\(\int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) [7]

Optimal result

Integrand size = 19, antiderivative size = 271 \[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\sqrt {2} b c \left (1+\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{a^2 \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} b c \left (1-\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{a^2 \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}+\frac {b \text {arctanh}(\cos (x))}{a^2}-\frac {\cot (x)}{a} \] Output:

2^(1/2)*b*c*(1+(-2*a*c+b^2)/b/(-4*a*c+b^2)^(1/2))*arctan(1/2*(2*c+(b-(-4*a 
*c+b^2)^(1/2))*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1 
/2))/a^2/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2)+2^(1/2)*b*c*(1-(-2*a*c 
+b^2)/b/(-4*a*c+b^2)^(1/2))*arctan(1/2*(2*c+(b+(-4*a*c+b^2)^(1/2))*tan(1/2 
*x))*2^(1/2)/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2))/a^2/(b^2-2*c*(a+c 
)+b*(-4*a*c+b^2)^(1/2))^(1/2)+b*arctanh(cos(x))/a^2-cot(x)/a
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.75 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.43 \[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\csc ^2(x) (-2 a-c+c \cos (2 x)-2 b \sin (x)) \left (-\frac {2 c \left (-i b^2+2 i a c+b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {2 i c \left (-b^2+2 a c+i b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}+a \cot \left (\frac {x}{2}\right )-2 b \log \left (\cos \left (\frac {x}{2}\right )\right )+2 b \log \left (\sin \left (\frac {x}{2}\right )\right )-a \tan \left (\frac {x}{2}\right )\right )}{4 a^2 \left (c+b \csc (x)+a \csc ^2(x)\right )} \] Input:

Integrate[Csc[x]^2/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

(Csc[x]^2*(-2*a - c + c*Cos[2*x] - 2*b*Sin[x])*((-2*c*((-I)*b^2 + (2*I)*a* 
c + b*Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b - I*Sqrt[-b^2 + 4*a*c])*Tan[x/2 
])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[-1/2 
*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]]) + ((2*I)*c 
*(-b^2 + 2*a*c + I*b*Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b + I*Sqrt[-b^2 + 
4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c] 
])])/(Sqrt[-1/2*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a* 
c]]) + a*Cot[x/2] - 2*b*Log[Cos[x/2]] + 2*b*Log[Sin[x/2]] - a*Tan[x/2]))/( 
4*a^2*(c + b*Csc[x] + a*Csc[x]^2))
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3737, 2009}

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.

\(\displaystyle \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (x)^2 \left (a+b \sin (x)+c \sin (x)^2\right )}dx\)

\(\Big \downarrow \) 3737

\(\displaystyle \int \left (\frac {b^2 \left (1-\frac {a c}{b^2}\right )+b c \sin (x)}{a^2 \left (a+b \sin (x)+c \sin ^2(x)\right )}-\frac {b \csc (x)}{a^2}+\frac {\csc ^2(x)}{a}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {2} b c \left (\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^2 \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\sqrt {2} b c \left (1-\frac {b^2-2 a c}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^2 \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {b \text {arctanh}(\cos (x))}{a^2}-\frac {\cot (x)}{a}\)

Input:

Int[Csc[x]^2/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

(Sqrt[2]*b*c*(1 + (b^2 - 2*a*c)/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b - 
Sqrt[b^2 - 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 
- 4*a*c]])])/(a^2*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]]) + (Sqrt[2 
]*b*c*(1 - (b^2 - 2*a*c)/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b + Sqrt[b^ 
2 - 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c 
]])])/(a^2*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + (b*ArcTanh[Cos 
[x]])/a^2 - Cot[x]/a
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3737

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n 
_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_), x_Symbol] :> Int[ExpandTr 
ig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x], x] / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ 
ersQ[m, n, p]
 

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {\tan \left (\frac {x}{2}\right )}{2 a}+\frac {-\frac {2 \left (-3 \sqrt {-4 a c +b^{2}}\, a b c +\sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+5 a \,b^{2} c -b^{4}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {2 \left (3 \sqrt {-4 a c +b^{2}}\, a b c -\sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+5 a \,b^{2} c -b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}}{a}-\frac {1}{2 a \tan \left (\frac {x}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\)	\(337\)
risch	\(\text {Expression too large to display}\)	\(2623\)

Input:

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

Output:

1/2/a*tan(1/2*x)+2/a*(-(-3*(-4*a*c+b^2)^(1/2)*a*b*c+(-4*a*c+b^2)^(1/2)*b^3 
-4*a^2*c^2+5*a*b^2*c-b^4)/a/(4*a*c-b^2)/(4*a*c-2*b^2+2*b*(-4*a*c+b^2)^(1/2 
)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*a*c-2*b^2+ 
2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))+(3*(-4*a*c+b^2)^(1/2)*a*b*c-(-4*a*c+b 
^2)^(1/2)*b^3-4*a^2*c^2+5*a*b^2*c-b^4)/a/(4*a*c-b^2)/(4*a*c-2*b^2-2*b*(-4* 
a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/( 
4*a*c-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)))-1/2/a/tan(1/2*x)-1/a^2*b 
*ln(tan(1/2*x))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral(csc(x)**2/(a + b*sin(x) + c*sin(x)**2), x)
 

Maxima [F]

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

integrate(csc(x)^2/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")
 

Output:

1/2*(2*(a^2*cos(2*x)^2 + a^2*sin(2*x)^2 - 2*a^2*cos(2*x) + a^2)*integrate( 
2*(2*b^2*c*cos(3*x)^2 + 2*b^2*c*cos(x)^2 + 2*b^2*c*sin(3*x)^2 + 2*b^2*c*si 
n(x)^2 + b*c^2*sin(x) + 4*(2*a*b^2 - a*c^2 - (2*a^2 - b^2)*c)*cos(2*x)^2 + 
 2*(2*b^3 + b*c^2)*cos(x)*sin(2*x) + 4*(2*a*b^2 - a*c^2 - (2*a^2 - b^2)*c) 
*sin(2*x)^2 - (b*c^2*sin(3*x) - b*c^2*sin(x) + 2*(b^2*c - a*c^2)*cos(2*x)) 
*cos(4*x) - 2*(2*b^2*c*cos(x) + (2*b^3 + b*c^2)*sin(2*x))*cos(3*x) - 2*(b^ 
2*c - a*c^2 + (2*b^3 + b*c^2)*sin(x))*cos(2*x) + (b*c^2*cos(3*x) - b*c^2*c 
os(x) - 2*(b^2*c - a*c^2)*sin(2*x))*sin(4*x) - (4*b^2*c*sin(x) + b*c^2 - 2 
*(2*b^3 + b*c^2)*cos(2*x))*sin(3*x))/(a^2*c^2*cos(4*x)^2 + 4*a^2*b^2*cos(3 
*x)^2 + 4*a^2*b^2*cos(x)^2 + a^2*c^2*sin(4*x)^2 + 4*a^2*b^2*sin(3*x)^2 + 4 
*a^2*b^2*sin(x)^2 + 4*a^2*b*c*sin(x) + a^2*c^2 + 4*(4*a^4 + 4*a^3*c + a^2* 
c^2)*cos(2*x)^2 + 8*(2*a^3*b + a^2*b*c)*cos(x)*sin(2*x) + 4*(4*a^4 + 4*a^3 
*c + a^2*c^2)*sin(2*x)^2 - 2*(2*a^2*b*c*sin(3*x) - 2*a^2*b*c*sin(x) - a^2* 
c^2 + 2*(2*a^3*c + a^2*c^2)*cos(2*x))*cos(4*x) - 8*(a^2*b^2*cos(x) + (2*a^ 
3*b + a^2*b*c)*sin(2*x))*cos(3*x) - 4*(2*a^3*c + a^2*c^2 + 2*(2*a^3*b + a^ 
2*b*c)*sin(x))*cos(2*x) + 4*(a^2*b*c*cos(3*x) - a^2*b*c*cos(x) - (2*a^3*c 
+ a^2*c^2)*sin(2*x))*sin(4*x) - 4*(2*a^2*b^2*sin(x) + a^2*b*c - 2*(2*a^3*b 
 + a^2*b*c)*cos(2*x))*sin(3*x)), x) + (b*cos(2*x)^2 + b*sin(2*x)^2 - 2*b*c 
os(2*x) + b)*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) - (b*cos(2*x)^2 + b*s 
in(2*x)^2 - 2*b*cos(2*x) + b)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) -...
 

Giac [F(-1)]

Timed out. \[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 45.70 (sec) , antiderivative size = 16102, normalized size of antiderivative = 59.42 \[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

int(1/(sin(x)^2*(a + c*sin(x)^2 + b*sin(x))),x)
 

Output:

tan(x/2)/(2*a) - 1/(2*a*tan(x/2)) - atan(((((-(b^8 + 8*a^3*c^5 + 8*a^4*c^4 
 + b^5*(-(4*a*c - b^2)^3)^(1/2) - b^6*c^2 + 8*a*b^4*c^3 - 18*a^2*b^2*c^4 + 
 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 10*a 
*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) + 2*a*b*c^3*(-(4*a*c - b^2)^ 
3)^(1/2) - 4*a*b^3*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^6*b^4 - a^4*b^6 + 16* 
a^6*c^4 + 32*a^7*c^3 + 16*a^8*c^2 + 10*a^5*b^4*c - 8*a^7*b^2*c + a^4*b^4*c 
^2 - 8*a^5*b^2*c^3 - 32*a^6*b^2*c^2)))^(1/2)*((32*(4*a^5*b^4 - 8*a^3*b^6 + 
 16*a^5*c^4 + 20*a^6*c^3 + 4*a^7*c^2 + 53*a^4*b^4*c - 17*a^6*b^2*c + 8*a^3 
*b^4*c^2 - 36*a^4*b^2*c^3 - 89*a^5*b^2*c^2))/a^3 - ((32*(4*a^5*b^5 - 3*a^7 
*b^3 + 16*a^6*b*c^3 - 25*a^6*b^3*c + 36*a^7*b*c^2 - 4*a^5*b^3*c^2 + 12*a^8 
*b*c))/a^3 - (32*tan(x/2)*(8*a^9*c - 16*a^4*b^6 + 17*a^6*b^4 - 2*a^8*b^2 + 
 192*a^6*c^4 + 384*a^7*c^3 + 200*a^8*c^2 + 144*a^5*b^4*c - 118*a^7*b^2*c + 
 16*a^4*b^4*c^2 - 112*a^5*b^2*c^3 - 416*a^6*b^2*c^2))/a^3)*(-(b^8 + 8*a^3* 
c^5 + 8*a^4*c^4 + b^5*(-(4*a*c - b^2)^3)^(1/2) - b^6*c^2 + 8*a*b^4*c^3 - 1 
8*a^2*b^2*c^4 + 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - b^3*c^2*(-(4*a*c - b^2)^ 
3)^(1/2) - 10*a*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) + 2*a*b*c^3*( 
-(4*a*c - b^2)^3)^(1/2) - 4*a*b^3*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^6*b^4 
- a^4*b^6 + 16*a^6*c^4 + 32*a^7*c^3 + 16*a^8*c^2 + 10*a^5*b^4*c - 8*a^7*b^ 
2*c + a^4*b^4*c^2 - 8*a^5*b^2*c^3 - 32*a^6*b^2*c^2)))^(1/2) + (32*tan(x/2) 
*(13*a^4*b^5 - 16*a^2*b^7 - 2*a^6*b^3 + 128*a^3*b^5*c + 128*a^4*b*c^4 +...
 

Reduce [F]

\[ \int \frac {\csc ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {too large to display} \] Input:

int(csc(x)^2/(a+b*sin(x)+c*sin(x)^2),x)
 

Output:

( - 6*cos(x)*a**4*c + 3*cos(x)*a**3*b**2 - 8*cos(x)*a**3*c**2 + 8*cos(x)*a 
**2*b**2*c - 48*cos(x)*a**2*c**3 + 72*cos(x)*a*b**2*c**2 - 64*cos(x)*a*c** 
4 - 16*cos(x)*b**4*c + 32*cos(x)*b**2*c**3 - 24*int(1/(6*tan(x/2)**6*a**4* 
c - 3*tan(x/2)**6*a**3*b**2 + 8*tan(x/2)**6*a**3*c**2 - 16*tan(x/2)**6*a** 
2*b**2*c + 4*tan(x/2)**6*a*b**4 + 12*tan(x/2)**5*a**3*b*c - 6*tan(x/2)**5* 
a**2*b**3 + 16*tan(x/2)**5*a**2*b*c**2 - 32*tan(x/2)**5*a*b**3*c + 8*tan(x 
/2)**5*b**5 + 12*tan(x/2)**4*a**4*c - 6*tan(x/2)**4*a**3*b**2 + 40*tan(x/2 
)**4*a**3*c**2 - 44*tan(x/2)**4*a**2*b**2*c + 32*tan(x/2)**4*a**2*c**3 + 8 
*tan(x/2)**4*a*b**4 - 64*tan(x/2)**4*a*b**2*c**2 + 16*tan(x/2)**4*b**4*c + 
 12*tan(x/2)**3*a**3*b*c - 6*tan(x/2)**3*a**2*b**3 + 16*tan(x/2)**3*a**2*b 
*c**2 - 32*tan(x/2)**3*a*b**3*c + 8*tan(x/2)**3*b**5 + 6*tan(x/2)**2*a**4* 
c - 3*tan(x/2)**2*a**3*b**2 + 8*tan(x/2)**2*a**3*c**2 - 16*tan(x/2)**2*a** 
2*b**2*c + 4*tan(x/2)**2*a*b**4),x)*sin(x)*a**6*b**2*c**2 - 144*int(1/(6*t 
an(x/2)**6*a**4*c - 3*tan(x/2)**6*a**3*b**2 + 8*tan(x/2)**6*a**3*c**2 - 16 
*tan(x/2)**6*a**2*b**2*c + 4*tan(x/2)**6*a*b**4 + 12*tan(x/2)**5*a**3*b*c 
- 6*tan(x/2)**5*a**2*b**3 + 16*tan(x/2)**5*a**2*b*c**2 - 32*tan(x/2)**5*a* 
b**3*c + 8*tan(x/2)**5*b**5 + 12*tan(x/2)**4*a**4*c - 6*tan(x/2)**4*a**3*b 
**2 + 40*tan(x/2)**4*a**3*c**2 - 44*tan(x/2)**4*a**2*b**2*c + 32*tan(x/2)* 
*4*a**2*c**3 + 8*tan(x/2)**4*a*b**4 - 64*tan(x/2)**4*a*b**2*c**2 + 16*tan( 
x/2)**4*b**4*c + 12*tan(x/2)**3*a**3*b*c - 6*tan(x/2)**3*a**2*b**3 + 16...