3.2.98 \(\int \frac {1}{(a+b \sin (x))^3} \, dx\) [198]

Optimal. Leaf size=102 \[ \frac {\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

[Out]

(2*a^2+b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)+1/2*b*cos(x)/(a^2-b^2)/(a+b*sin(x))^2+3/2
*a*b*cos(x)/(a^2-b^2)^2/(a+b*sin(x))

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {2743, 2833, 12, 2739, 632, 210} \begin {gather*} \frac {\left (2 a^2+b^2\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x])^(-3),x]

[Out]

((2*a^2 + b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b*Cos[x])/(2*(a^2 - b^2)*(a + b*
Sin[x])^2) + (3*a*b*Cos[x])/(2*(a^2 - b^2)^2*(a + b*Sin[x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2833

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sin (x))^3} \, dx &=\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\int \frac {-2 a+b \sin (x)}{(a+b \sin (x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\int \frac {2 a^2+b^2}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\left (2 a^2+b^2\right ) \int \frac {1}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\left (2 a^2+b^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {\left (2 \left (2 a^2+b^2\right )\right ) \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 )}{\left (a^2-b^2\right )^2}\\ &=\frac {\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 93, normalized size = 0.91 \begin {gather*} \frac {\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b \cos (x) \left (4 a^2-b^2+3 a b \sin (x)\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x])^(-3),x]

[Out]

((2*a^2 + b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b*Cos[x]*(4*a^2 - b^2 + 3*a*b*Si
n[x]))/(2*(a - b)^2*(a + b)^2*(a + b*Sin[x])^2)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(92)=184\).
time = 0.25, size = 248, normalized size = 2.43

method	result	size
default	\(\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{4}+7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 a^{2} b^{2}+2 b^{4}}}{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\)	\(248\)
risch	\(-\frac {i \left (-2 i b \,a^{2} {\mathrm e}^{3 i x}-i {\mathrm e}^{3 i x} b^{3}+10 i a^{2} b \,{\mathrm e}^{i x}-i b^{3} {\mathrm e}^{i x}+6 a^{3} {\mathrm e}^{2 i x}+3 b^{2} a \,{\mathrm e}^{2 i x}-3 a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\)	\(409\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)

[Out]

2*(1/2*b^2*(5*a^2-2*b^2)/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*x)^3+1/2*b*(4*a^4+7*a^2*b^2-2*b^4)/(a^4-2*a^2*b^2+b^4)/
a^2*tan(1/2*x)^2+1/2*b^2*(11*a^2-2*b^2)/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)+1/2*b*(4*a^2-b^2)/(a^4-2*a^2*b^2+b^4)
)/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^2+(2*a^2+b^2)/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*
x)+2*b)/(a^2-b^2)^(1/2))

________________________________________________________________________________________

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(1/(a+b*sin(x))^3,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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (92) = 184\).
time = 0.50, size = 516, normalized size = 5.06 \begin {gather*} \left [\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (x\right )\right )} \sqrt {-a^{2} + b^{2}} \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 ) + 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )}{4 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}, \frac {3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (x\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )}{2 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^3,x, algorithm="fricas")

[Out]

[1/4*(6*(a^3*b^2 - a*b^4)*cos(x)*sin(x) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^4)*cos(x)^2 + 2*(2*a^3*b +
a*b^3)*sin(x))*sqrt(-a^2 + b^2)*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*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*(4*a^4*b - 5*a^2*b^3 + b^5)*cos(x)
)/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3
+ 3*a^3*b^5 - a*b^7)*sin(x)), 1/2*(3*(a^3*b^2 - a*b^4)*cos(x)*sin(x) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2
+ b^4)*cos(x)^2 + 2*(2*a^3*b + a*b^3)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x)))
+ (4*a^4*b - 5*a^2*b^3 + b^5)*cos(x))/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 -
b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sin {\left (x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))**3,x)

[Out]

Integral((a + b*sin(x))**(-3), x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (92) = 184\).
time = 0.43, size = 215, normalized size = 2.11 \begin {gather*} \frac {{\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 )} {\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, x\right )^{2} + 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, x\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, x\right ) + 4 \, a^{4} b - a^{2} b^{3}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^3,x, algorithm="giac")

[Out]

(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*(2*a^2 + b^2)/((a^4 - 2*a^2*b^2
+ b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*tan(1/2*x)^3 - 2*a*b^4*tan(1/2*x)^3 + 4*a^4*b*tan(1/2*x)^2 + 7*a^2*b^3*t
an(1/2*x)^2 - 2*b^5*tan(1/2*x)^2 + 11*a^3*b^2*tan(1/2*x) - 2*a*b^4*tan(1/2*x) + 4*a^4*b - a^2*b^3)/((a^6 - 2*a
^4*b^2 + a^2*b^4)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2)

________________________________________________________________________________________

Mupad [B]
time = 7.32, size = 349, normalized size = 3.42 \begin {gather*} \frac {\frac {4\,a^2\,b-b^3}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^2\,b-b^3\right )\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (5\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (11\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^2+b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2+b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,a^2+b^2}\right )\,\left (2\,a^2+b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(x))^3,x)

[Out]

((4*a^2*b - b^3)/(a^4 + b^4 - 2*a^2*b^2) + (tan(x/2)^2*(4*a^2*b - b^3)*(a^2 + 2*b^2))/(a^2*(a^4 + b^4 - 2*a^2*
b^2)) + (b*tan(x/2)^3*(5*a^2*b - 2*b^3))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(x/2)*(11*a^2*b - 2*b^3))/(a*(a^4
+ b^4 - 2*a^2*b^2)))/(tan(x/2)^2*(2*a^2 + 4*b^2) + a^2 + a^2*tan(x/2)^4 + 4*a*b*tan(x/2) + 4*a*b*tan(x/2)^3)
+ (atan(((((2*a^2 + b^2)*(2*a^4*b + 2*b^5 - 4*a^2*b^3))/(2*(a + b)^(5/2)*(a - b)^(5/2)*(a^4 + b^4 - 2*a^2*b^2)
) + (a*tan(x/2)*(2*a^2 + b^2))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a^2*b^2))/(2*a^2 + b^2))*(2*a^2 +
b^2))/((a + b)^(5/2)*(a - b)^(5/2))

________________________________________________________________________________________