\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [573]

Optimal result

Integrand size = 25, antiderivative size = 126 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {C x}{b^2}+\frac {2 a \left (A b^2-a^2 C+2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

C*x/b^2+2*a*(A*b^2-C*a^2+2*C*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^2/(a+b)^(3/
2)/d-(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3101, 2814, 2738, 211} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {C x}{b^2} \]

[In]

Int[(A + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

(C*x)/b^2 + (2*a*(A*b^2 - a^2*C + 2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*
b^2*(a + b)^(3/2)*d) - ((A*b^2 + a^2*C)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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]

Rule 3101

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(
-(A*b^2 + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1
)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*b*(A + C)*(m + 1) - (A*b^2 + a^2*C + b^2*(A + C)*(m +
1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {-a b (A+C)-\left (a^2-b^2\right ) C \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {C x}{b^2}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (-a \left (a^2-b^2\right ) C+a b^2 (A+C)\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )} \\ & = \frac {C x}{b^2}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (2 \left (-a \left (a^2-b^2\right ) C+a b^2 (A+C)\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right ) d} \\ & = \frac {C x}{b^2}+\frac {2 a \left (A b^2-a^2 C+2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {C (c+d x)-\frac {2 a \left (-A b^2+\left (a^2-2 b^2\right ) C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-\frac {b \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{b^2 d} \]

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

(C*(c + d*x) - (2*a*(-(A*b^2) + (a^2 - 2*b^2)*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 +
b^2)^(3/2) - (b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])))/(b^2*d)

Maple [A] (verified)

Time = 1.69 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {2 \left (A \,b^{2}-a^{2} C +2 b^{2} C \right ) a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(164\)
default	\(\frac {\frac {-\frac {2 \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {2 \left (A \,b^{2}-a^{2} C +2 b^{2} C \right ) a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{b^{2}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(164\)
risch	\(\frac {C x}{b^{2}}+\frac {2 i \left (A \,b^{2}+a^{2} C \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{2} \left (-a^{2}+b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(579\)

[In]

int((A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/b^2*(-(A*b^2+C*a^2)*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)+
(A*b^2-C*a^2+2*C*b^2)*a/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+
2*C/b^2*arctan(tan(1/2*d*x+1/2*c)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 534, normalized size of antiderivative = 4.24 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {2 \, {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} d x - {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2} + {\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}}, \frac {{\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} d x - {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2} + {\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d}\right ] \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(2*(C*a^4*b - 2*C*a^2*b^3 + C*b^5)*d*x*cos(d*x + c) + 2*(C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*d*x - (C*a^4 - (A
+ 2*C)*a^2*b^2 + (C*a^3*b - (A + 2*C)*a*b^3)*cos(d*x + c))*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^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^5 + b
^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*d), ((C*a^4*b - 2*C*a^2*b^3 + C*b^5)*d*x*cos(d*x + c) + (C*
a^5 - 2*C*a^3*b^2 + C*a*b^4)*d*x - (C*a^4 - (A + 2*C)*a^2*b^2 + (C*a^3*b - (A + 2*C)*a*b^3)*cos(d*x + c))*sqrt
(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (C*a^4*b + (A - C)*a^2*b^3 - A*b^5)
*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.60 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (C a^{3} - A a b^{2} - 2 \, C a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (d x + c\right )} C}{b^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \]

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

(2*(C*a^3 - A*a*b^2 - 2*C*a*b^2)*(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)))/((a^2*b^2 - b^4)*sqrt(a^2 - b^2)) + (d*x + c)*C/b^2 - 2*(C*
a^2*tan(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*
x + 1/2*c)^2 + a + b)))/d

Mupad [B] (verification not implemented)

Time = 9.17 (sec) , antiderivative size = 3862, normalized size of antiderivative = 30.65 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int((A + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^2,x)

[Out]

(2*C*atan(((C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*
b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 +
4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 + (32*tan(c/2 + (d*x
)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^
4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (C*((C*((32*(A*a^4*b^5 - A*a
^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^
4 - a^3*b^3) + (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i
)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 - (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5
- 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))
/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2)/((64*(C^3*a^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*a^3*
b^2 + A*C^2*a*b^4 - A*C^2*a^4*b + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2 + A^2*C*a^2*b^3))/(a*b^5 + b^6 - a^2*b^4 - a
^3*b^3) + (C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b
^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 +
4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 + (32*tan(c/2 + (d*x)
/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4
*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 + (C*((C*((32*(A*a^4*b^5 - A
*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*
b^4 - a^3*b^3) + (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*3
2i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 - (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b
^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2
))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2)))/(b^2*d) - (2*tan(c/2 + (d*x)/2)*(A*b^2 + C*a^2))/(d*(a + b)*(
a*b - b^2)*(a + b + tan(c/2 + (d*x)/2)^2*(a - b))) + (a*atan(((a*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (
d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2
*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (a*((32*(A*a^4*b^5 - A*a^2*b^7
- A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3
*b^3) - (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 -
4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b
^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2
))*(A*b^2 - C*a^2 + 2*C*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) + (a*(-(a + b)^3*(a - b)^3)^(1/2)*((3
2*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^
3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (a*((32*(A*a^4*b^5
- A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 -
a^2*b^4 - a^3*b^3) + (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9
- 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*
b^6 + 3*a^4*b^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*
b^4 - a^6*b^2))*(A*b^2 - C*a^2 + 2*C*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))/((64*(C^3*a^5 + 2*C^3*a
*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*a^3*b^2 + A*C^2*a*b^4 - A*C^2*a^4*b + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2
+ A^2*C*a^2*b^3))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (a*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)
*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^
2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (a*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3
*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) -
(32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*
b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^
6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(A*b
^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) - (a*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2
+ (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*
C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (a*((32*(A*a^4*b^5 - A*a^2*b
^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 -
a^3*b^3) + (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^
8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^
4*b^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*
b^2))*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(A*b^2
- C*a^2 + 2*C*b^2)*2i)/(d*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))