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

Optimal. Leaf size=159 \[ \frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}-\frac{b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac{x}{a} \]

[Out]

x/a + (2*b^5*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) - (b^3*Sec[x])/(a^2 - b^2)^2 + (
b*Sec[x])/(a^2 - b^2) - (b*Sec[x]^3)/(3*(a^2 - b^2)) + (a*b^2*Tan[x])/(a^2 - b^2)^2 - (a*Tan[x])/(a^2 - b^2) +
(a*Tan[x]^3)/(3*(a^2 - b^2))

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Rubi [A]  time = 0.323515, antiderivative size = 205, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {3898, 2902, 2606, 3473, 8, 2735, 2660, 618, 206} \[ \frac{b^4 x}{a \left (a^2-b^2\right )^2}-\frac{a b^2 x}{\left (a^2-b^2\right )^2}+\frac{a x}{a^2-b^2}+\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}-\frac{b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^4/(a + b*Csc[x]),x]

[Out]

-((a*b^2*x)/(a^2 - b^2)^2) + (b^4*x)/(a*(a^2 - b^2)^2) + (a*x)/(a^2 - b^2) + (2*b^5*ArcTanh[(a + b*Tan[x/2])/S
qrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)) - (b^3*Sec[x])/(a^2 - b^2)^2 + (b*Sec[x])/(a^2 - b^2) - (b*Sec[x]^3)/(3
*(a^2 - b^2)) + (a*b^2*Tan[x])/(a^2 - b^2)^2 - (a*Tan[x])/(a^2 - b^2) + (a*Tan[x]^3)/(3*(a^2 - b^2))

Rule 3898

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])

Rule 2902

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 2606

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 3473

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 8

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

Rule 2735

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 2660

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 618

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 206

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

Rubi steps

\begin{align*} \int \frac{\tan ^4(x)}{a+b \csc (x)} \, dx &=\int \frac{\sin (x) \tan ^4(x)}{b+a \sin (x)} \, dx\\ &=\frac{a \int \tan ^4(x) \, dx}{a^2-b^2}-\frac{b \int \sec (x) \tan ^3(x) \, dx}{a^2-b^2}+\frac{b^2 \int \frac{\sin (x) \tan ^2(x)}{b+a \sin (x)} \, dx}{a^2-b^2}\\ &=\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac{\left (a b^2\right ) \int \tan ^2(x) \, dx}{\left (a^2-b^2\right )^2}-\frac{b^3 \int \sec (x) \tan (x) \, dx}{\left (a^2-b^2\right )^2}+\frac{b^4 \int \frac{\sin (x)}{b+a \sin (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{a \int \tan ^2(x) \, dx}{a^2-b^2}-\frac{b \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (x)\right )}{a^2-b^2}\\ &=\frac{b^4 x}{a \left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}+\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}-\frac{\left (a b^2\right ) \int 1 \, dx}{\left (a^2-b^2\right )^2}-\frac{b^3 \operatorname{Subst}(\int 1 \, dx,x,\sec (x))}{\left (a^2-b^2\right )^2}-\frac{b^5 \int \frac{1}{b+a \sin (x)} \, dx}{a \left (a^2-b^2\right )^2}+\frac{a \int 1 \, dx}{a^2-b^2}\\ &=-\frac{a b^2 x}{\left (a^2-b^2\right )^2}+\frac{b^4 x}{a \left (a^2-b^2\right )^2}+\frac{a x}{a^2-b^2}-\frac{b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}+\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}-\frac{\left (2 b^5\right ) \operatorname{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 )^2}\\ &=-\frac{a b^2 x}{\left (a^2-b^2\right )^2}+\frac{b^4 x}{a \left (a^2-b^2\right )^2}+\frac{a x}{a^2-b^2}-\frac{b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}+\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}+\frac{\left (4 b^5\right ) \operatorname{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 )^2}\\ &=-\frac{a b^2 x}{\left (a^2-b^2\right )^2}+\frac{b^4 x}{a \left (a^2-b^2\right )^2}+\frac{a x}{a^2-b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac{b^3 \sec (x)}{\left (a^2-b^2\right )^2}+\frac{b \sec (x)}{a^2-b^2}-\frac{b \sec ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a b^2 \tan (x)}{\left (a^2-b^2\right )^2}-\frac{a \tan (x)}{a^2-b^2}+\frac{a \tan ^3(x)}{3 \left (a^2-b^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.03132, size = 194, normalized size = 1.22 \[ \frac{\csc (x) (a \sin (x)+b) \left (\frac{\sec ^3(x) \left (3 a^2 b^2 \sin (x)+7 a^2 b^2 \sin (3 x)-6 a^2 b^2 x \cos (3 x)+6 a b \left (a^2-2 b^2\right ) \cos (2 x)+9 x \left (a^2-b^2\right )^2 \cos (x)+2 a^3 b-4 a^4 \sin (3 x)+3 a^4 x \cos (3 x)-8 a b^3+3 b^4 x \cos (3 x)\right )}{(a-b)^2 (a+b)^2}-\frac{24 b^5 \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}\right )}{12 a (a+b \csc (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^4/(a + b*Csc[x]),x]

[Out]

(Csc[x]*(b + a*Sin[x])*((-24*b^5*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (Sec[x]^3*(2*
a^3*b - 8*a*b^3 + 9*(a^2 - b^2)^2*x*Cos[x] + 6*a*b*(a^2 - 2*b^2)*Cos[2*x] + 3*a^4*x*Cos[3*x] - 6*a^2*b^2*x*Cos
[3*x] + 3*b^4*x*Cos[3*x] + 3*a^2*b^2*Sin[x] - 4*a^4*Sin[3*x] + 7*a^2*b^2*Sin[3*x]))/((a - b)^2*(a + b)^2)))/(1
2*a*(a + b*Csc[x]))

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Maple [A]  time = 0.077, size = 210, normalized size = 1.3 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{a}}-{\frac{64}{192\,a-192\,b} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+32\,{\frac{1}{ \left ( 64\,a-64\,b \right ) \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}+{\frac{a}{ \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{3\,b}{2\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{64}{192\,a+192\,b} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-32\,{\frac{1}{ \left ( 64\,a+64\,b \right ) \left ( \tan \left ( x/2 \right ) -1 \right ) ^{2}}}+{\frac{a}{ \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,b}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{b}^{5}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}a\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^4/(a+b*csc(x)),x)

[Out]

2/a*arctan(tan(1/2*x))-64/3/(tan(1/2*x)+1)^3/(64*a-64*b)+32/(64*a-64*b)/(tan(1/2*x)+1)^2+1/(a-b)^2/(tan(1/2*x)
+1)*a-3/2/(a-b)^2/(tan(1/2*x)+1)*b-64/3/(tan(1/2*x)-1)^3/(64*a+64*b)-32/(64*a+64*b)/(tan(1/2*x)-1)^2+1/(a+b)^2
/(tan(1/2*x)-1)*a+3/2/(a+b)^2/(tan(1/2*x)-1)*b-2/(a-b)^2/(a+b)^2*b^5/a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/
2*x)+2*a)/(-a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.608527, size = 1081, normalized size = 6.8 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - b^{2}} b^{5} \cos \left (x\right )^{3} \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^{5} b + 4 \, a^{3} b^{3} - 2 \, a b^{5} + 6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} x \cos \left (x\right )^{3} + 6 \,{\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} -{\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{3}}, \frac{3 \, \sqrt{-a^{2} + b^{2}} b^{5} \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 )^{3} - a^{5} b + 2 \, a^{3} b^{3} - a b^{5} + 3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} x \cos \left (x\right )^{3} + 3 \,{\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (x\right )^{2} +{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} -{\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (x\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(a^2 - b^2)*b^5*cos(x)^3*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^5*b + 4*a^3*b^3 - 2*a*b^5 + 6
*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cos(x)^3 + 6*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(x)^2 + 2*(a^6 - 2*a^4*b^
2 + a^2*b^4 - (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(x)
^3), 1/3*(3*sqrt(-a^2 + b^2)*b^5*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x)))*cos(x)^3 - a^5*
b + 2*a^3*b^3 - a*b^5 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cos(x)^3 + 3*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos
(x)^2 + (a^6 - 2*a^4*b^2 + a^2*b^4 - (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(x)^2)*sin(x))/((a^7 - 3*a^5*b^2 + 3*
a^3*b^4 - a*b^6)*cos(x)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (x \right )}}{a + b \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**4/(a+b*csc(x)),x)

[Out]

Integral(tan(x)**4/(a + b*csc(x)), x)

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Giac [A]  time = 1.37961, size = 292, normalized size = 1.84 \begin{align*} -\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^{5}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{x}{a} + \frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{5} + 3 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{4} - 10 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 16 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} - 12 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + 3 \, a^{3} \tan \left (\frac{1}{2} \, x\right ) - 6 \, a b^{2} \tan \left (\frac{1}{2} \, x\right ) - 2 \, a^{2} b + 5 \, b^{3}\right )}}{3 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^4/(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^5/((a^5 - 2*a^3*b^2 + a*b
^4)*sqrt(-a^2 + b^2)) + x/a + 2/3*(3*a^3*tan(1/2*x)^5 - 6*a*b^2*tan(1/2*x)^5 + 3*b^3*tan(1/2*x)^4 - 10*a^3*tan
(1/2*x)^3 + 16*a*b^2*tan(1/2*x)^3 + 6*a^2*b*tan(1/2*x)^2 - 12*b^3*tan(1/2*x)^2 + 3*a^3*tan(1/2*x) - 6*a*b^2*ta
n(1/2*x) - 2*a^2*b + 5*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*x)^2 - 1)^3)