3.2.77 \(\int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx\) [177]

Optimal. Leaf size=82 \[ \frac {\left (2 a^2+b^2\right ) x}{2 b^3}-\frac {2 a^3 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2872, 3102, 2814, 2739, 632, 210} \begin {gather*} \frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {2 a^3 \text {ArcTan}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cos (x)}{b^2}-\frac {\sin (x) \cos (x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + b*Sin[x]),x]

[Out]

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

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 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 2872

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 - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/
(d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a
*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n
- 2))*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] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m]
|| (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 78, normalized size = 0.95 \begin {gather*} \frac {4 a^2 x+2 b^2 x-\frac {8 a^3 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+4 a b \cos (x)-b^2 \sin (2 x)}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(a + b*Sin[x]),x]

[Out]

(4*a^2*x + 2*b^2*x - (8*a^3*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 4*a*b*Cos[x] - b^2*Sin
[2*x])/(4*b^3)

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Maple [A]
time = 0.15, size = 112, normalized size = 1.37

method	result	size
default	\(-\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}+\frac {\frac {2 \left (\frac {b^{2} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+a b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2}+a b \right )}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{3}}\)	\(112\)
risch	\(\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}+\frac {a \,{\mathrm e}^{i x}}{2 b^{2}}+\frac {a \,{\mathrm e}^{-i x}}{2 b^{2}}+\frac {i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}-\frac {i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}-\frac {\sin \left (2 x \right )}{4 b}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(sin(x)^3/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.37, size = 291, normalized size = 3.55 \begin {gather*} \left [-\frac {\sqrt {-a^{2} + b^{2}} a^{3} \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 ) + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} x - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )}}, \frac {2 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.45, size = 112, normalized size = 1.37 \begin {gather*} -\frac {2 \, {\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 )} a^{3}}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{3}} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right ) + 2 \, a}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 7.08, size = 1004, normalized size = 12.24 \begin {gather*} \frac {\frac {2\,a}{b^2}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{b}+\frac {2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {\mathrm {atan}\left (\frac {40\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b^2+40\,a^3+\frac {48\,a^5}{b^2}}+\frac {48\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{48\,a^5+40\,a^3\,b^2+8\,a\,b^4}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b+\frac {40\,a^3}{b}+\frac {48\,a^5}{b^3}}\right )\,\left (a^2\,2{}\mathrm {i}+b^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^3}+\frac {a^3\,\mathrm {atan}\left (\frac {\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}+\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}+\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}}+\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}-\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}-\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}}}{\frac {16\,\left (2\,a^7+a^5\,b^2\right )}{b^5}+\frac {16\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^8+8\,a^6\,b^2+2\,a^4\,b^4\right )}{b^6}+\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}+\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}+\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}-\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}-\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}-\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}}\right )\,2{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + b*sin(x)),x)

[Out]

((2*a)/b^2 - tan(x/2)/b + tan(x/2)^3/b + (2*a*tan(x/2)^2)/b^2)/(2*tan(x/2)^2 + tan(x/2)^4 + 1) - (atan((40*a^3
*tan(x/2))/(8*a*b^2 + 40*a^3 + (48*a^5)/b^2) + (48*a^5*tan(x/2))/(8*a*b^4 + 48*a^5 + 40*a^3*b^2) + (8*a*b*tan(
x/2))/(8*a*b + (40*a^3)/b + (48*a^5)/b^3))*(a^2*2i + b^2*1i)*1i)/b^3 + (a^3*atan(((a^3*((8*(a^2*b^6 + 4*a^4*b^
4 + 4*a^6*b^2))/b^5 + (8*tan(x/2)*(2*a*b^8 + 7*a^3*b^6 + 4*a^5*b^4 - 8*a^7*b^2))/b^6 + (a^3*(64*a^4*tan(x/2) +
(8*(2*a*b^8 + 2*a^3*b^6))/b^5 + (a^3*(32*a^2*b^3 + (8*tan(x/2)*(12*a*b^10 - 8*a^3*b^8))/b^6))/(b^3*(b^2 - a^2
)^(1/2))))/(b^3*(b^2 - a^2)^(1/2)))*1i)/(b^3*(b^2 - a^2)^(1/2)) + (a^3*((8*(a^2*b^6 + 4*a^4*b^4 + 4*a^6*b^2))/
b^5 + (8*tan(x/2)*(2*a*b^8 + 7*a^3*b^6 + 4*a^5*b^4 - 8*a^7*b^2))/b^6 - (a^3*(64*a^4*tan(x/2) + (8*(2*a*b^8 + 2
*a^3*b^6))/b^5 - (a^3*(32*a^2*b^3 + (8*tan(x/2)*(12*a*b^10 - 8*a^3*b^8))/b^6))/(b^3*(b^2 - a^2)^(1/2))))/(b^3*
(b^2 - a^2)^(1/2)))*1i)/(b^3*(b^2 - a^2)^(1/2)))/((16*(2*a^7 + a^5*b^2))/b^5 + (16*tan(x/2)*(8*a^8 + 2*a^4*b^4
+ 8*a^6*b^2))/b^6 + (a^3*((8*(a^2*b^6 + 4*a^4*b^4 + 4*a^6*b^2))/b^5 + (8*tan(x/2)*(2*a*b^8 + 7*a^3*b^6 + 4*a^
5*b^4 - 8*a^7*b^2))/b^6 + (a^3*(64*a^4*tan(x/2) + (8*(2*a*b^8 + 2*a^3*b^6))/b^5 + (a^3*(32*a^2*b^3 + (8*tan(x/
2)*(12*a*b^10 - 8*a^3*b^8))/b^6))/(b^3*(b^2 - a^2)^(1/2))))/(b^3*(b^2 - a^2)^(1/2))))/(b^3*(b^2 - a^2)^(1/2))
- (a^3*((8*(a^2*b^6 + 4*a^4*b^4 + 4*a^6*b^2))/b^5 + (8*tan(x/2)*(2*a*b^8 + 7*a^3*b^6 + 4*a^5*b^4 - 8*a^7*b^2))
/b^6 - (a^3*(64*a^4*tan(x/2) + (8*(2*a*b^8 + 2*a^3*b^6))/b^5 - (a^3*(32*a^2*b^3 + (8*tan(x/2)*(12*a*b^10 - 8*a
^3*b^8))/b^6))/(b^3*(b^2 - a^2)^(1/2))))/(b^3*(b^2 - a^2)^(1/2))))/(b^3*(b^2 - a^2)^(1/2))))*2i)/(b^3*(b^2 - a
^2)^(1/2))

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