\(\int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx\) [190]

Optimal result

Integrand size = 11, antiderivative size = 93 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {2881, 3080, 3855, 2739, 632, 210} \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))} \]

[In]

Int[Csc[x]/(a + b*Sin[x])^2,x]

[Out]

(-2*b*(2*a^2 - b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)) - ArcTanh[Cos[x]]/a^2 -
(b^2*Cos[x])/(a*(a^2 - b^2)*(a + b*Sin[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 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*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*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 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 {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc (x) \left (a^2-b^2-a b \sin (x)\right )}{a+b \sin (x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \csc (x) \, dx}{a^2}-\frac {\left (b \left (2 a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = -\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (2 b \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^2 \left (a^2-b^2\right )} \\ & = -\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\left (4 b \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 )}{a^2 \left (a^2-b^2\right )} \\ & = -\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {\text {arctanh}(\cos (x))}{a^2}-\frac {b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\frac {\frac {2 b \left (-2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {a b^2 \cos (x)}{(a-b) (a+b) (a+b \sin (x))}}{a^2} \]

[In]

Integrate[Csc[x]/(a + b*Sin[x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.32

method	result	size
default	\(-\frac {4 b \left (\frac {\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {a b}{2 a^{2}-2 b^{2}}}{a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\)	\(123\)
risch	\(\frac {2 i b \left (-i a \,{\mathrm e}^{i x}+b \right )}{a \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}+\frac {2 i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {2 i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{a^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{a^{2}}\)	\(380\)

[In]

int(csc(x)/(a+b*sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (87) = 174\).

Time = 0.51 (sec) , antiderivative size = 511, normalized size of antiderivative = 5.49 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\left [-\frac {{\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\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 (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}, \frac {2 \, {\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\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 ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}\right ] \]

[In]

integrate(csc(x)/(a+b*sin(x))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\int \frac {\csc {\left (x \right )}}{\left (a + b \sin {\left (x \right )}\right )^{2}}\, dx \]

[In]

integrate(csc(x)/(a+b*sin(x))**2,x)

[Out]

Integral(csc(x)/(a + b*sin(x))**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(x)/(a+b*sin(x))^2,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 = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (b^{3} \tan \left (\frac {1}{2} \, x\right ) + a b^{2}\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} \]

[In]

integrate(csc(x)/(a+b*sin(x))^2,x, algorithm="giac")

[Out]

-2*(2*a^2*b - b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4 - a^2*
b^2)*sqrt(a^2 - b^2)) - 2*(b^3*tan(1/2*x) + a*b^2)/((a^4 - a^2*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)) + l
og(abs(tan(1/2*x)))/a^2

Mupad [B] (verification not implemented)

Time = 7.61 (sec) , antiderivative size = 1356, normalized size of antiderivative = 14.58 \[ \int \frac {\csc (x)}{(a+b \sin (x))^2} \, dx=\text {Too large to display} \]

[In]

int(1/(sin(x)*(a + b*sin(x))^2),x)

[Out]

log(tan(x/2))/a^2 - ((2*b^2)/(a*(a^2 - b^2)) + (2*b^3*tan(x/2))/(a^2*(a^2 - b^2)))/(a + 2*b*tan(x/2) + a*tan(x
/2)^2) - (b*atan(((b*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(a^6 - 4*b^6 + 11*a^2*b^4 - 8*a^4
*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) - (2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) + (b*((2*(a^6*b - a^4*b^3))/(a^4
- a^2*b^2) - (2*tan(x/2)*(3*a^8 - 4*a^2*b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^
2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a
^6*b^2) - (b*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) - (2*tan(x/
2)*(a^6 - 4*b^6 + 11*a^2*b^4 - 8*a^4*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) + (b*((2*(a^6*b - a^4*b^3))/(a^4 - a^2*b^
2) - (2*tan(x/2)*(3*a^8 - 4*a^2*b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)*(-(a
+ b)^3*(a - b)^3)^(1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))
/((4*(2*a^2*b - b^3))/(a^4 - a^2*b^2) + (4*tan(x/2)*(2*b^4 - 4*a^2*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) + (b*(2*a^2
- b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(a^6 - 4*b^6 + 11*a^2*b^4 - 8*a^4*b^2))/(a*b^4 + a^5 - 2*a^3
*b^2) - (2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) + (b*((2*(a^6*b - a^4*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(3
*a^8 - 4*a^2*b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(
1/2))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) + (b*(2*a^2 - b^2)*(-(
a + b)^3*(a - b)^3)^(1/2)*((2*(3*a^4*b - 2*a^2*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(a^6 - 4*b^6 + 11*a^2*b^4 -
8*a^4*b^2))/(a*b^4 + a^5 - 2*a^3*b^2) + (b*((2*(a^6*b - a^4*b^3))/(a^4 - a^2*b^2) - (2*tan(x/2)*(3*a^8 - 4*a^
2*b^6 + 11*a^4*b^4 - 10*a^6*b^2))/(a*b^4 + a^5 - 2*a^3*b^2))*(2*a^2 - b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^8
- a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*(2*a^2 - b^2)*(-(a + b)^3*(a -
b)^3)^(1/2)*2i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)