\(\int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx\) [1]

Optimal result

Integrand size = 15, antiderivative size = 54 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (a+b \sin (x))}{b} \]

[Out]

B*ln(a+b*sin(x))/b+2*A*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4486, 2739, 632, 210, 2747, 31} \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (a+b \sin (x))}{b} \]

[In]

Int[(A + B*Cos[x])/(a + b*Sin[x]),x]

[Out]

(2*A*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (B*Log[a + b*Sin[x]])/b

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A}{a+b \sin (x)}+\frac {B \cos (x)}{a+b \sin (x)}\right ) \, dx \\ & = A \int \frac {1}{a+b \sin (x)} \, dx+B \int \frac {\cos (x)}{a+b \sin (x)} \, dx \\ & = (2 A) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {B \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (x)\right )}{b} \\ & = \frac {B \log (a+b \sin (x))}{b}-(4 A) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (a+b \sin (x))}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {B \log (a+b \sin (x))}{b} \]

[In]

Integrate[(A + B*Cos[x])/(a + b*Sin[x]),x]

[Out]

(2*A*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (B*Log[a + b*Sin[x]])/b

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98

method	result	size
parts	\(\frac {2 A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}+\frac {B \ln \left (a +b \sin \left (x \right )\right )}{b}\)	\(53\)
default	\(-\frac {B \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{b}+\frac {B \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 A b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b}\)	\(83\)
risch	\(\frac {i x B}{b}-\frac {2 i B x \,a^{2} b}{a^{2} b^{2}-b^{4}}+\frac {2 i B x \,b^{3}}{a^{2} b^{2}-b^{4}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A b +\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B \,a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i a A b +\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A b +\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{\left (a^{2}-b^{2}\right ) b}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A b -\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B \,a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i a A b -\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B}{a^{2}-b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A b -\sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) \sqrt {-A^{2} a^{2} b^{2}+A^{2} b^{4}}}{\left (a^{2}-b^{2}\right ) b}\)	\(456\)

[In]

int((A+B*cos(x))/(a+b*sin(x)),x,method=_RETURNVERBOSE)

[Out]

2*A/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+B*ln(a+b*sin(x))/b

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 246, normalized size of antiderivative = 4.56 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} A b \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2}} A b \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}\right ] \]

[In]

integrate((A+B*cos(x))/(a+b*sin(x)),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 552 vs. \(2 (44) = 88\).

Time = 14.07 (sec) , antiderivative size = 552, normalized size of antiderivative = 10.22 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\begin {cases} \tilde {\infty } \left (A \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} - B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tan {\left (\frac {x}{2} \right )} \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {A \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} - B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tan {\left (\frac {x}{2} \right )} \right )}}{b} & \text {for}\: a = 0 \\\frac {2 A}{b \tan {\left (\frac {x}{2} \right )} - b} + \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} - \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} - \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} + \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} & \text {for}\: a = - b \\- \frac {2 A}{b \tan {\left (\frac {x}{2} \right )} + b} + \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} + \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} - \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} - \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} & \text {for}\: a = b \\\frac {A x + B \sin {\left (x \right )}}{a} & \text {for}\: b = 0 \\- \frac {A b \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {A b \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac {B a^{2} \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{a^{2} b - b^{3}} + \frac {B a^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {B a^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {B b^{2} \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{a^{2} b - b^{3}} - \frac {B b^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac {B b^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate((A+B*cos(x))/(a+b*sin(x)),x)

[Out]

Piecewise((zoo*(A*log(tan(x/2)) - B*log(tan(x/2)**2 + 1) + B*log(tan(x/2))), Eq(a, 0) & Eq(b, 0)), ((A*log(tan
(x/2)) - B*log(tan(x/2)**2 + 1) + B*log(tan(x/2)))/b, Eq(a, 0)), (2*A/(b*tan(x/2) - b) + 2*B*log(tan(x/2) - 1)
*tan(x/2)/(b*tan(x/2) - b) - 2*B*log(tan(x/2) - 1)/(b*tan(x/2) - b) - B*log(tan(x/2)**2 + 1)*tan(x/2)/(b*tan(x
/2) - b) + B*log(tan(x/2)**2 + 1)/(b*tan(x/2) - b), Eq(a, -b)), (-2*A/(b*tan(x/2) + b) + 2*B*log(tan(x/2) + 1)
*tan(x/2)/(b*tan(x/2) + b) + 2*B*log(tan(x/2) + 1)/(b*tan(x/2) + b) - B*log(tan(x/2)**2 + 1)*tan(x/2)/(b*tan(x
/2) + b) - B*log(tan(x/2)**2 + 1)/(b*tan(x/2) + b), Eq(a, b)), ((A*x + B*sin(x))/a, Eq(b, 0)), (-A*b*sqrt(-a**
2 + b**2)*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + A*b*sqrt(-a**2 + b**2)*log(tan(x/2) + b
/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - B*a**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) + B*a**2*log(tan(x/2)
+ b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + B*a**2*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b
**3) + B*b**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) - B*b**2*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**2*b
- b**3) - B*b**2*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*cos(x))/(a+b*sin(x)),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 = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.63 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{b} - \frac {B \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{b} \]

[In]

integrate((A+B*cos(x))/(a+b*sin(x)),x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*A/sqrt(a^2 - b^2) + B*log(a*t
an(1/2*x)^2 + 2*b*tan(1/2*x) + a)/b - B*log(tan(1/2*x)^2 + 1)/b

Mupad [B] (verification not implemented)

Time = 18.26 (sec) , antiderivative size = 1779, normalized size of antiderivative = 32.94 \[ \int \frac {A+B \cos (x)}{a+b \sin (x)} \, dx=\text {Too large to display} \]

[In]

int((A + B*cos(x))/(a + b*sin(x)),x)

[Out]

(log((a + b*sin(x))/(cos(x) + 1))*(2*B*b^3 - 2*B*a^2*b))/(2*(b^4 - a^2*b^2)) - (2*A*atan((tan(x/2)*(a^2 - b^2)
^(3/2)*((((((A*(((2*B*b^3 - 2*B*a^2*b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*A*a*b^3 + 32*B*a*b^3)
)/(a^2 - b^2)^(1/2) + (A*(2*B*b^3 - 2*B*a^2*b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)*(a^2 - b^2)^(1/2)))
*(2*B*b^3 - 2*B*a^2*b))/(2*(b^4 - a^2*b^2)) - (A^3*(96*a*b^4 - 64*a^3*b^2))/(a^2 - b^2)^(3/2) + (A*(64*B^2*a^3
- 32*A^2*a*b^2 - 96*B^2*a*b^2 + ((2*B*b^3 - 2*B*a^2*b)*(((2*B*b^3 - 2*B*a^2*b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b
^4 - a^2*b^2)) + 64*A*a*b^3 + 32*B*a*b^3))/(2*(b^4 - a^2*b^2)) + 128*A*B*a*b^2))/(a^2 - b^2)^(1/2))*(2*A^2*b^4
+ 4*B^2*a^4 + 8*B^2*b^4 - A^2*a^2*b^2 - 12*B^2*a^2*b^2 + 8*A*B*b^4 - 8*A*B*a^2*b^2))/(a^3*(a^2 - b^2)^(1/2)*(
A^2*b^2 + 4*B^2*a^2 - 4*B^2*b^2)^2) - (2*b*(A + 2*B)*(A*b^2 - 2*B*a^2 + 2*B*b^2)*((A*((A*(((2*B*b^3 - 2*B*a^2*
b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*A*a*b^3 + 32*B*a*b^3))/(a^2 - b^2)^(1/2) + (A*(2*B*b^3 -
2*B*a^2*b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(a^2 - b^2)^(1/2) + 32*B^3*a*b - (
(2*B*b^3 - 2*B*a^2*b)*(64*B^2*a^3 - 32*A^2*a*b^2 - 96*B^2*a*b^2 + ((2*B*b^3 - 2*B*a^2*b)*(((2*B*b^3 - 2*B*a^2*
b)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*A*a*b^3 + 32*B*a*b^3))/(2*(b^4 - a^2*b^2)) + 128*A*B*a*b^
2))/(2*(b^4 - a^2*b^2)) - 64*A*B^2*a*b + 32*A^2*B*a*b + (A^2*(2*B*b^3 - 2*B*a^2*b)*(96*a*b^4 - 64*a^3*b^2))/(2
*(b^4 - a^2*b^2)*(a^2 - b^2))))/(a^3*(A^2*b^2 + 4*B^2*a^2 - 4*B^2*b^2)^2)))/(32*A*a) + ((a^2 - b^2)*((A*(((2*B
*b^3 - 2*B*a^2*b)*(32*A*a^2*b^2 + 32*B*a^2*b^2 + (16*a^2*b^3*(2*B*b^3 - 2*B*a^2*b))/(b^4 - a^2*b^2)))/(2*(b^4
- a^2*b^2)) - 32*B^2*a^2*b + 64*A*B*a^2*b))/(a^2 - b^2)^(1/2) + ((2*B*b^3 - 2*B*a^2*b)*((A*(32*A*a^2*b^2 + 32*
B*a^2*b^2 + (16*a^2*b^3*(2*B*b^3 - 2*B*a^2*b))/(b^4 - a^2*b^2)))/(a^2 - b^2)^(1/2) + (16*A*a^2*b^3*(2*B*b^3 -
2*B*a^2*b))/((b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(2*(b^4 - a^2*b^2)) - (32*A^3*a^2*b^3)/(a^2 - b^2)^(3/2))*(2
*A^2*b^4 + 4*B^2*a^4 + 8*B^2*b^4 - A^2*a^2*b^2 - 12*B^2*a^2*b^2 + 8*A*B*b^4 - 8*A*B*a^2*b^2))/(32*A*a^4*(A^2*b
^2 + 4*B^2*a^2 - 4*B^2*b^2)^2) - (b*(a^2 - b^2)^(3/2)*(A + 2*B)*(A*b^2 - 2*B*a^2 + 2*B*b^2)*(32*B^3*a^2 - 32*A
*B^2*a^2 - ((2*B*b^3 - 2*B*a^2*b)*(((2*B*b^3 - 2*B*a^2*b)*(32*A*a^2*b^2 + 32*B*a^2*b^2 + (16*a^2*b^3*(2*B*b^3
- 2*B*a^2*b))/(b^4 - a^2*b^2)))/(2*(b^4 - a^2*b^2)) - 32*B^2*a^2*b + 64*A*B*a^2*b))/(2*(b^4 - a^2*b^2)) + (A*(
(A*(32*A*a^2*b^2 + 32*B*a^2*b^2 + (16*a^2*b^3*(2*B*b^3 - 2*B*a^2*b))/(b^4 - a^2*b^2)))/(a^2 - b^2)^(1/2) + (16
*A*a^2*b^3*(2*B*b^3 - 2*B*a^2*b))/((b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(a^2 - b^2)^(1/2) + (16*A^2*a^2*b^3*(2
*B*b^3 - 2*B*a^2*b))/((b^4 - a^2*b^2)*(a^2 - b^2))))/(16*A*a^4*(A^2*b^2 + 4*B^2*a^2 - 4*B^2*b^2)^2)))/(a^2 - b
^2)^(1/2) - (B*log(1/(cos(x) + 1)))/b