\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx\) [567]

Optimal result

Integrand size = 33, antiderivative size = 95 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {A b \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3135, 3080, 3855, 2738, 211} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {A b \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \]

[In]

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

[Out]

(2*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*Sqrt[a - b]*Sqrt[a + b]*d) - (A*b*
ArcTanh[Sin[c + d*x]])/(a^2*d) + (A*Tan[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 3080

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

Rule 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(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)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 {A \tan (c+d x)}{a d}+\frac {\int \frac {(-A b+a C \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a} \\ & = \frac {A \tan (c+d x)}{a d}-\frac {(A b) \int \sec (c+d x) \, dx}{a^2}+\left (\frac {A b^2}{a^2}+C\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx \\ & = -\frac {A b \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d}+\frac {\left (2 \left (\frac {A b^2}{a^2}+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 )}{d} \\ & = \frac {2 \left (\frac {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 )}{\sqrt {a-b} \sqrt {a+b} d}-\frac {A b \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.03 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.22 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {2 i \left (A b^2+a^2 C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (\cos (c)-i \sin (c))}{\sqrt {\left (-a^2+b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac {a A \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {a A \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{a^2 d (2 A+C+C \cos (2 (c+d x)))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {\frac {2 \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{d}\)	\(135\)
default	\(\frac {\frac {2 \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{d}\)	\(135\)
risch	\(\frac {2 i A}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}\)	\(360\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 412, normalized size of antiderivative = 4.34 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {{\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(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 = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.73 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a} + \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2}}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.47 (sec) , antiderivative size = 1328, normalized size of antiderivative = 13.98 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(A*a*tan(c + d*x))/(d*(a^2 - b^2)) - (C*atan(((2*A^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*A^2*b^7*sin(
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + C^2*a^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*C^2*a^4*b*sin(c/2 + (d*x)/
2)*(b^2 - a^2)^(3/2) + C^2*a^6*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2
- a^2)^(1/2) - A^2*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(
1/2) + A^2*a^5*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - C
^2*a^5*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*A*C*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*A*C*a
^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*A*C*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A*C*a^4*b
^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A*C*a^5*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (
d*x)/2)*(a*b^2 - a^3)*(C^2*a^5 - A^2*a*b^4 + A^2*a^3*b^2 - C^2*a^3*b^2 - 2*A*C*a*b^4 + 2*A*C*a^3*b^2)))*(-(a +
b)*(a - b))^(1/2)*2i)/(d*(a^2 - b^2)) - (2*A*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2))
+ (2*A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^2*d*(a^2 - b^2)) - (A*b^2*tan(c + d*x))/(a*d*(a^2
- b^2)) - (A*b^2*atan(((2*A^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a
^2)^(1/2) + C^2*a^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*C^2*a^4*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) +
C^2*a^6*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^3*
b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + A^2*a^5*b^2*sin(
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - C^2*a^5*b^2*sin(c/2 + (d
*x)/2)*(b^2 - a^2)^(1/2) + 4*A*C*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*A*C*a^2*b^5*sin(c/2 + (d*x)/
2)*(b^2 - a^2)^(1/2) - 2*A*C*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A*C*a^4*b^3*sin(c/2 + (d*x)/2)*(
b^2 - a^2)^(1/2) + 2*A*C*a^5*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(
C^2*a^5 - A^2*a*b^4 + A^2*a^3*b^2 - C^2*a^3*b^2 - 2*A*C*a*b^4 + 2*A*C*a^3*b^2)))*(-(a + b)*(a - b))^(1/2)*2i)/
(a^2*d*(a^2 - b^2))