3.7.87 \(\int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [687]

Optimal. Leaf size=125 \[ \frac {A x}{a^2}-\frac {2 b \left (2 a^2 A-A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4146, 4004, 3916, 2738, 214} \begin {gather*} -\frac {2 b \left (2 a^2 A+a^2 C-A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {A x}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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]

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 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

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

Rule 4146

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.20, size = 270, normalized size = 2.16 \begin {gather*} \frac {2 (b+a \cos (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \left (A x (b+a \cos (c+d x))+\frac {2 b \left (-A b^2+a^2 (2 A+C)\right ) \text {ArcTan}\left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x)) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^{3/2} d \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {\left (A b^2+a^2 C\right ) (-b \sin (c)+a \sin (d x))}{(a-b) (a+b) d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{a^2 (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.17, size = 169, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 \left (A \,b^{2}+a^{2} C \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 \left (2 A \,a^{2}-A \,b^{2}+a^{2} C \right ) b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(169\)
default	\(\frac {\frac {-\frac {2 \left (A \,b^{2}+a^{2} C \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 \left (2 A \,a^{2}-A \,b^{2}+a^{2} C \right ) b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(169\)
risch	\(\frac {A x}{a^{2}}+\frac {2 i \left (A \,b^{2}+a^{2} C \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(579\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (116) = 232\).
time = 4.22, size = 543, normalized size = 4.34 \begin {gather*} \left [\frac {2 \, {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x - {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} + {\left ({\left (2 \, A + C\right )} a^{3} b - A 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 (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 ) + 2 \, {\left (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x - {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} + {\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \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 (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)/(a + b*sec(c + d*x))**2, x)

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Giac [A]
time = 0.48, size = 205, normalized size = 1.64 \begin {gather*} -\frac {\frac {2 \, {\left (2 \, A a^{2} b + C a^{2} b - A b^{3}\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^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (d x + c\right )} A}{a^{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^{3} - a b^{2}\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 11.22, size = 2500, normalized size = 20.00 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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