Integrand size = 38, antiderivative size = 134 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\left (2 a b B-2 a^2 C-b^2 C\right ) x}{2 b^3}+\frac {2 a^2 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}+\frac {(b B-a C) \sin (c+d x)}{b^2 d}+\frac {C \cos (c+d x) \sin (c+d x)}{2 b d} \]
-1/2*(2*B*a*b-2*C*a^2-C*b^2)*x/b^3+(B*b-C*a)*sin(d*x+c)/b^2/d+1/2*C*cos(d* x+c)*sin(d*x+c)/b/d+2*a^2*(B*b-C*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/ (a+b)^(1/2))/b^3/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 1.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 \left (-2 a b B+2 a^2 C+b^2 C\right ) (c+d x)+\frac {8 a^2 (-b B+a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+4 b (b B-a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]
(2*(-2*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) + (8*a^2*(-(b*B) + a*C)*ArcTanh[ ((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 4*b*(b*B - a*C)*Sin[c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*b^3*d)
Time = 0.78 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3508, 3042, 3469, 3042, 3502, 3042, 3214, 3042, 3138, 218}
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.
\(\displaystyle \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {\cos ^2(c+d x) (B+C \cos (c+d x))}{a+b \cos (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3469 |
\(\displaystyle \frac {\int \frac {2 (b B-a C) \cos ^2(c+d x)+b C \cos (c+d x)+a C}{a+b \cos (c+d x)}dx}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b C \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {\int \frac {a b C-\left (-2 C a^2+2 b B a-b^2 C\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a b C+\left (2 C a^2-2 b B a+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (b B-a C) \int \frac {1}{a+b \cos (c+d x)}dx}{b}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{b}}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (b B-a C) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{b}}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {\frac {4 a^2 (b B-a C) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{b}}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\frac {4 a^2 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{b}}{b}+\frac {2 (b B-a C) \sin (c+d x)}{b d}}{2 b}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d}\) |
(C*Cos[c + d*x]*Sin[c + d*x])/(2*b*d) + ((-(((2*a*b*B - 2*a^2*C - b^2*C)*x )/b) + (4*a^2*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b ]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b + (2*(b*B - a*C)*Sin[c + d*x])/(b*d)) /(2*b)
3.8.95.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin [e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !(IGt Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.94 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {\frac {2 a^{2} \left (B b -a C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-B \,b^{2}+C a b +\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}+C a b -\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 B a b -2 a^{2} C -b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}}{d}\) | \(169\) |
default | \(\frac {\frac {2 a^{2} \left (B b -a C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-B \,b^{2}+C a b +\frac {1}{2} b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}+C a b -\frac {1}{2} b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 B a b -2 a^{2} C -b^{2} C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}}{d}\) | \(169\) |
risch | \(-\frac {a B x}{b^{2}}+\frac {x \,a^{2} C}{b^{3}}+\frac {C x}{2 b}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 d b}+\frac {i a C \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a C}{2 b^{2} d}-\frac {a^{2} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {C \sin \left (2 d x +2 c \right )}{4 b d}\) | \(424\) |
1/d*(2*a^2*(B*b-C*a)/b^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2* c)/((a-b)*(a+b))^(1/2))-2/b^3*(((-B*b^2+C*a*b+1/2*b^2*C)*tan(1/2*d*x+1/2*c )^3+(-B*b^2+C*a*b-1/2*b^2*C)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^ 2+1/2*(2*B*a*b-2*C*a^2-C*b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.30 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.18 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} d x + {\left (C a^{3} - B a^{2} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} d x - 2 \, {\left (C a^{3} - B a^{2} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \]
[1/2*((2*C*a^4 - 2*B*a^3*b - C*a^2*b^2 + 2*B*a*b^3 - C*b^4)*d*x + (C*a^3 - B*a^2*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b ^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (2*C*a^3*b - 2*B*a^ 2*b^2 - 2*C*a*b^3 + 2*B*b^4 - (C*a^2*b^2 - C*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d), 1/2*((2*C*a^4 - 2*B*a^3*b - C*a^2*b^2 + 2*B*a*b^3 - C*b^4)*d*x - 2*(C*a^3 - B*a^2*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c ) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*C*a^3*b - 2*B*a^2*b^2 - 2*C*a* b^3 + 2*B*b^4 - (C*a^2*b^2 - C*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d)]
Timed out. \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.69 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (2 \, C a^{2} - 2 \, B a b + C b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (C a^{3} - B a^{2} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]
1/2*((2*C*a^2 - 2*B*a*b + C*b^2)*(d*x + c)/b^3 + 4*(C*a^3 - B*a^2*b)*(pi*f loor(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)))/(sqrt(a^2 - b^2)*b^3) - 2 *(2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/ 2*d*x + 1/2*c)^3 + 2*C*a*tan(1/2*d*x + 1/2*c) - 2*B*b*tan(1/2*d*x + 1/2*c) - C*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d
Time = 5.74 (sec) , antiderivative size = 3761, normalized size of antiderivative = 28.07 \[ \int \frac {\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)*(2*B*b - 2*C*a + C*b))/b^2 - (tan(c/2 + (d*x)/2)^3*(2 *C*a - 2*B*b + C*b))/b^2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^ 4 + 1)) - (atan(((((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^ 8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b ^6))/b^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) - (8*tan(c/2 + (d*x)/2 )*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B ^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13*C^2*a^3*b ^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C *a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4)*(C*a^2* 2i + C*b^2*1i - B*a*b*2i)*1i)/(2*b^3) - (((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4 *B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b ^9))/b^6 + (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) + (8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2 *a^2*b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C *a^5*b^2))/b^4)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*1i)/(2*b^3))/((16*(4*C^3* a^8 - 6*C^3*a^7*b + 4*B^3*a^4*b^4 - 4*B^3*a^5*b^3 - C^3*a^3*b^5 + 2*C^3...