\(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\) [566]

Optimal result

Integrand size = 31, antiderivative size = 88 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {C x}{b}-\frac {2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} b \sqrt {a+b} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3137, 2738, 211, 3855} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d \sqrt {a-b} \sqrt {a+b}}+\frac {A \text {arctanh}(\sin (c+d x))}{a d}+\frac {C x}{b} \]

[In]

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

[Out]

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

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 3137

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \left (A+C \cos ^2(c+d x)\right ) \left (\left (a C d x-A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}+2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))\right )}{a b d (2 A+C+C \cos (2 (c+d x))) \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)-i \sin (2 c))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 \left (-A \,b^{2}-a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(116\)
default	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 \left (-A \,b^{2}-a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(116\)
risch	\(\frac {C x}{b}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}\)	\(341\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 353, normalized size of antiderivative = 4.01 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (C a^{3} - C a b^{2}\right )} d x - {\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}, \frac {2 \, {\left (C a^{3} - C a b^{2}\right )} d x - 2 \, {\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}\right ] \]

[In]

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

[Out]

[1/2*(2*(C*a^3 - C*a*b^2)*d*x - (C*a^2 + A*b^2)*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*co
s(d*x + c) + a^2)) + (A*a^2*b - A*b^3)*log(sin(d*x + c) + 1) - (A*a^2*b - A*b^3)*log(-sin(d*x + c) + 1))/((a^3
*b - a*b^3)*d), 1/2*(2*(C*a^3 - C*a*b^2)*d*x - 2*(C*a^2 + A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/
(sqrt(a^2 - b^2)*sin(d*x + c))) + (A*a^2*b - A*b^3)*log(sin(d*x + c) + 1) - (A*a^2*b - A*b^3)*log(-sin(d*x + c
) + 1))/((a^3*b - a*b^3)*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.62 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} C}{b} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.69 (sec) , antiderivative size = 2862, normalized size of antiderivative = 32.52 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(2*A*atanh((16384*A^5*b^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*
b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C
^2*a*b^3 - (32768*A^4*C*b^5)/a) + (16384*A^5*b^5*tan(c/2 + (d*x)/2))/(16384*A^5*b^5 - 16384*A*C^4*a^5 + 32768*
A^4*C*b^5 - 16384*A^5*a*b^4 + 32768*A^2*C^3*a^2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^
3*C^2*a^3*b^2 + 16384*A*C^4*a^4*b - 32768*A^4*C*a*b^4) + (16384*A*C^4*a^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 +
16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384
*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (32768*A^4*C*b^4*tan(c/2 + (
d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 3276
8*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (32
768*A^4*C*b^5*tan(c/2 + (d*x)/2))/(16384*A^5*b^5 - 16384*A*C^4*a^5 + 32768*A^4*C*b^5 - 16384*A^5*a*b^4 + 32768
*A^2*C^3*a^2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^3*C^2*a^3*b^2 + 16384*A*C^4*a^4*b -
32768*A^4*C*a*b^4) - (16384*A*C^4*a^3*b*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^
4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^
3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) - (32768*A^2*C^3*a*b^3*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 163
84*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C
^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) - (32768*A^3*C^2*a*b^3*tan(c/2 + (
d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 3276
8*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (32
768*A^2*C^3*a^2*b^2*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a
+ 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b
^3 - (32768*A^4*C*b^5)/a) + (32768*A^3*C^2*a^2*b^2*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 3276
8*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^
2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a)))/(a*d) + (2*C*atan((16384*C^5*a^4*tan(c/2 + (d*x)/2)
)/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C
^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (16384*C^5
*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4*a^5 - 16384*A^4*C*b^5 - 16384*C^5*a^4*b - 32768*A^2*C^3*
a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b^3 + 32768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^
4*C*a*b^4) + (32768*A*C^4*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*
C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A
^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A*C^4*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4*a^5 -
16384*A^4*C*b^5 - 16384*C^5*a^4*b - 32768*A^2*C^3*a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b^3 + 32
768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^4*C*a*b^4) + (16384*A^4*C*b^4*tan(c/2 + (d*x)/2))/(16384*C^5
*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 -
16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) - (16384*A^4*C*a*b^3*tan
(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^
2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/
b) - (32768*A^2*C^3*a^3*b*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*
a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C
^2*a^3*b - (32768*A*C^4*a^5)/b) - (32768*A^3*C^2*a^3*b*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 +
16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 3276
8*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A^2*C^3*a^2*b^2*tan(c/2 + (d*x)/2))/(163
84*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2
*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A^3*C^2*a
^2*b^2*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*
C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A
*C^4*a^5)/b)))/(b*d) - (log(b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(-(a + b)*(a - b)
)^(1/2)*(A*b^2 + C*a^2))/(d*(a*b^3 - a^3*b)) - (log(a*tan(c/2 + (d*x)/2) - b*tan(c/2 + (d*x)/2) + (b^2 - a^2)^
(1/2))*(A*b^2*(b^2 - a^2)^(1/2) + C*a^2*(b^2 - a^2)^(1/2)))/(a*b*d*(a^2 - b^2))