Integrand size = 19, antiderivative size = 230 \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {x}{c}-\frac {\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}} \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}} \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{c \sqrt {b^2-4 a c}} \]
-x/c-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))*2^(1/2)*(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1 /2))^(1/2)/c/(-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))*2^(1/2)*(b^2- 2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2)/c/(-4*a*c+b^2)^(1/2)
Result contains complex when optimal does not.
Time = 0.58 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {-x+\frac {\left (i b^2-2 i c (a+c)+b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {\left (-i b^2+2 i c (a+c)+b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}}{c} \]
(-x + ((I*b^2 - (2*I)*c*(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*Sqr t[-b^2 + 4*a*c]])])/(Sqrt[-1/2*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) - I*b*S qrt[-b^2 + 4*a*c]]) + (((-I)*b^2 + (2*I)*c*(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]]))/c
Time = 0.75 (sec) , antiderivative size = 230, 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, 3747, 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.
| \(\displaystyle \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^2}{a+b \sin (x)+c \sin (x)^2}dx\) |
| \(\Big \downarrow \) 3747 |
| \(\displaystyle \int \left (\frac {a \left (\frac {c}{a}+1\right )+b \sin (x)}{c \left (a+b \sin (x)+c \sin ^2(x)\right )}-\frac {1}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2} \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2} \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c \sqrt {b^2-4 a c}}-\frac {x}{c}\) |
-(x/c) - (Sqrt[2]*Sqrt[b^2 - 2*c*(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]])])/(c*Sqrt[b^2 - 4*a*c]) + (Sqrt[2]*Sqrt[b^2 - 2*c*(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]])])/(c*Sqrt[b^2 - 4 *a*c])
3.1.10.3.1 Defintions of rubi rules used
Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n _.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_.), x_Symbol] :> Int[ExpandT rig[(1 - sin[d + e*x]^2)^(m/2)*(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] && IntegerQ[m/2] & & NeQ[b^2 - 4*a*c, 0] && IntegersQ[n, p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.91 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(-\frac {x}{c}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{2} c^{4}-8 a \,b^{2} c^{3}+b^{4} c^{2}\right ) \textit {\_Z}^{4}+\left (8 a^{2} c^{2}-6 a \,b^{2} c +8 a \,c^{3}+b^{4}-2 b^{2} c^{2}\right ) \textit {\_Z}^{2}+a^{2}+2 a c -b^{2}+c^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i x}+\left (\frac {8 c^{3} a}{b}-2 b \,c^{2}\right ) \textit {\_R}^{3}+\left (\frac {4 i c^{2} a}{b}-i b c \right ) \textit {\_R}^{2}+\left (\frac {2 a c}{b}-b +\frac {2 c^{2}}{b}\right ) \textit {\_R} +\frac {i a}{b}+\frac {i c}{b}\right )\right )\) | \(178\) |
| default | \(\frac {2 a \left (-\frac {\left (-\sqrt {-4 a c +b^{2}}\, b a +\sqrt {-4 a c +b^{2}}\, b c +4 a^{2} c -a \,b^{2}+4 a \,c^{2}-b^{2} c \right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {\left (\sqrt {-4 a c +b^{2}}\, b a -\sqrt {-4 a c +b^{2}}\, b c +4 a^{2} c -a \,b^{2}+4 a \,c^{2}-b^{2} c \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{c}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{c}\) | \(320\) |
-x/c+sum(_R*ln(exp(I*x)+(8*c^3/b*a-2*b*c^2)*_R^3+(4*I/b*c^2*a-I*b*c)*_R^2+ (2*a*c/b-b+2*c^2/b)*_R+I/b*a+I/b*c),_R=RootOf((16*a^2*c^4-8*a*b^2*c^3+b^4* c^2)*_Z^4+(8*a^2*c^2-6*a*b^2*c+8*a*c^3+b^4-2*b^2*c^2)*_Z^2+a^2+2*a*c-b^2+c ^2))
Leaf count of result is larger than twice the leaf count of optimal. 971 vs. \(2 (196) = 392\).
Time = 0.39 (sec) , antiderivative size = 971, normalized size of antiderivative = 4.22 \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\sqrt {2} c \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \log \left (\sqrt {2} {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \cos \left (x\right ) + b^{2} \sin \left (x\right ) + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sin \left (x\right ) + 2 \, b c\right ) - \sqrt {2} c \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \log \left (\sqrt {2} {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \cos \left (x\right ) - b^{2} \sin \left (x\right ) - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sin \left (x\right ) - 2 \, b c\right ) - \sqrt {2} c \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \log \left (\sqrt {2} {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \cos \left (x\right ) + b^{2} \sin \left (x\right ) - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sin \left (x\right ) + 2 \, b c\right ) + \sqrt {2} c \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \log \left (\sqrt {2} {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sqrt {-\frac {b^{2} - 2 \, a c - 2 \, c^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}}{b^{2} c^{2} - 4 \, a c^{3}}} \cos \left (x\right ) - b^{2} \sin \left (x\right ) + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {\frac {b^{2}}{b^{2} c^{4} - 4 \, a c^{5}}} \sin \left (x\right ) - 2 \, b c\right ) - 4 \, x}{4 \, c} \]
1/4*(sqrt(2)*c*sqrt(-(b^2 - 2*a*c - 2*c^2 + (b^2*c^2 - 4*a*c^3)*sqrt(b^2/( b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*log(sqrt(2)*(b^2*c^3 - 4*a*c^4)* sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sqrt(-(b^2 - 2*a*c - 2*c^2 + (b^2*c^2 - 4*a* c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*cos(x) + b^2*sin( x) + (b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sin(x) + 2*b*c) - s qrt(2)*c*sqrt(-(b^2 - 2*a*c - 2*c^2 + (b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2*c^ 4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*log(sqrt(2)*(b^2*c^3 - 4*a*c^4)*sqrt(b ^2/(b^2*c^4 - 4*a*c^5))*sqrt(-(b^2 - 2*a*c - 2*c^2 + (b^2*c^2 - 4*a*c^3)*s qrt(b^2/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*cos(x) - b^2*sin(x) - ( b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sin(x) - 2*b*c) - sqrt(2) *c*sqrt(-(b^2 - 2*a*c - 2*c^2 - (b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2*c^4 - 4* a*c^5)))/(b^2*c^2 - 4*a*c^3))*log(sqrt(2)*(b^2*c^3 - 4*a*c^4)*sqrt(b^2/(b^ 2*c^4 - 4*a*c^5))*sqrt(-(b^2 - 2*a*c - 2*c^2 - (b^2*c^2 - 4*a*c^3)*sqrt(b^ 2/(b^2*c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*cos(x) + b^2*sin(x) - (b^2*c^ 2 - 4*a*c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sin(x) + 2*b*c) + sqrt(2)*c*sqr t(-(b^2 - 2*a*c - 2*c^2 - (b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5) ))/(b^2*c^2 - 4*a*c^3))*log(sqrt(2)*(b^2*c^3 - 4*a*c^4)*sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sqrt(-(b^2 - 2*a*c - 2*c^2 - (b^2*c^2 - 4*a*c^3)*sqrt(b^2/(b^2 *c^4 - 4*a*c^5)))/(b^2*c^2 - 4*a*c^3))*cos(x) - b^2*sin(x) + (b^2*c^2 - 4* a*c^3)*sqrt(b^2/(b^2*c^4 - 4*a*c^5))*sin(x) - 2*b*c) - 4*x)/c
Timed out. \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\cos \left (x\right )^{2}}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \]
(c*integrate(2*(2*b^2*cos(3*x)^2 + 2*b^2*cos(x)^2 + 2*b^2*sin(3*x)^2 + 2*b ^2*sin(x)^2 + 4*(2*a^2 + 3*a*c + c^2)*cos(2*x)^2 + 2*(4*a*b + 3*b*c)*cos(x )*sin(2*x) + 4*(2*a^2 + 3*a*c + c^2)*sin(2*x)^2 + b*c*sin(x) - (b*c*sin(3* x) - b*c*sin(x) + 2*(a*c + c^2)*cos(2*x))*cos(4*x) - 2*(2*b^2*cos(x) + (4* a*b + 3*b*c)*sin(2*x))*cos(3*x) - 2*(a*c + c^2 + (4*a*b + 3*b*c)*sin(x))*c os(2*x) + (b*c*cos(3*x) - b*c*cos(x) - 2*(a*c + c^2)*sin(2*x))*sin(4*x) - (4*b^2*sin(x) + b*c - 2*(4*a*b + 3*b*c)*cos(2*x))*sin(3*x))/(c^3*cos(4*x)^ 2 + 4*b^2*c*cos(3*x)^2 + 4*b^2*c*cos(x)^2 + c^3*sin(4*x)^2 + 4*b^2*c*sin(3 *x)^2 + 4*b^2*c*sin(x)^2 + 4*b*c^2*sin(x) + c^3 + 4*(4*a^2*c + 4*a*c^2 + c ^3)*cos(2*x)^2 + 8*(2*a*b*c + b*c^2)*cos(x)*sin(2*x) + 4*(4*a^2*c + 4*a*c^ 2 + c^3)*sin(2*x)^2 - 2*(2*b*c^2*sin(3*x) - 2*b*c^2*sin(x) - c^3 + 2*(2*a* c^2 + c^3)*cos(2*x))*cos(4*x) - 8*(b^2*c*cos(x) + (2*a*b*c + b*c^2)*sin(2* x))*cos(3*x) - 4*(2*a*c^2 + c^3 + 2*(2*a*b*c + b*c^2)*sin(x))*cos(2*x) + 4 *(b*c^2*cos(3*x) - b*c^2*cos(x) - (2*a*c^2 + c^3)*sin(2*x))*sin(4*x) - 4*( 2*b^2*c*sin(x) + b*c^2 - 2*(2*a*b*c + b*c^2)*cos(2*x))*sin(3*x)), x) - x)/ c
Timed out. \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 29.24 (sec) , antiderivative size = 11164, normalized size of antiderivative = 48.54 \[ \int \frac {\cos ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
atan(((-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^ 2 - 6*a*b^2*c)/(2*(16*a^2*c^4 + b^4*c^2 - 8*a*b^2*c^3)))^(1/2)*(tan(x/2)*( 81920*a*b^4 + 139264*a*c^4 + 196608*a^4*c + 24576*a^5 - 98304*a^3*b^2 + 42 5984*a^2*c^3 + 458752*a^3*c^2 - 212992*a*b^2*c^2 - 327680*a^2*b^2*c) - 245 76*a^4*b + 32768*a^2*b^3 + (-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2*c)/(2*(16*a^2*c^4 + b^4*c^2 - 8*a*b^2*c^3 )))^(1/2)*((-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b ^2*c^2 - 6*a*b^2*c)/(2*(16*a^2*c^4 + b^4*c^2 - 8*a*b^2*c^3)))^(1/2)*((-(8* a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2 *c)/(2*(16*a^2*c^4 + b^4*c^2 - 8*a*b^2*c^3)))^(1/2)*((-(8*a*c^3 + b*(-(4*a *c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2*c)/(2*(16*a^2*c ^4 + b^4*c^2 - 8*a*b^2*c^3)))^(1/2)*(tan(x/2)*(524288*a^2*c^7 + 1179648*a^ 3*c^6 + 851968*a^4*c^5 + 196608*a^5*c^4 - 131072*a*b^2*c^6 + 139264*a*b^4* c^4 - 16384*a*b^6*c^2 - 851968*a^2*b^2*c^5 + 147456*a^2*b^4*c^3 - 540672*a ^3*b^2*c^4 + 16384*a^3*b^4*c^2 - 114688*a^4*b^2*c^3) - 32768*a*b^3*c^5 + 2 4576*a*b^5*c^3 + 131072*a^2*b*c^6 + 163840*a^3*b*c^5 + 98304*a^4*b*c^4 - 1 39264*a^2*b^3*c^4 - 24576*a^3*b^3*c^3) - tan(x/2)*(32768*a*b^5*c^2 - 32768 *a*b^3*c^4 + 131072*a^2*b*c^5 + 262144*a^3*b*c^4 + 131072*a^4*b*c^3 - 1966 08*a^2*b^3*c^3 - 32768*a^3*b^3*c^2) + 131072*a^2*c^6 + 163840*a^3*c^5 - 65 536*a^4*c^4 - 98304*a^5*c^3 - 32768*a*b^2*c^5 + 32768*a*b^4*c^3 - 17203...