Integrand size = 15, antiderivative size = 55 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\frac {\arctan \left (\frac {b+a \tan (x)}{\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log (a+b \sin (2 x))}{4 b} \]
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\frac {\arctan \left (\frac {b+a \tan (x)}{\sqrt {a^2-b^2}}\right )}{2 \sqrt {a^2-b^2}}-\frac {\log (a+b \sin (2 x))}{4 b} \]
Time = 0.39 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.65, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4889, 2142, 27, 240, 1142, 27, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)^2}{a+b \sin (2 x)}dx\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \int \frac {\tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (a \tan ^2(x)+a+2 b \tan (x)\right )}d\tan (x)\) |
| \(\Big \downarrow \) 2142 |
| \(\displaystyle \frac {\int -\frac {2 a b \tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{4 b^2}+\frac {\int \frac {2 b \tan (x)}{\tan ^2(x)+1}d\tan (x)}{4 b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan (x)}{\tan ^2(x)+1}d\tan (x)}{2 b}-\frac {a \int \frac {\tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{2 b}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \int \frac {\tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{2 b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \left (\frac {\int \frac {2 (b+a \tan (x))}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{2 a}-\frac {b \int \frac {1}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{a}\right )}{2 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \left (\frac {\int \frac {b+a \tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{a}-\frac {b \int \frac {1}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{a}\right )}{2 b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \left (\frac {2 b \int \frac {1}{-(2 b+2 a \tan (x))^2-4 \left (a^2-b^2\right )}d(2 b+2 a \tan (x))}{a}+\frac {\int \frac {b+a \tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{a}\right )}{2 b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \left (\frac {\int \frac {b+a \tan (x)}{a \tan ^2(x)+2 b \tan (x)+a}d\tan (x)}{a}-\frac {b \arctan \left (\frac {2 a \tan (x)+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}\right )}{2 b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\log \left (\tan ^2(x)+1\right )}{4 b}-\frac {a \left (\frac {\log \left (a \tan ^2(x)+a+2 b \tan (x)\right )}{2 a}-\frac {b \arctan \left (\frac {2 a \tan (x)+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}\right )}{2 b}\) |
Log[1 + Tan[x]^2]/(4*b) - (a*(-((b*ArcTan[(2*b + 2*a*Tan[x])/(2*Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2])) + Log[a + 2*b*Tan[x] + a*Tan[x]^2]/(2*a)))/( 2*b)
3.9.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (f_.)*(x_)^2)), x_Sym bol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2] , q = c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2}, Simp[1/q Int[(A*c^2*d - a *c*C*d + A*b^2*f - a*b*B*f - a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d + A*b*f - a*B*f)*x)/(a + b*x + c*x^2), x], x] + Simp[1/q Int[(c*C*d^2 + b*B*d*f - A*c*d*f - a*C*d*f + a*A*f^2 - f*(B*c*d - b*C*d + A*b*f - a*B*f)*x)/(d + f*x ^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, f}, x] && PolyQ[Px, x, 2]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 2.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45
| method | result | size |
| default | \(-\frac {a \left (\frac {\ln \left (a \tan \left (x \right )^{2}+2 b \tan \left (x \right )+a \right )}{2 a}-\frac {b \arctan \left (\frac {2 a \tan \left (x \right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}}\right )}{2 b}+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{4 b}\) | \(80\) |
| risch | \(-\frac {i x}{2 b}+\frac {i x \,a^{2} b}{a^{2} b^{2}-b^{4}}-\frac {i x \,b^{3}}{a^{2} b^{2}-b^{4}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {-i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right ) a^{2}}{4 \left (a^{2}-b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{2 i x}-\frac {-i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right )}{4 a^{2}-4 b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {-i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right ) \sqrt {-a^{2} b^{2}+b^{4}}}{4 \left (a^{2}-b^{2}\right ) b}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right ) a^{2}}{4 \left (a^{2}-b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{2 i x}+\frac {i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right )}{4 a^{2}-4 b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i a b +\sqrt {-a^{2} b^{2}+b^{4}}}{b^{2}}\right ) \sqrt {-a^{2} b^{2}+b^{4}}}{4 \left (a^{2}-b^{2}\right ) b}\) | \(369\) |
-1/2*a/b*(1/2/a*ln(a*tan(x)^2+2*b*tan(x)+a)-b/a/(a^2-b^2)^(1/2)*arctan(1/2 *(2*a*tan(x)+2*b)/(a^2-b^2)^(1/2)))+1/4/b*ln(1+tan(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (47) = 94\).
Time = 0.28 (sec) , antiderivative size = 320, normalized size of antiderivative = 5.82 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} b \log \left (-\frac {4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{4} - 4 \, a b \cos \left (x\right ) \sin \left (x\right ) - 4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} - 2 \, b^{2} + 2 \, {\left (2 \, b \cos \left (x\right )^{2} + 2 \, {\left (2 \, a \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sin \left (x\right ) - b\right )} \sqrt {-a^{2} + b^{2}}}{4 \, b^{2} \cos \left (x\right )^{4} - 4 \, b^{2} \cos \left (x\right )^{2} - 4 \, a b \cos \left (x\right ) \sin \left (x\right ) - a^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \left (x\right )^{4} + 4 \, b^{2} \cos \left (x\right )^{2} + 4 \, a b \cos \left (x\right ) \sin \left (x\right ) + a^{2}\right )}{8 \, {\left (a^{2} b - b^{3}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {{\left (2 \, a \cos \left (x\right ) \sin \left (x\right ) + b\right )} \sqrt {a^{2} - b^{2}}}{2 \, {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \left (x\right )^{4} + 4 \, b^{2} \cos \left (x\right )^{2} + 4 \, a b \cos \left (x\right ) \sin \left (x\right ) + a^{2}\right )}{8 \, {\left (a^{2} b - b^{3}\right )}}\right ] \]
[-1/8*(sqrt(-a^2 + b^2)*b*log(-(4*(2*a^2 - b^2)*cos(x)^4 - 4*a*b*cos(x)*si n(x) - 4*(2*a^2 - b^2)*cos(x)^2 + a^2 - 2*b^2 + 2*(2*b*cos(x)^2 + 2*(2*a*c os(x)^3 - a*cos(x))*sin(x) - b)*sqrt(-a^2 + b^2))/(4*b^2*cos(x)^4 - 4*b^2* cos(x)^2 - 4*a*b*cos(x)*sin(x) - a^2)) + (a^2 - b^2)*log(-4*b^2*cos(x)^4 + 4*b^2*cos(x)^2 + 4*a*b*cos(x)*sin(x) + a^2))/(a^2*b - b^3), -1/8*(2*sqrt( a^2 - b^2)*b*arctan(-(2*a*cos(x)*sin(x) + b)*sqrt(a^2 - b^2)/(2*(a^2 - b^2 )*cos(x)^2 - a^2 + b^2)) + (a^2 - b^2)*log(-4*b^2*cos(x)^4 + 4*b^2*cos(x)^ 2 + 4*a*b*cos(x)*sin(x) + a^2))/(a^2*b - b^3)]
Time = 2.91 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.25 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=- \begin {cases} \frac {\log {\left (\frac {a}{b} + \sin {\left (2 x \right )} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {\sin {\left (2 x \right )}}{4 a} & \text {otherwise} \end {cases} + \begin {cases} \tilde {\infty } \log {\left (\tan {\left (x \right )} \right )} & \text {for}\: a = 0 \wedge b = 0 \\\frac {\log {\left (\tan {\left (x \right )} \right )}}{4 b} & \text {for}\: a = 0 \\\frac {1}{2 b \tan {\left (x \right )} - 2 b} & \text {for}\: a = - b \\- \frac {1}{2 b \tan {\left (x \right )} + 2 b} & \text {for}\: a = b \\\frac {\log {\left (\tan {\left (x \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{4 \sqrt {- a^{2} + b^{2}}} - \frac {\log {\left (\tan {\left (x \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{4 \sqrt {- a^{2} + b^{2}}} & \text {otherwise} \end {cases} \]
-Piecewise((log(a/b + sin(2*x))/(4*b), Ne(b, 0)), (sin(2*x)/(4*a), True)) + Piecewise((zoo*log(tan(x)), Eq(a, 0) & Eq(b, 0)), (log(tan(x))/(4*b), Eq (a, 0)), (1/(2*b*tan(x) - 2*b), Eq(a, -b)), (-1/(2*b*tan(x) + 2*b), Eq(a, b)), (log(tan(x) + b/a - sqrt(-a**2 + b**2)/a)/(4*sqrt(-a**2 + b**2)) - lo g(tan(x) + b/a + sqrt(-a**2 + b**2)/a)/(4*sqrt(-a**2 + b**2)), True))
Exception generated. \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\frac {\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )}{2 \, \sqrt {a^{2} - b^{2}}} - \frac {\log \left (a \tan \left (x\right )^{2} + 2 \, b \tan \left (x\right ) + a\right )}{4 \, b} + \frac {\log \left (\tan \left (x\right )^{2} + 1\right )}{4 \, b} \]
1/2*(pi*floor(x/pi + 1/2)*sgn(a) + arctan((a*tan(x) + b)/sqrt(a^2 - b^2))) /sqrt(a^2 - b^2) - 1/4*log(a*tan(x)^2 + 2*b*tan(x) + a)/b + 1/4*log(tan(x) ^2 + 1)/b
Time = 28.41 (sec) , antiderivative size = 1108, normalized size of antiderivative = 20.15 \[ \int \frac {\sin ^2(x)}{a+b \sin (2 x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (x\right )}^2+1\right )}{4\,b}+\frac {\mathrm {atan}\left (\frac {2\,\mathrm {tan}\left (x\right )\,{\left (a^2-b^2\right )}^{3/2}\,\left (\frac {\left (4\,a^2\,b-2\,b^3\right )\,\left (2\,a\,b-\frac {\frac {8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{8\,\sqrt {a^2-b^2}\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (4\,a^3-16\,a\,b^2+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}-\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{32\,\left (a^2-b^2\right )\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^3\,{\left (4\,a^2-3\,b^2\right )}^2}-\frac {\left (4\,a^4-5\,a^2\,b^2+2\,b^4\right )\,\left (\frac {4\,a^3-16\,a\,b^2+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}-\frac {96\,a\,b^4-64\,a^3\,b^2}{64\,{\left (a^2-b^2\right )}^{3/2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (\frac {8\,a\,b^3+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}}{4\,\sqrt {a^2-b^2}}+\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{8\,\sqrt {a^2-b^2}\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^3\,\sqrt {a^2-b^2}\,{\left (4\,a^2-3\,b^2\right )}^2}\right )}{a}+\frac {2\,\left (a^2-b^2\right )\,\left (\frac {6\,a^2\,b-\frac {8\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{{\left (16\,b^4-16\,a^2\,b^2\right )}^2}}{4\,\sqrt {a^2-b^2}}+\frac {a^2\,b^3}{2\,{\left (a^2-b^2\right )}^{3/2}}-\frac {4\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{\sqrt {a^2-b^2}\,{\left (16\,b^4-16\,a^2\,b^2\right )}^2}\right )\,\left (4\,a^4-5\,a^2\,b^2+2\,b^4\right )}{a^4\,{\left (4\,a^2-3\,b^2\right )}^2}-\frac {2\,\left (4\,a^2\,b-2\,b^3\right )\,{\left (a^2-b^2\right )}^{3/2}\,\left (\frac {\left (8\,a^2\,b-8\,b^3\right )\,\left (6\,a^2\,b-\frac {8\,a^2\,b^3\,{\left (8\,a^2\,b-8\,b^3\right )}^2}{{\left (16\,b^4-16\,a^2\,b^2\right )}^2}\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )}-a^2+\frac {3\,a^2\,b^3\,\left (8\,a^2\,b-8\,b^3\right )}{\left (a^2-b^2\right )\,\left (16\,b^4-16\,a^2\,b^2\right )}\right )}{a^4\,{\left (4\,a^2-3\,b^2\right )}^2}\right )}{2\,\sqrt {a^2-b^2}}+\frac {\ln \left (a\,{\mathrm {tan}\left (x\right )}^2+2\,b\,\mathrm {tan}\left (x\right )+a\right )\,\left (8\,a^2\,b-8\,b^3\right )}{2\,\left (16\,b^4-16\,a^2\,b^2\right )} \]
log(tan(x)^2 + 1)/(4*b) + atan((2*tan(x)*(a^2 - b^2)^(3/2)*(((4*a^2*b - 2* b^3)*(2*a*b - ((8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*( 16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2)) + ((8*a^2*b - 8*b^3)*(96*a*b^ 4 - 64*a^3*b^2))/(8*(a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^ 2)^(1/2)) + ((8*a^2*b - 8*b^3)*(4*a^3 - 16*a*b^2 + ((8*a^2*b - 8*b^3)*(8*a *b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2*b^2 ))))/(2*(16*b^4 - 16*a^2*b^2))))/(2*(16*b^4 - 16*a^2*b^2)) - ((8*a^2*b - 8 *b^3)*(96*a*b^4 - 64*a^3*b^2))/(32*(a^2 - b^2)*(16*b^4 - 16*a^2*b^2))))/(a ^3*(4*a^2 - 3*b^2)^2) - ((4*a^4 + 2*b^4 - 5*a^2*b^2)*((4*a^3 - 16*a*b^2 + ((8*a^2*b - 8*b^3)*(8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/ (2*(16*b^4 - 16*a^2*b^2))))/(2*(16*b^4 - 16*a^2*b^2)))/(4*(a^2 - b^2)^(1/2 )) - (96*a*b^4 - 64*a^3*b^2)/(64*(a^2 - b^2)^(3/2)) + ((8*a^2*b - 8*b^3)*( (8*a*b^3 + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(16*b^4 - 16*a^2 *b^2)))/(4*(a^2 - b^2)^(1/2)) + ((8*a^2*b - 8*b^3)*(96*a*b^4 - 64*a^3*b^2) )/(8*(a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2))))/(2*(16*b^4 - 16*a^2*b^2))) )/(a^3*(a^2 - b^2)^(1/2)*(4*a^2 - 3*b^2)^2)))/a + (2*(a^2 - b^2)*((6*a^2*b - (8*a^2*b^3*(8*a^2*b - 8*b^3)^2)/(16*b^4 - 16*a^2*b^2)^2)/(4*(a^2 - b^2) ^(1/2)) + (a^2*b^3)/(2*(a^2 - b^2)^(3/2)) - (4*a^2*b^3*(8*a^2*b - 8*b^3)^2 )/((a^2 - b^2)^(1/2)*(16*b^4 - 16*a^2*b^2)^2))*(4*a^4 + 2*b^4 - 5*a^2*b^2) )/(a^4*(4*a^2 - 3*b^2)^2) - (2*(4*a^2*b - 2*b^3)*(a^2 - b^2)^(3/2)*(((8...