\(\int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [707]

Optimal result

Integrand size = 30, antiderivative size = 107 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {x}{a}-\frac {2 b \left (3 a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{3/2} (a+b)^{3/2} d}+\frac {2 b^2 \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4127, 4008, 4004, 3916, 2738, 214} \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {2 b \left (3 a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{3/2} (a+b)^{3/2}}+\frac {2 b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {x}{a} \]

[In]

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^3,x]

[Out]

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

Rule 214

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 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]

Rule 4008

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

Rule 4127

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

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

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 b \left (-3 a^2+b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a \left (a^2-b^2\right ) (c+d x) \cos (c+d x)+b \left (\left (a^2-b^2\right ) (c+d x)+2 a b \sin (c+d x)\right )}{b+a \cos (c+d x)}}{a (a-b) (a+b) d} \]

[In]

Integrate[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 b \left (-\frac {2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (3 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a}}{d}\)	\(151\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 b \left (-\frac {2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (3 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a}}{d}\)	\(151\)
risch	\(\frac {x}{a}+\frac {4 i b^{2} \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}\)	\(409\)

[In]

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (98) = 196\).

Time = 0.28 (sec) , antiderivative size = 481, normalized size of antiderivative = 4.50 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x + {\left (3 \, a^{2} b^{2} - b^{4} + {\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 4 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d\right )}}, \frac {{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x - {\left (3 \, a^{2} b^{2} - b^{4} + {\left (3 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d}\right ] \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {a - b \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate((a**2-b**2*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,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 = 0.36 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.64 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {\frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{2} - b^{2}\right )}} - \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {d x + c}{a}}{d} \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 21.85 (sec) , antiderivative size = 2739, normalized size of antiderivative = 25.60 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

int((a^2 - b^2/cos(c + d*x)^2)/(a + b/cos(c + d*x))^3,x)

[Out]

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