\(\int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx\) [445]

Optimal result

Integrand size = 12, antiderivative size = 97 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {a x}{a^2+b^2}+\frac {2 a c \text {arctanh}\left (\frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt {a^2+b^2-c^2}}+\frac {b \log (c+a \cos (x)+b \sin (x))}{a^2+b^2} \]

[Out]

a*x/(a^2+b^2)+b*ln(c+a*cos(x)+b*sin(x))/(a^2+b^2)+2*a*c*arctanh((b-(a-c)*tan(1/2*x))/(a^2+b^2-c^2)^(1/2))/(a^2
+b^2)/(a^2+b^2-c^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3238, 3217, 3203, 632, 212} \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {2 a c \text {arctanh}\left (\frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt {a^2+b^2-c^2}}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2} \]

[In]

Int[(a + c*Sec[x] + b*Tan[x])^(-1),x]

[Out]

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

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 3203

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[2*(f/e), Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 3217

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)]), x_Symbol] :> Simp[b*B*((d + e*x)/(e*(b^2 + c^2))), x] + (Dist[(A*(b^2 + c^2) - a*b*B)/(b^2 + c^2), Int[
1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] + Simp[c*B*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 +
c^2))), x]) /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*b*B, 0]

Rule 3238

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {a x+\frac {2 a c \text {arctanh}\left (\frac {b+(-a+c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}}+b \log (c+a \cos (x)+b \sin (x))}{a^2+b^2} \]

[In]

Integrate[(a + c*Sec[x] + b*Tan[x])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.76

method	result	size
default	\(\frac {\frac {2 \left (a b -c b \right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -c \tan \left (\frac {x}{2}\right )^{2}-2 b \tan \left (\frac {x}{2}\right )-a -c \right )}{2 a -2 c}+\frac {2 \left (a c -b^{2}+\frac {\left (a b -c b \right ) b}{a -c}\right ) \arctan \left (\frac {2 \left (a -c \right ) \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}-b^{2}+c^{2}}}}{a^{2}+b^{2}}+\frac {-b \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+2 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}+b^{2}}\)	\(171\)
risch	\(\text {Expression too large to display}\)	\(1314\)

[In]

int(1/(a+c*sec(x)+b*tan(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (93) = 186\).

Time = 0.32 (sec) , antiderivative size = 553, normalized size of antiderivative = 5.70 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\left [\frac {\sqrt {a^{2} + b^{2} - c^{2}} a c \log \left (\frac {2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (a^{2} - b^{2}\right )} c^{2} + 2 \, {\left (a^{3} + a b^{2}\right )} c \cos \left (x\right ) - {\left (a^{4} - b^{4} - 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \left (x\right )^{2} + 2 \, {\left ({\left (a^{2} b + b^{3}\right )} c - {\left (a^{3} b + a b^{3} - 2 \, a b c^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) + 2 \, {\left (2 \, a b c \cos \left (x\right )^{2} - a b c + {\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) - {\left (a^{3} + a b^{2} + {\left (a^{2} - b^{2}\right )} c \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt {a^{2} + b^{2} - c^{2}}}{2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )}\right ) + 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x + {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}, -\frac {2 \, \sqrt {-a^{2} - b^{2} + c^{2}} a c \arctan \left (\frac {{\left (a c \cos \left (x\right ) + b c \sin \left (x\right ) + a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}}{{\left (a^{2} b + b^{3} - b c^{2}\right )} \cos \left (x\right ) - {\left (a^{3} + a b^{2} - a c^{2}\right )} \sin \left (x\right )}\right ) - 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x - {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}\right ] \]

[In]

integrate(1/(a+c*sec(x)+b*tan(x)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\int \frac {1}{a + b \tan {\left (x \right )} + c \sec {\left (x \right )}}\, dx \]

[In]

integrate(1/(a+c*sec(x)+b*tan(x)),x)

[Out]

Integral(1/(a + b*tan(x) + c*sec(x)), x)

Maxima [F(-2)]

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

[In]

integrate(1/(a+c*sec(x)+b*tan(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(c^2-b^2-a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, c\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - c \tan \left (\frac {1}{2} \, x\right ) - b}{\sqrt {-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{{\left (a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}} + \frac {a x}{a^{2} + b^{2}} + \frac {b \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a + c\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{a^{2} + b^{2}} \]

[In]

integrate(1/(a+c*sec(x)+b*tan(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 42.59 (sec) , antiderivative size = 988, normalized size of antiderivative = 10.19 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {\ln \left (32\,a\,c-32\,c^2+32\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )+\frac {\left (32\,a^2\,b-32\,b\,c^2+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )\,\left (-a^2+2\,a\,c+3\,b^2-2\,c^2\right )-\frac {\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (32\,a^4-64\,a^3\,c-64\,a^2\,b^2+32\,a^2\,c^2-32\,b^2\,c^2+96\,a\,b^2\,c+32\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )\,\left (4\,a^2-4\,c\,a+b^2\right )+\frac {32\,\left (a-c\right )\,\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (3\,a^3\,b-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2+a^2\,b\,c+2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2+3\,a\,b^3-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^2\,c-4\,a\,b\,c^2+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^4+b^3\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (b\,\left (a^2-c^2\right )+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )}{c^2\,\left (a^2+b^2-c^2\right )+{\left (a^2+b^2-c^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}{b+a\,1{}\mathrm {i}}+\frac {\ln \left (32\,a\,c-32\,c^2+32\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )+\frac {\left (32\,a^2\,b-32\,b\,c^2+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )\,\left (-a^2+2\,a\,c+3\,b^2-2\,c^2\right )-\frac {\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (32\,a^4-64\,a^3\,c-64\,a^2\,b^2+32\,a^2\,c^2-32\,b^2\,c^2+96\,a\,b^2\,c+32\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-c\right )\,\left (4\,a^2-4\,c\,a+b^2\right )+\frac {32\,\left (a-c\right )\,\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (3\,a^3\,b-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2+a^2\,b\,c+2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2+3\,a\,b^3-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^2\,c-4\,a\,b\,c^2+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^4+b^3\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (b\,\left (a^2-c^2\right )+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )}{c^2\,\left (a^2+b^2-c^2\right )+{\left (a^2+b^2-c^2\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a+b\,1{}\mathrm {i}} \]

[In]

int(1/(a + b*tan(x) + c/cos(x)),x)

[Out]

(log(32*a*c - 32*c^2 + 32*b*tan(x/2)*(a - c) + ((32*a^2*b - 32*b*c^2 + 32*tan(x/2)*(a - c)*(2*a*c - a^2 + 3*b^
2 - 2*c^2) - ((a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a^4 - 64*a^3*c - 64*a^2*b^2 + 32*a^2*c^2
- 32*b^2*c^2 + 96*a*b^2*c + 32*b*tan(x/2)*(a - c)*(4*a^2 - 4*a*c + b^2) + (32*(a - c)*(a^2*b - b*c^2 + b^3 +
a*c*(a^2 + b^2 - c^2)^(1/2))*(3*b^4*tan(x/2) + 3*a*b^3 + 3*a^3*b + b^3*c + 3*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(
x/2) - 2*b^2*c^2*tan(x/2) - 2*a^3*c*tan(x/2) - 4*a*b*c^2 + a^2*b*c - 2*a*b^2*c*tan(x/2)))/((a^2 + b^2)*(a^2 +
b^2 - c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^2)))*(a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2)))/((a^2 + b^
2)*(a^2 + b^2 - c^2)))*(b*(a^2 - c^2) + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2)))/(c^2*(a^2 + b^2 - c^2) + (a^2 + b^
2 - c^2)^2) - log(tan(x/2) + 1i)/(a*1i + b) - (log(tan(x/2) - 1i)*1i)/(a + b*1i) + (log(32*a*c - 32*c^2 + 32*b
*tan(x/2)*(a - c) + ((32*a^2*b - 32*b*c^2 + 32*tan(x/2)*(a - c)*(2*a*c - a^2 + 3*b^2 - 2*c^2) - ((a^2*b - b*c^
2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a^4 - 64*a^3*c - 64*a^2*b^2 + 32*a^2*c^2 - 32*b^2*c^2 + 96*a*b^2*c
+ 32*b*tan(x/2)*(a - c)*(4*a^2 - 4*a*c + b^2) + (32*(a - c)*(a^2*b - b*c^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2)
)*(3*b^4*tan(x/2) + 3*a*b^3 + 3*a^3*b + b^3*c + 3*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(x/2) - 2*b^2*c^2*tan(x/2) -
2*a^3*c*tan(x/2) - 4*a*b*c^2 + a^2*b*c - 2*a*b^2*c*tan(x/2)))/((a^2 + b^2)*(a^2 + b^2 - c^2))))/((a^2 + b^2)*
(a^2 + b^2 - c^2)))*(a^2*b - b*c^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2)))/((a^2 + b^2)*(a^2 + b^2 - c^2)))*(b*(
a^2 - c^2) + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2)))/(c^2*(a^2 + b^2 - c^2) + (a^2 + b^2 - c^2)^2)