Integrand size = 21, antiderivative size = 110 \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\left (a^2+2 b^2\right ) x}{2 a^3}-\frac {2 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {b \sin (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \]
1/2*(a^2+2*b^2)*x/a^3-b*sin(d*x+c)/a^2/d+1/2*cos(d*x+c)*sin(d*x+c)/a/d-2*b ^3*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/d/(a-b)^(1/2)/( a+b)^(1/2)
Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \left (a^2+2 b^2\right ) (c+d x)+\frac {8 b^3 \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 a b \sin (c+d x)+a^2 \sin (2 (c+d x))}{4 a^3 d} \]
(2*(a^2 + 2*b^2)*(c + d*x) + (8*b^3*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sq rt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 4*a*b*Sin[c + d*x] + a^2*Sin[2*(c + d*x) ])/(4*a^3*d)
Time = 0.79 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4340, 25, 3042, 4592, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4340 |
| \(\displaystyle \frac {\int -\frac {\cos (c+d x) \left (-b \sec ^2(c+d x)-a \sec (c+d x)+2 b\right )}{a+b \sec (c+d x)}dx}{2 a}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-b \sec ^2(c+d x)-a \sec (c+d x)+2 b\right )}{a+b \sec (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a \csc \left (c+d x+\frac {\pi }{2}\right )+2 b}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\int \frac {a^2+b \sec (c+d x) a+2 b^2}{a+b \sec (c+d x)}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\int \frac {a^2+b \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 b^2}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^2 \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^2 \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a}}{2 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {4 b^2 \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {2 b \sin (c+d x)}{a d}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {4 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
(Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - (-((((a^2 + 2*b^2)*x)/a - (4*b^3*Arc Tanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/a) + (2*b*Sin[c + d*x])/(a*d))/(2*a)
3.5.94.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n)), x] - Sim p[1/(a*d*n) Int[((d*Csc[e + f*x])^(n + 1)/(a + b*Csc[e + f*x]))*Simp[b*n - a*(n + 1)*Csc[e + f*x] - b*(n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Time = 0.69 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2}-a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2}-a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\) | \(138\) |
| default | \(\frac {-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2}-a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2}-a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (a^{2}+2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\) | \(138\) |
| risch | \(\frac {x}{2 a}+\frac {x \,b^{2}}{a^{3}}+\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {\sin \left (2 d x +2 c \right )}{4 d a}\) | \(218\) |
1/d*(-2*b^3/a^3/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b )*(a+b))^(1/2))+2/a^3*(((-1/2*a^2-a*b)*tan(1/2*d*x+1/2*c)^3+(1/2*a^2-a*b)* tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(a^2+2*b^2)*arctan(tan( 1/2*d*x+1/2*c))))
Time = 0.32 (sec) , antiderivative size = 334, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {\sqrt {a^{2} - b^{2}} b^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} d x - {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}, -\frac {2 \, \sqrt {-a^{2} + b^{2}} b^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} d x + {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \]
[1/2*(sqrt(a^2 - b^2)*b^3*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2 )/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + (a^4 + a^2*b^2 - 2*b^ 4)*d*x - (2*a^3*b - 2*a*b^3 - (a^4 - a^2*b^2)*cos(d*x + c))*sin(d*x + c))/ ((a^5 - a^3*b^2)*d), -1/2*(2*sqrt(-a^2 + b^2)*b^3*arctan(-sqrt(-a^2 + b^2) *(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (a^4 + a^2*b^2 - 2*b^4 )*d*x + (2*a^3*b - 2*a*b^3 - (a^4 - a^2*b^2)*cos(d*x + c))*sin(d*x + c))/( (a^5 - a^3*b^2)*d)]
\[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{3}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \]
-1/2*(4*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan (1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*b^3/(sqrt(- a^2 + b^2)*a^3) - (a^2 + 2*b^2)*(d*x + c)/a^3 + 2*(a*tan(1/2*d*x + 1/2*c)^ 3 + 2*b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) + 2*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d
Time = 14.29 (sec) , antiderivative size = 592, normalized size of antiderivative = 5.38 \[ \int \frac {\cos ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {a\,\left (\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {\sin \left (2\,c+2\,d\,x\right )}{4}\right )}{d\,\left (a^2-b^2\right )}-\frac {b\,\sin \left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}+\frac {b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}}{a\,d\,\left (a^2-b^2\right )}+\frac {b^3\,\sin \left (c+d\,x\right )}{a^2\,d\,\left (a^2-b^2\right )}-\frac {2\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d\,\left (a^2-b^2\right )}-\frac {b^3\,\mathrm {atan}\left (\frac {\left (8\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+8\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-8\,a^2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^4\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+3\,a^5\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,a^6\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-2\,a^7\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^8\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (4\,b^5\,\left (a^2-b^2\right )+2\,a\,b^6-a^7+4\,b^7-2\,a^2\,b^5+a^3\,b^4-2\,a^4\,b^3-2\,a^5\,b^2+2\,a^2\,b^3\,\left (a^2-b^2\right )+2\,a\,b^4\,\left (a^2-b^2\right )\right )}\right )\,2{}\mathrm {i}}{a^3\,d\,\sqrt {a^2-b^2}} \]
(a*(atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + sin(2*c + 2*d*x)/4))/(d* (a^2 - b^2)) - (b*sin(c + d*x))/(d*(a^2 - b^2)) + (b^2*atan(sin(c/2 + (d*x )/2)/cos(c/2 + (d*x)/2)) - (b^2*sin(2*c + 2*d*x))/4)/(a*d*(a^2 - b^2)) - ( b^3*atan(((8*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) - a^9*sin(c/2 + (d*x )/2)*(a^2 - b^2)^(1/2) + 8*b^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 8*a^ 2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 3*a^4*b^5*sin(c/2 + (d*x)/2)* (a^2 - b^2)^(1/2) + 3*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*a^6 *b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 2*a^7*b^2*sin(c/2 + (d*x)/2)*( a^2 - b^2)^(1/2) + a^8*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))*1i)/(cos(c/ 2 + (d*x)/2)*(a*b^2 - a^3)*(4*b^5*(a^2 - b^2) + 2*a*b^6 - a^7 + 4*b^7 - 2* a^2*b^5 + a^3*b^4 - 2*a^4*b^3 - 2*a^5*b^2 + 2*a^2*b^3*(a^2 - b^2) + 2*a*b^ 4*(a^2 - b^2))))*2i)/(a^3*d*(a^2 - b^2)^(1/2)) + (b^3*sin(c + d*x))/(a^2*d *(a^2 - b^2)) - (2*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^3*d *(a^2 - b^2))