3.1.32 \(\int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx\) [32]

3.1.32.1 Optimal result
3.1.32.2 Mathematica [A] (verified)
3.1.32.3 Rubi [A] (verified)
3.1.32.4 Maple [A] (verified)
3.1.32.5 Fricas [B] (verification not implemented)
3.1.32.6 Sympy [F]
3.1.32.7 Maxima [F(-2)]
3.1.32.8 Giac [B] (verification not implemented)
3.1.32.9 Mupad [B] (verification not implemented)

3.1.32.1 Optimal result

Integrand size = 13, antiderivative size = 110 \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=\frac {2 b^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )} \]

output
2*b^4*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)-1 
/3*(3*b^3+a*(2*a^2-5*b^2)*cos(x))*csc(x)/(a^2-b^2)^2+1/3*(b-a*cos(x))*csc( 
x)^3/(a^2-b^2)
 
3.1.32.2 Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=-\frac {2 b^4 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {\left (\left (-6 a^3+9 a b^2\right ) \cos (x)+6 b^3 \cos (2 x)+\left (2 a^2-5 b^2\right ) (2 b+a \cos (3 x))\right ) \csc ^3(x)}{12 (a-b)^2 (a+b)^2} \]

input
Integrate[Csc[x]^4/(a + b*Cos[x]),x]
 
output
(-2*b^4*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 
 (((-6*a^3 + 9*a*b^2)*Cos[x] + 6*b^3*Cos[2*x] + (2*a^2 - 5*b^2)*(2*b + a*C 
os[3*x]))*Csc[x]^3)/(12*(a - b)^2*(a + b)^2)
 
3.1.32.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 3175, 25, 3042, 3345, 27, 3042, 3138, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos \left (x-\frac {\pi }{2}\right )^4 \left (a-b \sin \left (x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3175

\(\displaystyle \frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}-\frac {\int -\frac {\left (2 a^2+2 b \cos (x) a-3 b^2\right ) \csc ^2(x)}{a+b \cos (x)}dx}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (2 a^2+2 b \cos (x) a-3 b^2\right ) \csc ^2(x)}{a+b \cos (x)}dx}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 a^2-2 b \sin \left (x-\frac {\pi }{2}\right ) a-3 b^2}{\cos \left (x-\frac {\pi }{2}\right )^2 \left (a-b \sin \left (x-\frac {\pi }{2}\right )\right )}dx}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3345

\(\displaystyle \frac {-\frac {\int -\frac {3 b^4}{a+b \cos (x)}dx}{a^2-b^2}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 b^4 \int \frac {1}{a+b \cos (x)}dx}{a^2-b^2}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 b^4 \int \frac {1}{a+b \sin \left (x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {6 b^4 \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {6 b^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}+\frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}\)

input
Int[Csc[x]^4/(a + b*Cos[x]),x]
 
output
((b - a*Cos[x])*Csc[x]^3)/(3*(a^2 - b^2)) + ((6*b^4*ArcTan[(Sqrt[a - b]*Ta 
n[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)) - ((3*b^3 + a* 
(2*a^2 - 5*b^2)*Cos[x])*Csc[x])/(a^2 - b^2))/(3*(a^2 - b^2))
 

3.1.32.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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 3175
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^ 
(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2* 
(a^2 - b^2)*(p + 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m* 
(a^2*(p + 2) - b^2*(m + p + 2) + a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegersQ 
[2*m, 2*p]
 

rule 3345
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co 
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 
Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 
 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 
 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], 
x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[p, -1] && IntegerQ[2*m]
 
3.1.32.4 Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15

method result size
default \(\frac {\frac {a \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}-\frac {b \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}+3 a \tan \left (\frac {x}{2}\right )-5 b \tan \left (\frac {x}{2}\right )}{8 \left (a -b \right )^{2}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {x}{2}\right )^{3}}-\frac {3 a +5 b}{8 \left (a +b \right )^{2} \tan \left (\frac {x}{2}\right )}+\frac {2 b^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}\) \(127\)
risch \(-\frac {2 i \left (3 b^{3} {\mathrm e}^{5 i x}-3 a \,b^{2} {\mathrm e}^{4 i x}+4 a^{2} b \,{\mathrm e}^{3 i x}-10 b^{3} {\mathrm e}^{3 i x}-6 a^{3} {\mathrm e}^{2 i x}+12 a \,b^{2} {\mathrm e}^{2 i x}+3 b^{3} {\mathrm e}^{i x}+2 a^{3}-5 a \,b^{2}\right )}{3 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i x}-1\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) \(267\)

input
int(csc(x)^4/(a+cos(x)*b),x,method=_RETURNVERBOSE)
 
output
1/8/(a-b)^2*(1/3*a*tan(1/2*x)^3-1/3*b*tan(1/2*x)^3+3*a*tan(1/2*x)-5*b*tan( 
1/2*x))-1/24/(a+b)/tan(1/2*x)^3-1/8*(3*a+5*b)/(a+b)^2/tan(1/2*x)+2/(a-b)^2 
/(a+b)^2*b^4/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b))^(1/ 
2))
 
3.1.32.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (97) = 194\).

Time = 0.28 (sec) , antiderivative size = 459, normalized size of antiderivative = 4.17 \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=\left [\frac {2 \, a^{4} b - 10 \, a^{2} b^{3} + 8 \, b^{5} + 2 \, {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (x\right )^{3} + 3 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) + 6 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2} - 6 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (x\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}, \frac {a^{4} b - 5 \, a^{2} b^{3} + 4 \, b^{5} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (x\right )^{3} - 3 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) + 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2} - 3 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (x\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}\right ] \]

input
integrate(csc(x)^4/(a+b*cos(x)),x, algorithm="fricas")
 
output
[1/6*(2*a^4*b - 10*a^2*b^3 + 8*b^5 + 2*(2*a^5 - 7*a^3*b^2 + 5*a*b^4)*cos(x 
)^3 + 3*(b^4*cos(x)^2 - b^4)*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - 
 b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/( 
b^2*cos(x)^2 + 2*a*b*cos(x) + a^2))*sin(x) + 6*(a^2*b^3 - b^5)*cos(x)^2 - 
6*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cos(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 
- (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^2)*sin(x)), 1/3*(a^4*b - 5*a^ 
2*b^3 + 4*b^5 + (2*a^5 - 7*a^3*b^2 + 5*a*b^4)*cos(x)^3 - 3*(b^4*cos(x)^2 - 
 b^4)*sqrt(a^2 - b^2)*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x)))*sin 
(x) + 3*(a^2*b^3 - b^5)*cos(x)^2 - 3*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cos(x))/( 
(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*c 
os(x)^2)*sin(x))]
 
3.1.32.6 Sympy [F]

\[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=\int \frac {\csc ^{4}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \]

input
integrate(csc(x)**4/(a+b*cos(x)),x)
 
output
Integral(csc(x)**4/(a + b*cos(x)), x)
 
3.1.32.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=\text {Exception raised: ValueError} \]

input
integrate(csc(x)^4/(a+b*cos(x)),x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 
3.1.32.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (97) = 194\).

Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.87 \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, x\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) - 24 \, a b \tan \left (\frac {1}{2} \, x\right ) + 15 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {9 \, a \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, b \tan \left (\frac {1}{2} \, x\right )^{2} + a + b}{24 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac {1}{2} \, x\right )^{3}} \]

input
integrate(csc(x)^4/(a+b*cos(x)),x, algorithm="giac")
 
output
-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*t 
an(1/2*x))/sqrt(a^2 - b^2)))*b^4/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) 
 + 1/24*(a^2*tan(1/2*x)^3 - 2*a*b*tan(1/2*x)^3 + b^2*tan(1/2*x)^3 + 9*a^2* 
tan(1/2*x) - 24*a*b*tan(1/2*x) + 15*b^2*tan(1/2*x))/(a^3 - 3*a^2*b + 3*a*b 
^2 - b^3) - 1/24*(9*a*tan(1/2*x)^2 + 15*b*tan(1/2*x)^2 + a + b)/((a^2 + 2* 
a*b + b^2)*tan(1/2*x)^3)
 
3.1.32.9 Mupad [B] (verification not implemented)

Time = 13.37 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx=\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {4}{8\,a-8\,b}-\frac {8\,a+8\,b}{{\left (8\,a-8\,b\right )}^2}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{3\,\left (8\,a-8\,b\right )}-\frac {\frac {a^2-2\,a\,b+b^2}{3\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (-3\,a^3+a^2\,b+7\,a\,b^2-5\,b^3\right )}{{\left (a+b\right )}^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (8\,a^2-16\,a\,b+8\,b^2\right )}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

input
int(1/(sin(x)^4*(a + b*cos(x))),x)
 
output
tan(x/2)*(4/(8*a - 8*b) - (8*a + 8*b)/(8*a - 8*b)^2) + tan(x/2)^3/(3*(8*a 
- 8*b)) - ((a^2 - 2*a*b + b^2)/(3*(a + b)) - (tan(x/2)^2*(7*a*b^2 + a^2*b 
- 3*a^3 - 5*b^3))/(a + b)^2)/(tan(x/2)^3*(8*a^2 - 16*a*b + 8*b^2)) + (2*b^ 
4*atan((tan(x/2)*(a^4 + b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a - b)^(3/2))))/ 
((a + b)^(5/2)*(a - b)^(5/2))