Integrand size = 31, antiderivative size = 84 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\frac {C x}{c}+\frac {2 (A c-a C) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{c \sqrt {a^2-c^2} e}+\frac {B \log (a+c \sin (d+e x))}{c e} \]
C*x/c+B*ln(a+c*sin(e*x+d))/c/e+2*(A*c-C*a)*arctan((c+a*tan(1/2*e*x+1/2*d)) /(a^2-c^2)^(1/2))/c/e/(a^2-c^2)^(1/2)
Time = 3.67 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\frac {C (d+e x)+\frac {2 (A c-a C) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{\sqrt {a^2-c^2}}+B \log (a+c \sin (d+e x))}{c e} \]
(C*(d + e*x) + (2*(A*c - a*C)*ArcTan[(c + a*Tan[(d + e*x)/2])/Sqrt[a^2 - c ^2]])/Sqrt[a^2 - c^2] + B*Log[a + c*Sin[d + e*x]])/(c*e)
Time = 0.46 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4876, 3042, 3147, 16, 3214, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)}dx\) |
| \(\Big \downarrow \) 4876 |
| \(\displaystyle \int \frac {A+C \sin (d+e x)}{a+c \sin (d+e x)}dx+B \int \frac {\cos (d+e x)}{a+c \sin (d+e x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin (d+e x)}{a+c \sin (d+e x)}dx+B \int \frac {\cos (d+e x)}{a+c \sin (d+e x)}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \int \frac {A+C \sin (d+e x)}{a+c \sin (d+e x)}dx+\frac {B \int \frac {1}{a+c \sin (d+e x)}d(c \sin (d+e x))}{c e}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \int \frac {A+C \sin (d+e x)}{a+c \sin (d+e x)}dx+\frac {B \log (a+c \sin (d+e x))}{c e}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {(A c-a C) \int \frac {1}{a+c \sin (d+e x)}dx}{c}+\frac {B \log (a+c \sin (d+e x))}{c e}+\frac {C x}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A c-a C) \int \frac {1}{a+c \sin (d+e x)}dx}{c}+\frac {B \log (a+c \sin (d+e x))}{c e}+\frac {C x}{c}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {2 (A c-a C) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (d+e x)\right )+2 c \tan \left (\frac {1}{2} (d+e x)\right )+a}d\tan \left (\frac {1}{2} (d+e x)\right )}{c e}+\frac {B \log (a+c \sin (d+e x))}{c e}+\frac {C x}{c}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {4 (A c-a C) \int \frac {1}{-\left (2 c+2 a \tan \left (\frac {1}{2} (d+e x)\right )\right )^2-4 \left (a^2-c^2\right )}d\left (2 c+2 a \tan \left (\frac {1}{2} (d+e x)\right )\right )}{c e}+\frac {B \log (a+c \sin (d+e x))}{c e}+\frac {C x}{c}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 (A c-a C) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (d+e x)\right )+2 c}{2 \sqrt {a^2-c^2}}\right )}{c e \sqrt {a^2-c^2}}+\frac {B \log (a+c \sin (d+e x))}{c e}+\frac {C x}{c}\) |
(C*x)/c + (2*(A*c - a*C)*ArcTan[(2*c + 2*a*Tan[(d + e*x)/2])/(2*Sqrt[a^2 - c^2])])/(c*Sqrt[a^2 - c^2]*e) + (B*Log[a + c*Sin[d + e*x]])/(c*e)
3.6.62.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && 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[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] : > With[{e = FreeFactors[Sin[c*(a + b*x)], x]}, Int[ActivateTrig[u*v], x] + Simp[d Int[ActivateTrig[u]*Cos[c*(a + b*x)]^n, x], x] /; FunctionOfQ[Sin[ c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] && !FreeQ[v, x] && Intege rQ[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Cos] || EqQ[F, cos])
Time = 1.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14
| method | result | size |
| parts | \(\frac {\frac {2 \left (c A -C a \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{c \sqrt {a^{2}-c^{2}}}+\frac {2 C \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c}}{e}+\frac {B \ln \left (a +c \sin \left (e x +d \right )\right )}{c e}\) | \(96\) |
| derivativedivides | \(\frac {\frac {-B \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )+2 C \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c}+\frac {B \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a \right )+\frac {2 \left (c A -C a \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\sqrt {a^{2}-c^{2}}}}{c}}{e}\) | \(128\) |
| default | \(\frac {\frac {-B \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )+2 C \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{c}+\frac {B \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a \right )+\frac {2 \left (c A -C a \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\sqrt {a^{2}-c^{2}}}}{c}}{e}\) | \(128\) |
| risch | \(\frac {i x B}{c}+\frac {C x}{c}-\frac {2 i B \,a^{2} c \,e^{2} x}{a^{2} c^{2} e^{2}-c^{4} e^{2}}+\frac {2 i B \,c^{3} e^{2} x}{a^{2} c^{2} e^{2}-c^{4} e^{2}}-\frac {2 i B \,a^{2} c d e}{a^{2} c^{2} e^{2}-c^{4} e^{2}}+\frac {2 i B \,c^{3} d e}{a^{2} c^{2} e^{2}-c^{4} e^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}+\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) B \,a^{2}}{\left (a^{2}-c^{2}\right ) e c}-\frac {c \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}+\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) B}{\left (a^{2}-c^{2}\right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}+\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) \sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{\left (a^{2}-c^{2}\right ) e c}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}-\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) B \,a^{2}}{\left (a^{2}-c^{2}\right ) e c}-\frac {c \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}-\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) B}{\left (a^{2}-c^{2}\right ) e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a A c -i C \,a^{2}-\sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{c \left (c A -C a \right )}\right ) \sqrt {-A^{2} a^{2} c^{2}+A^{2} c^{4}+2 A C \,a^{3} c -2 A C a \,c^{3}-C^{2} a^{4}+C^{2} a^{2} c^{2}}}{\left (a^{2}-c^{2}\right ) e c}\) | \(939\) |
1/e*(2*(A*c-C*a)/c/(a^2-c^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*e*x+1/2*d)+2*c) /(a^2-c^2)^(1/2))+2*C/c*arctan(tan(1/2*e*x+1/2*d)))+B*ln(a+c*sin(e*x+d))/c /e
Time = 0.28 (sec) , antiderivative size = 346, normalized size of antiderivative = 4.12 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\left [\frac {2 \, {\left (C a^{2} - C c^{2}\right )} e x + {\left (C a - A c\right )} \sqrt {-a^{2} + c^{2}} \log \left (\frac {{\left (2 \, a^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2} + 2 \, {\left (a \cos \left (e x + d\right ) \sin \left (e x + d\right ) + c \cos \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}}}{c^{2} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2}}\right ) + {\left (B a^{2} - B c^{2}\right )} \log \left (-c^{2} \cos \left (e x + d\right )^{2} + 2 \, a c \sin \left (e x + d\right ) + a^{2} + c^{2}\right )}{2 \, {\left (a^{2} c - c^{3}\right )} e}, \frac {2 \, {\left (C a^{2} - C c^{2}\right )} e x + 2 \, {\left (C a - A c\right )} \sqrt {a^{2} - c^{2}} \arctan \left (-\frac {a \sin \left (e x + d\right ) + c}{\sqrt {a^{2} - c^{2}} \cos \left (e x + d\right )}\right ) + {\left (B a^{2} - B c^{2}\right )} \log \left (-c^{2} \cos \left (e x + d\right )^{2} + 2 \, a c \sin \left (e x + d\right ) + a^{2} + c^{2}\right )}{2 \, {\left (a^{2} c - c^{3}\right )} e}\right ] \]
[1/2*(2*(C*a^2 - C*c^2)*e*x + (C*a - A*c)*sqrt(-a^2 + c^2)*log(((2*a^2 - c ^2)*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2 + 2*(a*cos(e*x + d)*si n(e*x + d) + c*cos(e*x + d))*sqrt(-a^2 + c^2))/(c^2*cos(e*x + d)^2 - 2*a*c *sin(e*x + d) - a^2 - c^2)) + (B*a^2 - B*c^2)*log(-c^2*cos(e*x + d)^2 + 2* a*c*sin(e*x + d) + a^2 + c^2))/((a^2*c - c^3)*e), 1/2*(2*(C*a^2 - C*c^2)*e *x + 2*(C*a - A*c)*sqrt(a^2 - c^2)*arctan(-(a*sin(e*x + d) + c)/(sqrt(a^2 - c^2)*cos(e*x + d))) + (B*a^2 - B*c^2)*log(-c^2*cos(e*x + d)^2 + 2*a*c*si n(e*x + d) + a^2 + c^2))/((a^2*c - c^3)*e)]
Leaf count of result is larger than twice the leaf count of optimal. 1110 vs. \(2 (70) = 140\).
Time = 13.61 (sec) , antiderivative size = 1110, normalized size of antiderivative = 13.21 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*x*(A + B*cos(d) + C*sin(d))/sin(d), Eq(a, 0) & Eq(c, 0) & E q(e, 0)), ((A*log(tan(d/2 + e*x/2))/e - B*log(tan(d/2 + e*x/2)**2 + 1)/e + B*log(tan(d/2 + e*x/2))/e + C*x)/c, Eq(a, 0)), (2*A/(c*e*tan(d/2 + e*x/2) - c*e) + 2*B*log(tan(d/2 + e*x/2) - 1)*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e* x/2) - c*e) - 2*B*log(tan(d/2 + e*x/2) - 1)/(c*e*tan(d/2 + e*x/2) - c*e) - B*log(tan(d/2 + e*x/2)**2 + 1)*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) - c *e) + B*log(tan(d/2 + e*x/2)**2 + 1)/(c*e*tan(d/2 + e*x/2) - c*e) + C*e*x* tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) - c*e) - C*e*x/(c*e*tan(d/2 + e*x/2 ) - c*e) + 2*C/(c*e*tan(d/2 + e*x/2) - c*e), Eq(a, -c)), (-2*A/(c*e*tan(d/ 2 + e*x/2) + c*e) + 2*B*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(c*e*ta n(d/2 + e*x/2) + c*e) + 2*B*log(tan(d/2 + e*x/2) + 1)/(c*e*tan(d/2 + e*x/2 ) + c*e) - B*log(tan(d/2 + e*x/2)**2 + 1)*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) - B*log(tan(d/2 + e*x/2)**2 + 1)/(c*e*tan(d/2 + e*x/2) + c*e ) + C*e*x*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) + C*e*x/(c*e*tan(d /2 + e*x/2) + c*e) + 2*C/(c*e*tan(d/2 + e*x/2) + c*e), Eq(a, c)), ((A*x + B*sin(d + e*x)/e - C*cos(d + e*x)/e)/a, Eq(c, 0)), (x*(A + B*cos(d) + C*si n(d))/(a + c*sin(d)), Eq(e, 0)), (-A*c*sqrt(-a**2 + c**2)*log(tan(d/2 + e* x/2) + c/a - sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) + A*c*sqrt(-a**2 + c**2)*log(tan(d/2 + e*x/2) + c/a + sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3* e) - B*a**2*log(tan(d/2 + e*x/2)**2 + 1)/(a**2*c*e - c**3*e) + B*a**2*l...
Exception generated. \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\text {Exception raised: ValueError} \]
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*c^2-4*a^2>0)', see `assume?` f or more de
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\frac {\frac {{\left (e x + d\right )} C}{c} + \frac {B \log \left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a\right )}{c} - \frac {B \log \left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 1\right )}{c} - \frac {2 \, {\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - c^{2}}}\right )\right )} {\left (C a - A c\right )}}{\sqrt {a^{2} - c^{2}} c}}{e} \]
((e*x + d)*C/c + B*log(a*tan(1/2*e*x + 1/2*d)^2 + 2*c*tan(1/2*e*x + 1/2*d) + a)/c - B*log(tan(1/2*e*x + 1/2*d)^2 + 1)/c - 2*(pi*floor(1/2*(e*x + d)/ pi + 1/2)*sgn(a) + arctan((a*tan(1/2*e*x + 1/2*d) + c)/sqrt(a^2 - c^2)))*( C*a - A*c)/(sqrt(a^2 - c^2)*c))/e
Time = 34.85 (sec) , antiderivative size = 1143, normalized size of antiderivative = 13.61 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{a+c \sin (d+e x)} \, dx=\frac {\ln \left (32\,B^3\,a^2-32\,A\,B^2\,a^2+32\,A\,C^2\,a^2+32\,B\,C^2\,a^2+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (c\,A^2\,B-2\,c\,A\,B^2-2\,a\,A\,B\,C-2\,c\,A\,C^2+c\,B^3+2\,a\,B^2\,C+2\,c\,B\,C^2+2\,a\,C^3\right )-32\,A^2\,C\,a\,c+32\,B^2\,C\,a\,c-\frac {\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,C^2\,a^2\,c-32\,B^2\,a^2\,c-128\,B\,C\,a^3+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-A^2\,c^2+4\,A\,B\,c^2+2\,A\,C\,a\,c+2\,B^2\,a^2-3\,B^2\,c^2-4\,B\,C\,a\,c-2\,C^2\,a^2+2\,C^2\,c^2\right )+64\,A\,B\,a^2\,c+64\,B\,C\,a\,c^2+\frac {\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,A\,a^2\,c^2+32\,B\,a^2\,c^2-32\,C\,a\,c^3+32\,a\,c^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c-2\,C\,a+B\,c\right )+\frac {32\,a\,c\,\left (-2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,a^2+a\,c+3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,c^2\right )\,\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )}{a^2-c^2}\right )}{c\,\left (a^2-c^2\right )}\right )}{c\,\left (a^2-c^2\right )}\right )\,\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )}{c\,e\,\left (a^2-c^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{c\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )}{c\,e}+\frac {\ln \left (32\,B^3\,a^2-32\,A\,B^2\,a^2+32\,A\,C^2\,a^2+32\,B\,C^2\,a^2+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (c\,A^2\,B-2\,c\,A\,B^2-2\,a\,A\,B\,C-2\,c\,A\,C^2+c\,B^3+2\,a\,B^2\,C+2\,c\,B\,C^2+2\,a\,C^3\right )-32\,A^2\,C\,a\,c+32\,B^2\,C\,a\,c-\frac {\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,C^2\,a^2\,c-32\,B^2\,a^2\,c-128\,B\,C\,a^3+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-A^2\,c^2+4\,A\,B\,c^2+2\,A\,C\,a\,c+2\,B^2\,a^2-3\,B^2\,c^2-4\,B\,C\,a\,c-2\,C^2\,a^2+2\,C^2\,c^2\right )+64\,A\,B\,a^2\,c+64\,B\,C\,a\,c^2+\frac {\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,A\,a^2\,c^2+32\,B\,a^2\,c^2-32\,C\,a\,c^3+32\,a\,c^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c-2\,C\,a+B\,c\right )+\frac {32\,a\,c\,\left (-2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,a^2+a\,c+3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,c^2\right )\,\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )}{a^2-c^2}\right )}{c\,\left (a^2-c^2\right )}\right )}{c\,\left (a^2-c^2\right )}\right )\,\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )}{c\,e\,\left (a^2-c^2\right )} \]
(log(32*B^3*a^2 - 32*A*B^2*a^2 + 32*A*C^2*a^2 + 32*B*C^2*a^2 + 32*a*tan(d/ 2 + (e*x)/2)*(2*C^3*a + B^3*c - 2*A*B^2*c + A^2*B*c + 2*B^2*C*a - 2*A*C^2* c + 2*B*C^2*c - 2*A*B*C*a) - 32*A^2*C*a*c + 32*B^2*C*a*c - ((B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2))*(32*C^2*a^2*c - 32*B^2*a ^2*c - 128*B*C*a^3 + 32*a*tan(d/2 + (e*x)/2)*(2*B^2*a^2 - A^2*c^2 - 2*C^2* a^2 - 3*B^2*c^2 + 2*C^2*c^2 + 4*A*B*c^2 + 2*A*C*a*c - 4*B*C*a*c) + 64*A*B* a^2*c + 64*B*C*a*c^2 + ((B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2))*(32*A*a^2*c^2 + 32*B*a^2*c^2 - 32*C*a*c^3 + 32*a*c^2*tan(d/2 + (e*x)/2)*(2*A*c - 2*C*a + B*c) + (32*a*c*(a*c - 2*a^2*tan(d/2 + (e*x)/2 ) + 3*c^2*tan(d/2 + (e*x)/2))*(B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a *(c^2 - a^2)^(1/2)))/(a^2 - c^2)))/(c*(a^2 - c^2))))/(c*(a^2 - c^2)))*(B*a ^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2)))/(c*e*(a^2 - c ^2)) - (log(tan(d/2 + (e*x)/2) + 1i)*(B - C*1i))/(c*e) - (log(tan(d/2 + (e *x)/2) - 1i)*(B + C*1i))/(c*e) + (log(32*B^3*a^2 - 32*A*B^2*a^2 + 32*A*C^2 *a^2 + 32*B*C^2*a^2 + 32*a*tan(d/2 + (e*x)/2)*(2*C^3*a + B^3*c - 2*A*B^2*c + A^2*B*c + 2*B^2*C*a - 2*A*C^2*c + 2*B*C^2*c - 2*A*B*C*a) - 32*A^2*C*a*c + 32*B^2*C*a*c - ((B*a^2 - B*c^2 - A*c*(c^2 - a^2)^(1/2) + C*a*(c^2 - a^2 )^(1/2))*(32*C^2*a^2*c - 32*B^2*a^2*c - 128*B*C*a^3 + 32*a*tan(d/2 + (e*x) /2)*(2*B^2*a^2 - A^2*c^2 - 2*C^2*a^2 - 3*B^2*c^2 + 2*C^2*c^2 + 4*A*B*c^2 + 2*A*C*a*c - 4*B*C*a*c) + 64*A*B*a^2*c + 64*B*C*a*c^2 + ((B*a^2 - B*c^2...