\(\int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx\) [20]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=-\frac {x}{a}+\frac {2 b^3 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}}-\frac {b \sec (x)}{a^2-b^2}+\frac {a \tan (x)}{a^2-b^2} \]

[Out]

-x/a+2*b^3*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a/(a^2-b^2)^(3/2)-b*sec(x)/(a^2-b^2)+a*tan(x)/(a^2-b^2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3983, 2981, 2686, 8, 3554, 2814, 2739, 632, 212} \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=\frac {2 b^3 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}}+\frac {b^2 x}{a \left (a^2-b^2\right )}-\frac {a x}{a^2-b^2}+\frac {a \tan (x)}{a^2-b^2}-\frac {b \sec (x)}{a^2-b^2} \]

[In]

Int[Tan[x]^2/(a + b*Csc[x]),x]

[Out]

-((a*x)/(a^2 - b^2)) + (b^2*x)/(a*(a^2 - b^2)) + (2*b^3*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b
^2)^(3/2)) - (b*Sec[x])/(a^2 - b^2) + (a*Tan[x])/(a^2 - b^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 2814

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

Rule 2981

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[a*(d^2/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[b*(d/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[a^2*(d^2/(g^2*(a^2 - b^2
))), Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3983

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[Cos[c + d*x]^m
*((b + a*Sin[c + d*x])^n/Sin[c + d*x]^(m + n)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin (x) \tan ^2(x)}{b+a \sin (x)} \, dx \\ & = \frac {a \int \tan ^2(x) \, dx}{a^2-b^2}-\frac {b \int \sec (x) \tan (x) \, dx}{a^2-b^2}+\frac {b^2 \int \frac {\sin (x)}{b+a \sin (x)} \, dx}{a^2-b^2} \\ & = \frac {b^2 x}{a \left (a^2-b^2\right )}+\frac {a \tan (x)}{a^2-b^2}-\frac {a \int 1 \, dx}{a^2-b^2}-\frac {b \text {Subst}(\int 1 \, dx,x,\sec (x))}{a^2-b^2}-\frac {b^3 \int \frac {1}{b+a \sin (x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {a x}{a^2-b^2}+\frac {b^2 x}{a \left (a^2-b^2\right )}-\frac {b \sec (x)}{a^2-b^2}+\frac {a \tan (x)}{a^2-b^2}-\frac {\left (2 b^3\right ) \text {Subst}\left (\int \frac {1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )} \\ & = -\frac {a x}{a^2-b^2}+\frac {b^2 x}{a \left (a^2-b^2\right )}-\frac {b \sec (x)}{a^2-b^2}+\frac {a \tan (x)}{a^2-b^2}+\frac {\left (4 b^3\right ) \text {Subst}\left (\int \frac {1}{4 \left (a^2-b^2\right )-x^2} \, dx,x,2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2-b^2\right )} \\ & = -\frac {a x}{a^2-b^2}+\frac {b^2 x}{a \left (a^2-b^2\right )}+\frac {2 b^3 \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}}-\frac {b \sec (x)}{a^2-b^2}+\frac {a \tan (x)}{a^2-b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=-\frac {-2 b^3 \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+\sqrt {-a^2+b^2} \left (-a^2 x+b^2 x-a b \sec (x)+a^2 \tan (x)\right )}{a \left (-a^2+b^2\right )^{3/2}} \]

[In]

Integrate[Tan[x]^2/(a + b*Csc[x]),x]

[Out]

-((-2*b^3*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + Sqrt[-a^2 + b^2]*(-(a^2*x) + b^2*x - a*b*Sec[x] + a^2*Ta
n[x]))/(a*(-a^2 + b^2)^(3/2)))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {16}{\left (16 a +16 b \right ) \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {16}{\left (16 a -16 b \right ) \left (\tan \left (\frac {x}{2}\right )+1\right )}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a}-\frac {2 b^{3} \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}}\)	\(106\)
risch	\(-\frac {x}{a}+\frac {-2 i a +2 b \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}+1\right ) \left (-a^{2}+b^{2}\right )}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) a}\)	\(195\)

[In]

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

[Out]

-16/(16*a+16*b)/(tan(1/2*x)-1)-16/(16*a-16*b)/(tan(1/2*x)+1)-2/a*arctan(tan(1/2*x))-2/(a+b)/(a-b)*b^3/a/(-a^2+
b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.52 \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=\left [-\frac {\sqrt {a^{2} - b^{2}} b^{3} \cos \left (x\right ) \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} - 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b - 2 \, a b^{3} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cos \left (x\right ) - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )}, \frac {\sqrt {-a^{2} + b^{2}} b^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right ) - a^{3} b + a b^{3} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cos \left (x\right ) + {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )}\right ] \]

[In]

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

[Out]

[-1/2*(sqrt(a^2 - b^2)*b^3*cos(x)*log(-((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 - 2*(b*cos(x)*sin(x)
+ a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*a^3*b - 2*a*b^3 + 2*(a^4 - 2*a^2*
b^2 + b^4)*x*cos(x) - 2*(a^4 - a^2*b^2)*sin(x))/((a^5 - 2*a^3*b^2 + a*b^4)*cos(x)), (sqrt(-a^2 + b^2)*b^3*arct
an(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x)))*cos(x) - a^3*b + a*b^3 - (a^4 - 2*a^2*b^2 + b^4)*x*c
os(x) + (a^4 - a^2*b^2)*sin(x))/((a^5 - 2*a^3*b^2 + a*b^4)*cos(x))]

Sympy [F]

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

[In]

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

[Out]

Integral(tan(x)**2/(a + b*csc(x)), x)

Maxima [F(-2)]

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

[In]

integrate(tan(x)^2/(a+b*csc(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*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.23 \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {x}{a} - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 20.92 (sec) , antiderivative size = 2341, normalized size of antiderivative = 27.87 \[ \int \frac {\tan ^2(x)}{a+b \csc (x)} \, dx=\text {Too large to display} \]

[In]

int(tan(x)^2/(a + b/sin(x)),x)

[Out]

((2*b)/(a^2 - b^2) - (2*a*tan(x/2))/(a^2 - b^2))/(tan(x/2)^2 - 1) + (2*atan((64*a^11*b*tan(x/2))/(192*a*b^11 -
64*a^11*b - 768*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3) - (192*a*b^11*tan(x/2))/(192*a*b^11 - 64*
a^11*b - 768*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3) + (768*a^3*b^9*tan(x/2))/(192*a*b^11 - 64*a^1
1*b - 768*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3) - (1216*a^5*b^7*tan(x/2))/(192*a*b^11 - 64*a^11*
b - 768*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3) + (960*a^7*b^5*tan(x/2))/(192*a*b^11 - 64*a^11*b -
768*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3) - (384*a^9*b^3*tan(x/2))/(192*a*b^11 - 64*a^11*b - 76
8*a^3*b^9 + 1216*a^5*b^7 - 960*a^7*b^5 + 384*a^9*b^3)))/a + (b^3*atan(((b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x
/2)*(64*a^12*b + 64*b^13 - 320*a^2*b^11 + 736*a^4*b^9 - 992*a^6*b^7 + 800*a^8*b^5 - 352*a^10*b^3) - 32*a*b^12
+ 160*a^3*b^10 - 320*a^5*b^8 + 320*a^7*b^6 - 160*a^9*b^4 + 32*a^11*b^2 + (b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan
(x/2)*(64*a^2*b^12 - 256*a^4*b^10 + 384*a^6*b^8 - 256*a^8*b^6 + 64*a^10*b^4) - 32*a^13*b + 64*a^3*b^11 - 288*a
^5*b^9 + 512*a^7*b^7 - 448*a^9*b^5 + 192*a^11*b^3 + (b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(96*a^14*b + 64
*a^2*b^13 - 416*a^4*b^11 + 1120*a^6*b^9 - 1600*a^8*b^7 + 1280*a^10*b^5 - 544*a^12*b^3) - 32*a^3*b^12 + 160*a^5
*b^10 - 320*a^7*b^8 + 320*a^9*b^6 - 160*a^11*b^4 + 32*a^13*b^2))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)))/(a*b^
6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2))*1i)/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2) + (b^3*((a + b)^3*(a - b)^3)^(1/2)
*(tan(x/2)*(64*a^12*b + 64*b^13 - 320*a^2*b^11 + 736*a^4*b^9 - 992*a^6*b^7 + 800*a^8*b^5 - 352*a^10*b^3) - 32*
a*b^12 + 160*a^3*b^10 - 320*a^5*b^8 + 320*a^7*b^6 - 160*a^9*b^4 + 32*a^11*b^2 - (b^3*((a + b)^3*(a - b)^3)^(1/
2)*(tan(x/2)*(64*a^2*b^12 - 256*a^4*b^10 + 384*a^6*b^8 - 256*a^8*b^6 + 64*a^10*b^4) - 32*a^13*b + 64*a^3*b^11
- 288*a^5*b^9 + 512*a^7*b^7 - 448*a^9*b^5 + 192*a^11*b^3 - (b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(96*a^14
*b + 64*a^2*b^13 - 416*a^4*b^11 + 1120*a^6*b^9 - 1600*a^8*b^7 + 1280*a^10*b^5 - 544*a^12*b^3) - 32*a^3*b^12 +
160*a^5*b^10 - 320*a^7*b^8 + 320*a^9*b^6 - 160*a^11*b^4 + 32*a^13*b^2))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2))
)/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2))*1i)/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2))/(2*tan(x/2)*(64*b^12 - 256
*a^2*b^10 + 384*a^4*b^8 - 256*a^6*b^6 + 64*a^8*b^4) - 64*a*b^11 + 192*a^3*b^9 - 192*a^5*b^7 + 64*a^7*b^5 + (b^
3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(64*a^12*b + 64*b^13 - 320*a^2*b^11 + 736*a^4*b^9 - 992*a^6*b^7 + 800*
a^8*b^5 - 352*a^10*b^3) - 32*a*b^12 + 160*a^3*b^10 - 320*a^5*b^8 + 320*a^7*b^6 - 160*a^9*b^4 + 32*a^11*b^2 + (
b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(64*a^2*b^12 - 256*a^4*b^10 + 384*a^6*b^8 - 256*a^8*b^6 + 64*a^10*b^
4) - 32*a^13*b + 64*a^3*b^11 - 288*a^5*b^9 + 512*a^7*b^7 - 448*a^9*b^5 + 192*a^11*b^3 + (b^3*((a + b)^3*(a - b
)^3)^(1/2)*(tan(x/2)*(96*a^14*b + 64*a^2*b^13 - 416*a^4*b^11 + 1120*a^6*b^9 - 1600*a^8*b^7 + 1280*a^10*b^5 - 5
44*a^12*b^3) - 32*a^3*b^12 + 160*a^5*b^10 - 320*a^7*b^8 + 320*a^9*b^6 - 160*a^11*b^4 + 32*a^13*b^2))/(a*b^6 -
a^7 - 3*a^3*b^4 + 3*a^5*b^2)))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2) -
(b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(64*a^12*b + 64*b^13 - 320*a^2*b^11 + 736*a^4*b^9 - 992*a^6*b^7 +
800*a^8*b^5 - 352*a^10*b^3) - 32*a*b^12 + 160*a^3*b^10 - 320*a^5*b^8 + 320*a^7*b^6 - 160*a^9*b^4 + 32*a^11*b^2
- (b^3*((a + b)^3*(a - b)^3)^(1/2)*(tan(x/2)*(64*a^2*b^12 - 256*a^4*b^10 + 384*a^6*b^8 - 256*a^8*b^6 + 64*a^1
0*b^4) - 32*a^13*b + 64*a^3*b^11 - 288*a^5*b^9 + 512*a^7*b^7 - 448*a^9*b^5 + 192*a^11*b^3 - (b^3*((a + b)^3*(a
- b)^3)^(1/2)*(tan(x/2)*(96*a^14*b + 64*a^2*b^13 - 416*a^4*b^11 + 1120*a^6*b^9 - 1600*a^8*b^7 + 1280*a^10*b^5
- 544*a^12*b^3) - 32*a^3*b^12 + 160*a^5*b^10 - 320*a^7*b^8 + 320*a^9*b^6 - 160*a^11*b^4 + 32*a^13*b^2))/(a*b^
6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^
2)))*((a + b)^3*(a - b)^3)^(1/2)*2i)/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)