\(\int \frac {1}{(a+b \csc (c+d x))^2} \, dx\) [49]

Optimal result

Integrand size = 12, antiderivative size = 108 \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\frac {x}{a^2}+\frac {2 b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}-\frac {b^2 \cot (c+d x)}{a \left (a^2-b^2\right ) d (a+b \csc (c+d x))} \]

[Out]

x/a^2+2*b*(2*a^2-b^2)*arctanh((a+b*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(3/2)/d-b^2*cot(d*x+c)/a
/(a^2-b^2)/d/(a+b*csc(d*x+c))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3870, 4004, 3916, 2739, 632, 212} \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\frac {2 b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac {b^2 \cot (c+d x)}{a d \left (a^2-b^2\right ) (a+b \csc (c+d x))}+\frac {x}{a^2} \]

[In]

Int[(a + b*Csc[c + d*x])^(-2),x]

[Out]

x/a^2 + (2*b*(2*a^2 - b^2)*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)*d) - (b^2
*Cot[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Csc[c + d*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 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 3870

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[c + d*x]*((a + b*Csc[c + d*x])^(n +
1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

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

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

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\frac {\csc (c+d x) \left (\frac {a b^2 \cot (c+d x)}{(-a+b) (a+b)}+(c+d x) (a+b \csc (c+d x))-\frac {2 b \left (-2 a^2+b^2\right ) \arctan \left (\frac {a+b \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right ) (a+b \csc (c+d x))}{\left (-a^2+b^2\right )^{3/2}}\right ) (b+a \sin (c+d x))}{a^2 d (a+b \csc (c+d x))^2} \]

[In]

Integrate[(a + b*Csc[c + d*x])^(-2),x]

[Out]

(Csc[c + d*x]*((a*b^2*Cot[c + d*x])/((-a + b)*(a + b)) + (c + d*x)*(a + b*Csc[c + d*x]) - (2*b*(-2*a^2 + b^2)*
ArcTan[(a + b*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*(a + b*Csc[c + d*x]))/(-a^2 + b^2)^(3/2))*(b + a*Sin[c + d*x
]))/(a^2*d*(a + b*Csc[c + d*x])^2)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.56

method	result	size
derivativedivides	\(\frac {-\frac {2 b \left (\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {a b}{2 a^{2}-2 b^{2}}}{\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}{2}+a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {b}{2}}+\frac {2 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (2 a^{2}-2 b^{2}\right ) \sqrt {-a^{2}+b^{2}}}\right )}{a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(168\)
default	\(\frac {-\frac {2 b \left (\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-2 b^{2}}+\frac {a b}{2 a^{2}-2 b^{2}}}{\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}{2}+a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {b}{2}}+\frac {2 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (2 a^{2}-2 b^{2}\right ) \sqrt {-a^{2}+b^{2}}}\right )}{a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(168\)
risch	\(\frac {x}{a^{2}}-\frac {2 i b^{2} \left (i a +b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} \left (-a^{2}+b^{2}\right ) d \left (2 b \,{\mathrm e}^{i \left (d x +c \right )}-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \right )}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\)	\(403\)

[In]

int(1/(a+b*csc(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/a^2*b*((1/2*a^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)+1/2*a*b/(a^2-b^2))/(1/2*tan(1/2*d*x+1/2*c)^2*b+a*tan(1/2*
d*x+1/2*c)+1/2*b)+2*(2*a^2-b^2)/(2*a^2-2*b^2)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d*x+1/2*c)+2*a)/(-a^2+b
^2)^(1/2)))+2/a^2*arctan(tan(1/2*d*x+1/2*c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (103) = 206\).

Time = 0.28 (sec) , antiderivative size = 493, normalized size of antiderivative = 4.56 \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \sin \left (d x + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x + {\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \sin \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x + {\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )}\right ) - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]

[In]

integrate(1/(a+b*csc(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\int \frac {1}{\left (a + b \csc {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(1/(a+b*csc(d*x+c))**2,x)

[Out]

Integral((a + b*csc(c + d*x))**(-2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*csc(d*x+c))^2,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 = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b\right )}} - \frac {d x + c}{a^{2}}}{d} \]

[In]

integrate(1/(a+b*csc(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*(2*a^2*b - b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*x + 1/2*c) + a)/sqrt(-a^2 +
b^2)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) + 2*(a*b*tan(1/2*d*x + 1/2*c) + b^2)/((a^3 - a*b^2)*(b*tan(1/2*d*x
+ 1/2*c)^2 + 2*a*tan(1/2*d*x + 1/2*c) + b)) - (d*x + c)/a^2)/d

Mupad [B] (verification not implemented)

Time = 22.69 (sec) , antiderivative size = 2677, normalized size of antiderivative = 24.79 \[ \int \frac {1}{(a+b \csc (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(1/(a + b/sin(c + d*x))^2,x)

[Out]

(b*atan(((b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a*b^6 - 2*a^3*b^4 + a^5*b^2))/(a^6 + a^2*b^4 - 2*a
^4*b^2) - (32*tan(c/2 + (d*x)/2)*(2*a*b^7 - 2*a^7*b - 8*a^3*b^5 + 9*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b
*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a^8*b - a^6*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 +
(d*x)/2)*(2*a^4*b^6 - 6*a^6*b^4 + 4*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a -
b)^3)^(1/2)*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(3*a^11
*b - 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)
))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) - (b*(2*a^2 - b^2)*((a
+ b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(2*a*b^7 - 2*a^7*b - 8*a^3*b^5 + 9*a^5*b^3))/(a^7 + a^3*b^4 -
2*a^5*b^2) - (32*(a*b^6 - 2*a^3*b^4 + a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a
- b)^3)^(1/2)*((32*(a^8*b - a^6*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(2*a^4*b^6 - 6*a^6*
b^4 + 4*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a^5*b^6 - 2
*a^7*b^4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(3*a^11*b - 2*a^5*b^7 + 7*a^7*b^5 -
8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4
- 3*a^6*b^2))*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))/((64*(b^5 - 2*a^2*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2
) + (64*tan(c/2 + (d*x)/2)*(2*b^6 - 6*a^2*b^4 + 4*a^4*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b*(2*a^2 - b^2)*((a
+ b)^3*(a - b)^3)^(1/2)*((32*(a*b^6 - 2*a^3*b^4 + a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) - (32*tan(c/2 + (d*x)
/2)*(2*a*b^7 - 2*a^7*b - 8*a^3*b^5 + 9*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a
- b)^3)^(1/2)*((32*(a^8*b - a^6*b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(2*a^4*b^6 - 6*a^6*
b^4 + 4*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a^5*b^6 - 2
*a^7*b^4 + a^9*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(3*a^11*b - 2*a^5*b^7 + 7*a^7*b^5 -
8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4
- 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*ta
n(c/2 + (d*x)/2)*(2*a*b^7 - 2*a^7*b - 8*a^3*b^5 + 9*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (32*(a*b^6 - 2*a^3
*b^4 + a^5*b^2))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a^8*b - a^6*
b^3))/(a^6 + a^2*b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(2*a^4*b^6 - 6*a^6*b^4 + 4*a^8*b^2))/(a^7 + a^3*b^4
- 2*a^5*b^2) - (b*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*((32*(a^5*b^6 - 2*a^7*b^4 + a^9*b^2))/(a^6 + a^2*
b^4 - 2*a^4*b^2) + (32*tan(c/2 + (d*x)/2)*(3*a^11*b - 2*a^5*b^7 + 7*a^7*b^5 - 8*a^9*b^3))/(a^7 + a^3*b^4 - 2*a
^5*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))/(a^8 - a^2*b^6 +
3*a^4*b^4 - 3*a^6*b^2)))*(2*a^2 - b^2)*((a + b)^3*(a - b)^3)^(1/2)*2i)/(d*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*
b^2)) - ((2*b*tan(c/2 + (d*x)/2))/(a^2 - b^2) + (2*b^2)/(a*(a^2 - b^2)))/(d*(b + 2*a*tan(c/2 + (d*x)/2) + b*ta
n(c/2 + (d*x)/2)^2)) - (2*atan((64*a^5*b*tan(c/2 + (d*x)/2))/((64*a^3*b^9)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (192*
a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (128*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (64*a^9*b^3)/(a^6 + a^2*b^4
- 2*a^4*b^2) - (64*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2)) - (64*a*b^5*tan(c/2 + (d*x)/2))/((64*a^3*b^9)/(a^6 +
a^2*b^4 - 2*a^4*b^2) - (192*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (128*a^7*b^5)/(a^6 + a^2*b^4 - 2*a^4*b^2) +
(64*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (64*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b^2)) + (64*a^3*b^3*tan(c/2 + (
d*x)/2))/((64*a^3*b^9)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (192*a^5*b^7)/(a^6 + a^2*b^4 - 2*a^4*b^2) + (128*a^7*b^5)
/(a^6 + a^2*b^4 - 2*a^4*b^2) + (64*a^9*b^3)/(a^6 + a^2*b^4 - 2*a^4*b^2) - (64*a^11*b)/(a^6 + a^2*b^4 - 2*a^4*b
^2))))/(a^2*d)