x)]^2 + 35*Sec[2*(a + b*x)]^3)*(c*Tan[a + b*x]*Tan[2* (a + b*x)])^(3/2))/(315*b)
Time = 1.19 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4897, 3042, 4301, 27, 2011, 3042, 4290, 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.
| \(\displaystyle \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (2 a+2 b x)^4 (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \sec ^4(2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^4 \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4301 |
| \(\displaystyle \frac {2}{9} c \int \frac {17 \sec ^4(2 a+2 b x) (c-c \sec (2 a+2 b x))}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {17}{9} c \int \frac {\sec ^4(2 a+2 b x) (c-c \sec (2 a+2 b x))}{\sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \int \sec ^4(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^4 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \sec ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\right )\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+3 c\right )dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\) |
(c^2*Sec[2*a + 2*b*x]^4*Tan[2*a + 2*b*x])/(9*b*Sqrt[-c + c*Sec[2*a + 2*b*x ]]) - (17*c*((c*Sec[2*a + 2*b*x]^3*Tan[2*a + 2*b*x])/(7*b*Sqrt[-c + c*Sec[ 2*a + 2*b*x]]) - (6*(((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5 *b*c) + ((7*c^2*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (2 *c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b))/(5*c)))/7))/9
3.7.11.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + 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_)]*(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] + Simp[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] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[b/(m + n - 1) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
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)]
Time = 6.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.49
| method | result | size |
| default | \(\frac {\sqrt {2}\, \cot \left (x b +a \right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (2176 \cos \left (x b +a \right )^{8}-4896 \cos \left (x b +a \right )^{6}+4284 \cos \left (x b +a \right )^{4}-1785 \cos \left (x b +a \right )^{2}+315\right ) c \sqrt {4}}{315 b \left (2 \cos \left (x b +a \right )^{2}-1\right )^{4}}\) | \(101\) |
1/315*2^(1/2)/b*cot(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(2176 *cos(b*x+a)^8-4896*cos(b*x+a)^6+4284*cos(b*x+a)^4-1785*cos(b*x+a)^2+315)*c /(2*cos(b*x+a)^2-1)^4*4^(1/2)
Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {2 \, \sqrt {2} {\left (315 \, c \tan \left (b x + a\right )^{8} - 525 \, c \tan \left (b x + a\right )^{6} + 819 \, c \tan \left (b x + a\right )^{4} - 423 \, c \tan \left (b x + a\right )^{2} + 94 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{315 \, {\left (b \tan \left (b x + a\right )^{9} - 4 \, b \tan \left (b x + a\right )^{7} + 6 \, b \tan \left (b x + a\right )^{5} - 4 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \]
2/315*sqrt(2)*(315*c*tan(b*x + a)^8 - 525*c*tan(b*x + a)^6 + 819*c*tan(b*x + a)^4 - 423*c*tan(b*x + a)^2 + 94*c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a )^2 - 1))/(b*tan(b*x + a)^9 - 4*b*tan(b*x + a)^7 + 6*b*tan(b*x + a)^5 - 4* b*tan(b*x + a)^3 + b*tan(b*x + a))
Timed out. \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
\[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int { \left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (2 \, b x + 2 \, a\right )^{4} \,d x } \]
-8/315*(630*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*((b*c*cos(4*b*x + 4*a)^4 + b*c*sin(4*b*x + 4*a)^4 + 4*b*c*cos(4 *b*x + 4*a)^3 + 6*b*c*cos(4*b*x + 4*a)^2 + 4*b*c*cos(4*b*x + 4*a) + 2*(b*c *cos(4*b*x + 4*a)^2 + 2*b*c*cos(4*b*x + 4*a) + b*c)*sin(4*b*x + 4*a)^2 + b *c)*integrate(-(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4* a) + 1)^(1/4)*(((cos(20*b*x + 20*a)*cos(4*b*x + 4*a) + 4*cos(16*b*x + 16*a )*cos(4*b*x + 4*a) + 6*cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 4*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(20*b*x + 20*a)*sin(4*b*x + 4*a) + 4*sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 6*sin(12*b*x + 12*a)*sin (4*b*x + 4*a) + 4*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)* cos(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + (cos(4*b*x + 4 *a)*sin(20*b*x + 20*a) + 4*cos(4*b*x + 4*a)*sin(16*b*x + 16*a) + 6*cos(4*b *x + 4*a)*sin(12*b*x + 12*a) + 4*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(2 0*b*x + 20*a)*sin(4*b*x + 4*a) - 4*cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 6 *cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 4*cos(8*b*x + 8*a)*sin(4*b*x + 4*a) )*sin(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*cos(7/2*arcta n2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(20*b*x + 20*a) + 4*cos(4*b*x + 4*a)*sin(16*b*x + 16*a) + 6*cos(4*b*x + 4*a)*sin(12* b*x + 12*a) + 4*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(20*b*x + 20*a)*sin (4*b*x + 4*a) - 4*cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 6*cos(12*b*x + ...
Timed out. \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Time = 34.92 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.86 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {\left (\frac {c\,16{}\mathrm {i}}{9\,b}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,16{}\mathrm {i}}{9\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^4}-\frac {\left (\frac {c\,40{}\mathrm {i}}{7\,b}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,88{}\mathrm {i}}{63\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {c\,24{}\mathrm {i}}{5\,b}-\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,176{}\mathrm {i}}{105\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^2}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}\,272{}\mathrm {i}}{315\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}\,136{}\mathrm {i}}{315\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )} \]
(((c*16i)/(9*b) + (c*exp(a*2i + b*x*2i)*16i)/(9*b))*((c*(exp(a*2i + b*x*2i )*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a* 4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1)^4) - (((c*40i)/(7*b) + (c*exp(a*2i + b*x*2i)*88i)/(63*b))*((c*(exp(a*2 i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1)^3) + (((c*24i)/(5*b) - (c*exp(a*2i + b*x*2i)*176i)/(105*b))* ((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1) *(exp(a*4i + b*x*4i) + 1)^2) + (c*exp(a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2 i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a *4i + b*x*4i) + 1)))^(1/2)*272i)/(315*b*(exp(a*2i + b*x*2i) - 1)) + (c*exp (a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2)*136i)/(315 *b*(exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1))