3.8.99 \(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [799]

Optimal. Leaf size=134 \[ \frac {\left (a^2 B+2 b^2 B-2 a b C\right ) x}{2 a^3}-\frac {2 b^2 (b B-a C) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {(b B-a C) \sin (c+d x)}{a^2 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 a d} \]

[Out]

1/2*(B*a^2+2*B*b^2-2*C*a*b)*x/a^3-(B*b-C*a)*sin(d*x+c)/a^2/d+1/2*B*cos(d*x+c)*sin(d*x+c)/a/d-2*b^2*(B*b-C*a)*a
rctanh((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)

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Rubi [A]
time = 0.35, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4119, 4189, 4004, 3916, 2738, 214} \begin {gather*} -\frac {2 b^2 (b B-a C) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {(b B-a C) \sin (c+d x)}{a^2 d}+\frac {x \left (a^2 B-2 a b C+2 b^2 B\right )}{2 a^3}+\frac {B \sin (c+d x) \cos (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

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

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 4119

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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]
+ Dist[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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4189

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] + Dist[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]

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 121, normalized size = 0.90 \begin {gather*} \frac {2 \left (a^2 B+2 b^2 B-2 a b C\right ) (c+d x)+\frac {8 b^2 (b B-a C) \tanh ^{-1}\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 B+a C) \sin (c+d x)+a^2 B \sin (2 (c+d x))}{4 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

(2*(a^2*B + 2*b^2*B - 2*a*b*C)*(c + d*x) + (8*b^2*(b*B - a*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/Sqrt[a^2 - b^2] + 4*a*(-(b*B) + a*C)*Sin[c + d*x] + a^2*B*Sin[2*(c + d*x)])/(4*a^3*d)

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Maple [A]
time = 0.22, size = 168, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2} B -a b B +a^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2} B -a b B +a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2} B +2 b^{2} B -2 a b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{2} \left (b B -a C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(168\)
default	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2} B -a b B +a^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2} B -a b B +a^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2} B +2 b^{2} B -2 a b C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{2} \left (b B -a C \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(168\)
risch	\(\frac {B x}{2 a}+\frac {x \,b^{2} B}{a^{3}}-\frac {b x C}{a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )} b B}{2 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b B}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {B \sin \left (2 d x +2 c \right )}{4 a d}\)	\(424\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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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(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),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 [A]
time = 2.72, size = 427, normalized size = 3.19 \begin {gather*} \left [\frac {{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x - {\left (C a b^{2} - B b^{3}\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 ) + {\left (2 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} + {\left (B a^{4} - B 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 {{\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} d x + 2 \, {\left (C a b^{2} - B b^{3}\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 (2 \, C a^{4} - 2 \, B a^{3} b - 2 \, C a^{2} b^{2} + 2 \, B a b^{3} + {\left (B a^{4} - B 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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*((B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2*B*b^4)*d*x - (C*a*b^2 - B*b^3)*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^4 - 2*B*a^3*b - 2*C*a^2*b^2 + 2*B*a*b^3 + (B*a^4
- B*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d), 1/2*((B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3
- 2*B*b^4)*d*x + 2*(C*a*b^2 - B*b^3)*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))) + (2*C*a^4 - 2*B*a^3*b - 2*C*a^2*b^2 + 2*B*a*b^3 + (B*a^4 - B*a^2*b^2)*cos(d*x + c))*sin(d*
x + c))/((a^5 - a^3*b^2)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)

[Out]

Integral((B + C*sec(c + d*x))*cos(c + d*x)**3*sec(c + d*x)/(a + b*sec(c + d*x)), x)

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Giac [A]
time = 0.47, size = 227, normalized size = 1.69 \begin {gather*} \frac {\frac {{\left (B a^{2} - 2 \, C a b + 2 \, B b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {4 \, {\left (C a b^{2} - B 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 )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {2 \, {\left (B a \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 )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 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 )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 8.02, size = 2500, normalized size = 18.66 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)

[Out]

((tan(c/2 + (d*x)/2)*(B*a - 2*B*b + 2*C*a))/a^2 - (tan(c/2 + (d*x)/2)^3*(B*a + 2*B*b - 2*C*a))/a^2)/(d*(2*tan(
c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^8*
b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9*b))/a^6 - (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i - C
*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) + (8*tan(c/2 + (d*
x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7
*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b -
32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*1i)/
(2*a^3) - (((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b -
4*C*a^9*b))/a^6 + (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/
a^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) - (8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3
*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16*C^2*a^3*b^4
- 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*
b^3 + 12*B*C*a^5*b^2))/a^4)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*1i)/(2*a^3))/((16*(4*B^3*b^8 - 6*B^3*a*b^7 + 6*B^
3*a^2*b^6 - 5*B^3*a^3*b^5 + 2*B^3*a^4*b^4 - B^3*a^5*b^3 - 4*C^3*a^3*b^5 + 4*C^3*a^4*b^4 - 12*B^2*C*a*b^7 + 12*
B*C^2*a^2*b^6 - 14*B*C^2*a^3*b^5 + 6*B*C^2*a^4*b^4 - 4*B*C^2*a^5*b^3 + 16*B^2*C*a^2*b^6 - 12*B^2*C*a^3*b^5 + 9
*B^2*C*a^4*b^4 - 2*B^2*C*a^5*b^3 + B^2*C*a^6*b^2))/a^6 + (((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^
8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9*b))/a^6 - (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i -
C*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) + (8*tan(c/2 + (
d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 +
7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b
- 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(
2*a^3) + (((((8*(2*B*a^10 + 4*B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b -
4*C*a^9*b))/a^6 + (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a
^7)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3) - (8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B^2*a*b^6 - 3*
B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16*C^2*a^3*b^4
- 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 - 20*B*C*a^4*b
^3 + 12*B*C*a^5*b^2))/a^4)*(B*a^2*1i + B*b^2*2i - C*a*b*2i))/(2*a^3)))*(B*a^2*1i + B*b^2*2i - C*a*b*2i)*1i)/(a
^3*d) - (b^2*atan(((b^2*((a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16*B
^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 + 16
*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4 -
20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4 + (b^2*((a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*(2*B*a^10 + 4*B*a^6*b^4
- 6*B*a^7*b^3 + 2*B*a^8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9*b))/a^6 - (8*b^2*tan(c/2 + (d*x)
/2)*((a + b)*(a - b))^(1/2)*(B*b - C*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3
*b^2))*1i)/(a^5 - a^3*b^2) + (b^2*((a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*
b^7 + 16*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^
2*b^5 + 16*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C
*a^3*b^4 - 20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4 - (b^2*((a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*(2*B*a^10 + 4*
B*a^6*b^4 - 6*B*a^7*b^3 + 2*B*a^8*b^2 - 4*C*a^7*b^3 + 8*C*a^8*b^2 - 2*B*a^9*b - 4*C*a^9*b))/a^6 + (8*b^2*tan(c
/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(B*b - C*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/
(a^5 - a^3*b^2))*1i)/(a^5 - a^3*b^2))/((16*(4*B^3*b^8 - 6*B^3*a*b^7 + 6*B^3*a^2*b^6 - 5*B^3*a^3*b^5 + 2*B^3*a^
4*b^4 - B^3*a^5*b^3 - 4*C^3*a^3*b^5 + 4*C^3*a^4*b^4 - 12*B^2*C*a*b^7 + 12*B*C^2*a^2*b^6 - 14*B*C^2*a^3*b^5 + 6
*B*C^2*a^4*b^4 - 4*B*C^2*a^5*b^3 + 16*B^2*C*a^2*b^6 - 12*B^2*C*a^3*b^5 + 9*B^2*C*a^4*b^4 - 2*B^2*C*a^5*b^3 + B
^2*C*a^6*b^2))/a^6 + (b^2*((a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*a^7 - 8*B^2*b^7 + 16
*B^2*a*b^6 - 3*B^2*a^6*b - 16*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 13*B^2*a^4*b^3 + 7*B^2*a^5*b^2 - 8*C^2*a^2*b^5 +
16*C^2*a^3*b^4 - 12*C^2*a^4*b^3 + 4*C^2*a^5*b^2 + 16*B*C*a*b^6 - 4*B*C*a^6*b - 32*B*C*a^2*b^5 + 28*B*C*a^3*b^4
- 20*B*C*a^4*b^3 + 12*B*C*a^5*b^2))/a^4 + (b^2...

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