Integrand size = 22, antiderivative size = 107 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {8 \cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {32 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {64 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \]
-8/15*cos(b*x+a)/b/sin(2*b*x+2*a)^(5/2)-1/9*csc(b*x+a)^3/b/sin(2*b*x+2*a)^ (3/2)+32/45*sin(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-64/45*cos(b*x+a)/b/sin(2*b*x +2*a)^(1/2)
Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (113 \csc (a+b x)+17 \csc ^3(a+b x)+5 \csc ^5(a+b x)-15 \sec (a+b x) \tan (a+b x)\right )}{180 b} \]
-1/180*(Sqrt[Sin[2*(a + b*x)]]*(113*Csc[a + b*x] + 17*Csc[a + b*x]^3 + 5*C sc[a + b*x]^5 - 15*Sec[a + b*x]*Tan[a + b*x]))/b
Time = 0.59 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4788, 3042, 4796, 3042, 4791, 3042, 4792, 3042, 4779}
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.
| \(\displaystyle \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sin (2 a+2 b x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4788 |
| \(\displaystyle \frac {4}{3} \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4}{3} \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{5/2}}dx-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 4796 |
| \(\displaystyle \frac {8}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)}dx-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{7/2}}dx-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 4791 |
| \(\displaystyle \frac {8}{3} \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{3} \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 4792 |
| \(\displaystyle \frac {8}{3} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{3} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
| \(\Big \downarrow \) 4779 |
| \(\displaystyle \frac {8}{3} \left (\frac {4}{5} \left (\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{9 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
(8*((4*(Sin[a + b*x]/(3*b*Sin[2*a + 2*b*x]^(3/2)) - (2*Cos[a + b*x])/(3*b* Sqrt[Sin[2*a + 2*b*x]])))/5 - Cos[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2))))/ 3 - Csc[a + b*x]^3/(9*b*Sin[2*a + 2*b*x]^(3/2))
3.2.22.3.1 Defintions of rubi rules used
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(-(e*Cos[a + b*x])^m)*((g*Sin[c + d*x])^(p + 1)/(b*g *m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[ d/b, 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p _), x_Symbol] :> Simp[(e*Sin[a + b*x])^m*((g*Sin[c + d*x])^(p + 1)/(2*b*g*( m + p + 1))), x] + Simp[(m + 2*p + 2)/(e^2*(m + p + 1)) Int[(e*Sin[a + b* x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] & & EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[Cos[a + b*x]*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Simp [(2*p + 3)/(2*g*(p + 1)) Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x], x ] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !Int egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-Sin[a + b*x])*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + S imp[(2*p + 3)/(2*g*(p + 1)) Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1), x] , x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && ! IntegerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[2*g Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && IntegerQ[2*p]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 139.42 (sec) , antiderivative size = 560, normalized size of antiderivative = 5.23
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (5 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{10}+192 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-96 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-7 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}+2 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+96 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+2 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-96 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-7 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+5 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\right )}{2880 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{5} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\) | \(560\) |
-1/2880*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/tan(1/2*a+1/2 *x*b)^5*(5*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1 /2*x*b)^10+192*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/ 2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b) )^(1/2)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2* x*b)^4-96*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1 /2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/ 2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^ 4-7*tan(1/2*a+1/2*x*b)^8*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/ 2)+2*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b )^6+96*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^ 6+2*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b) ^4-96*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^4 -7*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^ 2+5*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2))/(tan(1/2*a+1/2*x* b)^3-tan(1/2*a+1/2*x*b))^(1/2)/b
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 288 \, \cos \left (b x + a\right )^{4} + 180 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{180 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]
-1/180*(sqrt(2)*(128*cos(b*x + a)^6 - 288*cos(b*x + a)^4 + 180*cos(b*x + a )^2 - 15)*sqrt(cos(b*x + a)*sin(b*x + a)) + 128*(cos(b*x + a)^6 - 2*cos(b* x + a)^4 + cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)*sin(b*x + a))
Timed out. \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
Time = 26.08 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.58 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {2\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,16{}\mathrm {i}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}+\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^5}+\frac {64\,{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{45\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {98{}\mathrm {i}}{45\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,38{}\mathrm {i}}{45\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2} \]
(8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1 i)/2)^(1/2))/(9*b*(exp(a*2i + b*x*2i)*1i - 1i)^5) - (exp(a*1i + b*x*1i)*(( exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*16i)/(9*b*(e xp(a*2i + b*x*2i)*1i - 1i)^4) - (2*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2 i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(15*b*(exp(a*2i + b*x*2i)*1i - 1i)^3) + (64*exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(45*b*(exp(a*2i + b*x*2i) + 1)*(exp(a*2i + b*x*2i )*1i - 1i)) - (exp(a*1i + b*x*1i)*(98i/(45*b) + (exp(a*2i + b*x*2i)*38i)/( 45*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(( exp(a*2i + b*x*2i) + 1)^2*(exp(a*2i + b*x*2i)*1i - 1i)^2)