3.1.20 \(\int \frac {\csc ^4(x)}{(a+a \sin (x))^2} \, dx\) [20]

Optimal. Leaf size=65 \[ \frac {5 \tanh ^{-1}(\cos (x))}{a^2}-\frac {4 \cot (x)}{a^2}-\frac {\cot ^3(x)}{3 a^2}+\frac {\cot (x) \csc (x)}{a^2}-\frac {\cos (x)}{3 a^2 (1+\sin (x))^2}-\frac {13 \cos (x)}{3 a^2 (1+\sin (x))} \]

[Out]

5*arctanh(cos(x))/a^2-4*cot(x)/a^2-1/3*cot(x)^3/a^2+cot(x)*csc(x)/a^2-1/3*cos(x)/a^2/(1+sin(x))^2-13/3*cos(x)/
a^2/(1+sin(x))

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Rubi [A]
time = 0.11, antiderivative size = 71, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2845, 3057, 2827, 3852, 3853, 3855} \begin {gather*} -\frac {4 \cot ^3(x)}{a^2}-\frac {12 \cot (x)}{a^2}+\frac {5 \tanh ^{-1}(\cos (x))}{a^2}+\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (\sin (x)+1)}+\frac {\cot (x) \csc ^2(x)}{3 (a \sin (x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + a*Sin[x])^2,x]

[Out]

(5*ArcTanh[Cos[x]])/a^2 - (12*Cot[x])/a^2 - (4*Cot[x]^3)/a^2 + (5*Cot[x]*Csc[x])/a^2 + (10*Cot[x]*Csc[x]^2)/(3
*a^2*(1 + Sin[x])) + (Cot[x]*Csc[x]^2)/(3*(a + a*Sin[x])^2)

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 3057

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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*} \int \frac {\csc ^4(x)}{(a+a \sin (x))^2} \, dx &=\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}+\frac {\int \frac {\csc ^4(x) (6 a-4 a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}+\frac {\int \csc ^4(x) \left (36 a^2-30 a^2 \sin (x)\right ) \, dx}{3 a^4}\\ &=\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}-\frac {10 \int \csc ^3(x) \, dx}{a^2}+\frac {12 \int \csc ^4(x) \, dx}{a^2}\\ &=\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}-\frac {5 \int \csc (x) \, dx}{a^2}-\frac {12 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (x)\right )}{a^2}\\ &=\frac {5 \tanh ^{-1}(\cos (x))}{a^2}-\frac {12 \cot (x)}{a^2}-\frac {4 \cot ^3(x)}{a^2}+\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(65)=130\).
time = 2.50, size = 238, normalized size = 3.66 \begin {gather*} \frac {\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \left (-\cos \left (\frac {x}{2}\right ) \left (1+\cot \left (\frac {x}{2}\right )\right )^3+16 \sin \left (\frac {x}{2}\right )+6 \left (1+\cot \left (\frac {x}{2}\right )\right )^3 \sin \left (\frac {x}{2}\right )-8 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+208 \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2-44 \cot \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3+120 \log \left (\cos \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3-120 \log \left (\sin \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3+44 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3 \tan \left (\frac {x}{2}\right )-6 \cos \left (\frac {x}{2}\right ) \left (1+\tan \left (\frac {x}{2}\right )\right )^3+\sin \left (\frac {x}{2}\right ) \left (1+\tan \left (\frac {x}{2}\right )\right )^3\right )}{24 a^2 (1+\sin (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + a*Sin[x])^2,x]

[Out]

((Cos[x/2] + Sin[x/2])*(-(Cos[x/2]*(1 + Cot[x/2])^3) + 16*Sin[x/2] + 6*(1 + Cot[x/2])^3*Sin[x/2] - 8*(Cos[x/2]
+ Sin[x/2]) + 208*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 - 44*Cot[x/2]*(Cos[x/2] + Sin[x/2])^3 + 120*Log[Cos[x/2]]*
(Cos[x/2] + Sin[x/2])^3 - 120*Log[Sin[x/2]]*(Cos[x/2] + Sin[x/2])^3 + 44*(Cos[x/2] + Sin[x/2])^3*Tan[x/2] - 6*
Cos[x/2]*(1 + Tan[x/2])^3 + Sin[x/2]*(1 + Tan[x/2])^3))/(24*a^2*(1 + Sin[x])^2)

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Maple [A]
time = 0.18, size = 90, normalized size = 1.38

method	result	size
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}-2 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+15 \tan \left (\frac {x}{2}\right )-\frac {1}{3 \tan \left (\frac {x}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {x}{2}\right )^{2}}-\frac {15}{\tan \left (\frac {x}{2}\right )}-40 \ln \left (\tan \left (\frac {x}{2}\right )\right )-\frac {32}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {16}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {80}{\tan \left (\frac {x}{2}\right )+1}}{8 a^{2}}\)	\(90\)
risch	\(-\frac {2 \left (-85 \,{\mathrm e}^{6 i x}+45 i {\mathrm e}^{7 i x}+153 \,{\mathrm e}^{4 i x}-135 i {\mathrm e}^{5 i x}+15 \,{\mathrm e}^{8 i x}-99 \,{\mathrm e}^{2 i x}+155 i {\mathrm e}^{3 i x}+24-57 i {\mathrm e}^{i x}\right )}{3 \left ({\mathrm e}^{2 i x}-1\right )^{3} \left ({\mathrm e}^{i x}+i\right )^{3} a^{2}}+\frac {5 \ln \left ({\mathrm e}^{i x}+1\right )}{a^{2}}-\frac {5 \ln \left ({\mathrm e}^{i x}-1\right )}{a^{2}}\)	\(114\)
norman	\(\frac {-\frac {1}{24 a}+\frac {\tan \left (\frac {x}{2}\right )}{8 a}-\frac {5 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4 a}+\frac {5 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4 a}-\frac {\tan ^{8}\left (\frac {x}{2}\right )}{8 a}+\frac {\tan ^{9}\left (\frac {x}{2}\right )}{24 a}-\frac {115 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{6 a}-\frac {145 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4 a}-\frac {85 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a}}{\tan \left (\frac {x}{2}\right )^{3} a \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {5 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^4/(a+a*sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

1/8/a^2*(1/3*tan(1/2*x)^3-2*tan(1/2*x)^2+15*tan(1/2*x)-1/3/tan(1/2*x)^3+2/tan(1/2*x)^2-15/tan(1/2*x)-40*ln(tan
(1/2*x))-32/3/(tan(1/2*x)+1)^3+16/(tan(1/2*x)+1)^2-80/(tan(1/2*x)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (59) = 118\).
time = 0.39, size = 178, normalized size = 2.74 \begin {gather*} \frac {\frac {3 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {30 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {342 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {561 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {285 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} - 1}{24 \, {\left (\frac {a^{2} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {3 \, a^{2} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {a^{2} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} + \frac {\frac {45 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {6 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {\sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}}{24 \, a^{2}} - \frac {5 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(a+a*sin(x))^2,x, algorithm="maxima")

[Out]

1/24*(3*sin(x)/(cos(x) + 1) - 30*sin(x)^2/(cos(x) + 1)^2 - 342*sin(x)^3/(cos(x) + 1)^3 - 561*sin(x)^4/(cos(x)
+ 1)^4 - 285*sin(x)^5/(cos(x) + 1)^5 - 1)/(a^2*sin(x)^3/(cos(x) + 1)^3 + 3*a^2*sin(x)^4/(cos(x) + 1)^4 + 3*a^2
*sin(x)^5/(cos(x) + 1)^5 + a^2*sin(x)^6/(cos(x) + 1)^6) + 1/24*(45*sin(x)/(cos(x) + 1) - 6*sin(x)^2/(cos(x) +
1)^2 + sin(x)^3/(cos(x) + 1)^3)/a^2 - 5*log(sin(x)/(cos(x) + 1))/a^2

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (59) = 118\).
time = 0.40, size = 266, normalized size = 4.09 \begin {gather*} -\frac {48 \, \cos \left (x\right )^{5} - 18 \, \cos \left (x\right )^{4} - 108 \, \cos \left (x\right )^{3} + 22 \, \cos \left (x\right )^{2} - 15 \, {\left (\cos \left (x\right )^{5} + 2 \, \cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{3} - 4 \, \cos \left (x\right )^{2} + {\left (\cos \left (x\right )^{4} - \cos \left (x\right )^{3} - 3 \, \cos \left (x\right )^{2} + \cos \left (x\right ) + 2\right )} \sin \left (x\right ) + \cos \left (x\right ) + 2\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (x\right )^{5} + 2 \, \cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{3} - 4 \, \cos \left (x\right )^{2} + {\left (\cos \left (x\right )^{4} - \cos \left (x\right )^{3} - 3 \, \cos \left (x\right )^{2} + \cos \left (x\right ) + 2\right )} \sin \left (x\right ) + \cos \left (x\right ) + 2\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 2 \, {\left (24 \, \cos \left (x\right )^{4} + 33 \, \cos \left (x\right )^{3} - 21 \, \cos \left (x\right )^{2} - 32 \, \cos \left (x\right ) - 1\right )} \sin \left (x\right ) + 62 \, \cos \left (x\right ) - 2}{6 \, {\left (a^{2} \cos \left (x\right )^{5} + 2 \, a^{2} \cos \left (x\right )^{4} - 2 \, a^{2} \cos \left (x\right )^{3} - 4 \, a^{2} \cos \left (x\right )^{2} + a^{2} \cos \left (x\right ) + 2 \, a^{2} + {\left (a^{2} \cos \left (x\right )^{4} - a^{2} \cos \left (x\right )^{3} - 3 \, a^{2} \cos \left (x\right )^{2} + a^{2} \cos \left (x\right ) + 2 \, a^{2}\right )} \sin \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(a+a*sin(x))^2,x, algorithm="fricas")

[Out]

-1/6*(48*cos(x)^5 - 18*cos(x)^4 - 108*cos(x)^3 + 22*cos(x)^2 - 15*(cos(x)^5 + 2*cos(x)^4 - 2*cos(x)^3 - 4*cos(
x)^2 + (cos(x)^4 - cos(x)^3 - 3*cos(x)^2 + cos(x) + 2)*sin(x) + cos(x) + 2)*log(1/2*cos(x) + 1/2) + 15*(cos(x)
^5 + 2*cos(x)^4 - 2*cos(x)^3 - 4*cos(x)^2 + (cos(x)^4 - cos(x)^3 - 3*cos(x)^2 + cos(x) + 2)*sin(x) + cos(x) +
2)*log(-1/2*cos(x) + 1/2) - 2*(24*cos(x)^4 + 33*cos(x)^3 - 21*cos(x)^2 - 32*cos(x) - 1)*sin(x) + 62*cos(x) - 2
)/(a^2*cos(x)^5 + 2*a^2*cos(x)^4 - 2*a^2*cos(x)^3 - 4*a^2*cos(x)^2 + a^2*cos(x) + 2*a^2 + (a^2*cos(x)^4 - a^2*
cos(x)^3 - 3*a^2*cos(x)^2 + a^2*cos(x) + 2*a^2)*sin(x))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\csc ^{4}{\left (x \right )}}{\sin ^{2}{\left (x \right )} + 2 \sin {\left (x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**4/(a+a*sin(x))**2,x)

[Out]

Integral(csc(x)**4/(sin(x)**2 + 2*sin(x) + 1), x)/a**2

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Giac [A]
time = 0.58, size = 114, normalized size = 1.75 \begin {gather*} -\frac {5 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac {110 \, \tan \left (\frac {1}{2} \, x\right )^{6} + 45 \, \tan \left (\frac {1}{2} \, x\right )^{5} - 231 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 232 \, \tan \left (\frac {1}{2} \, x\right )^{3} - 30 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, x\right ) - 1}{24 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + \tan \left (\frac {1}{2} \, x\right )\right )}^{3} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 45 \, a^{4} \tan \left (\frac {1}{2} \, x\right )}{24 \, a^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(a+a*sin(x))^2,x, algorithm="giac")

[Out]

-5*log(abs(tan(1/2*x)))/a^2 + 1/24*(110*tan(1/2*x)^6 + 45*tan(1/2*x)^5 - 231*tan(1/2*x)^4 - 232*tan(1/2*x)^3 -
30*tan(1/2*x)^2 + 3*tan(1/2*x) - 1)/((tan(1/2*x)^2 + tan(1/2*x))^3*a^2) + 1/24*(a^4*tan(1/2*x)^3 - 6*a^4*tan(
1/2*x)^2 + 45*a^4*tan(1/2*x))/a^6

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Mupad [B]
time = 6.40, size = 101, normalized size = 1.55 \begin {gather*} \frac {15\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4\,a^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24\,a^2}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2}-\frac {\frac {95\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{8}+\frac {187\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{8}+\frac {57\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{4}+\frac {5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{8}+\frac {1}{24}}{a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^4*(a + a*sin(x))^2),x)

[Out]

(15*tan(x/2))/(8*a^2) - tan(x/2)^2/(4*a^2) + tan(x/2)^3/(24*a^2) - (5*log(tan(x/2)))/a^2 - ((5*tan(x/2)^2)/4 -
tan(x/2)/8 + (57*tan(x/2)^3)/4 + (187*tan(x/2)^4)/8 + (95*tan(x/2)^5)/8 + 1/24)/(a^2*tan(x/2)^3*(tan(x/2) + 1
)^3)

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