Integrand size = 19, antiderivative size = 388 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {x}{2 c}+\frac {\left (b^2-c (a+2 c)\right ) x}{c^3}+\frac {2 \left (b^2 \left (b^2-2 c (a+c)\right )-b \sqrt {b^2-4 a c} \left (b^2-2 c (a+c)\right )-2 c \left (a b^2-c (a+c)^2\right )\right ) \arctan \left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}-\frac {2 \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}}-\frac {b \sin (x)}{c^2}+\frac {\cos (x) \sin (x)}{2 c} \]
1/2*x/c+(b^2-c*(a+2*c))*x/c^3-b*sin(x)/c^2+1/2*cos(x)*sin(x)/c-2*arctan((b -2*c-(-4*a*c+b^2)^(1/2))^(1/2)*tan(1/2*x)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2) )*(b*(b^2-2*c*(a+c))+(-b^4-2*c^2*(a+c)^2+2*b^2*c*(2*a+c))/(-4*a*c+b^2)^(1/ 2))/c^3/(b-2*c-(-4*a*c+b^2)^(1/2))^(1/2)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2)- 2*arctan((b-2*c+(-4*a*c+b^2)^(1/2))^(1/2)*tan(1/2*x)/(b+2*c+(-4*a*c+b^2)^( 1/2))^(1/2))*(b^4+2*c^2*(a+c)^2-2*b^2*c*(2*a+c)+b^3*(-4*a*c+b^2)^(1/2)-2*b *c*(a+c)*(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)/(b-2*c+(-4*a*c+b^2)^(1 /2))^(1/2)/(b+2*c+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.65 (sec) , antiderivative size = 374, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\frac {4 b^2 x-2 c (2 a+3 c) x+\frac {4 \sqrt {2} \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {4 \sqrt {2} \left (-b^4-2 c^2 (a+c)^2+2 b^2 c (2 a+c)+b^3 \sqrt {b^2-4 a c}-2 b c (a+c) \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}-4 b c \sin (x)+c^2 \sin (2 x)}{4 c^3} \]
(4*b^2*x - 2*c*(2*a + 3*c)*x + (4*Sqrt[2]*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c *(2*a + c) + b^3*Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*ArcT anh[((b - 2*c + Sqrt[b^2 - 4*a*c])*Tan[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) - 2 *b*Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) - b*Sqr t[b^2 - 4*a*c]]) - (4*Sqrt[2]*(-b^4 - 2*c^2*(a + c)^2 + 2*b^2*c*(2*a + c) + b^3*Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*ArcTanh[((-b + 2*c + Sqrt[b^2 - 4*a*c])*Tan[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) + 2*b*Sqrt[b^ 2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) + b*Sqrt[b^2 - 4* a*c]]) - 4*b*c*Sin[x] + c^2*Sin[2*x])/(4*c^3)
Time = 7.08 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3748, 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.
| \(\displaystyle \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)^4}{a+b \cos (x)+c \cos (x)^2}dx\) |
| \(\Big \downarrow \) 3748 |
| \(\displaystyle \int \left (\frac {b^2-c (a+2 c)}{c^3}+\frac {-a b^2 \left (1-\frac {c (a+c)^2}{a b^2}\right )-\left (b^3 \cos (x) \left (1-\frac {2 c (a+c)}{b^2}\right )\right )}{c^3 \left (a+b \cos (x)+c \cos ^2(x)\right )}-\frac {b \cos (x)}{c^2}+\frac {\cos ^2(x)}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (-2 b^2 c (a+c)-b \sqrt {b^2-4 a c} \left (b^2-2 c (a+c)\right )-2 c \left (a b^2-c (a+c)^2\right )+b^4\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}-\frac {2 \left (-2 b^2 c (2 a+c)-2 b c (a+c) \sqrt {b^2-4 a c}+b^3 \sqrt {b^2-4 a c}+2 c^2 (a+c)^2+b^4\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}+\frac {x \left (b^2-c (a+2 c)\right )}{c^3}-\frac {b \sin (x)}{c^2}+\frac {x}{2 c}+\frac {\sin (x) \cos (x)}{2 c}\) |
x/(2*c) + ((b^2 - c*(a + 2*c))*x)/c^3 + (2*(b^4 - 2*b^2*c*(a + c) - b*Sqrt [b^2 - 4*a*c]*(b^2 - 2*c*(a + c)) - 2*c*(a*b^2 - c*(a + c)^2))*ArcTan[(Sqr t[b - 2*c - Sqrt[b^2 - 4*a*c]]*Tan[x/2])/Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]] ])/(c^3*Sqrt[b^2 - 4*a*c]*Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) - (2*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c) + b^3 *Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Tan[x/2])/Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]])/(c^3*S qrt[b^2 - 4*a*c]*Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c + Sqrt[b^2 - 4*a*c]]) - (b*Sin[x])/c^2 + (Cos[x]*Sin[x])/(2*c)
3.1.6.3.1 Defintions of rubi rules used
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]^(n_.)*(b_.) + cos[(d_.) + (e_.)*(x_)]^ (n2_.)*(c_.))^(p_.)*sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandT rig[(1 - cos[d + e*x]^2)^(m/2)*(a + b*cos[d + e*x]^n + c*cos[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]
Time = 5.47 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {2 \left (a -b +c \right ) \left (\frac {\left (\sqrt {-4 a c +b^{2}}\, a c -\sqrt {-4 a c +b^{2}}\, b^{2}-\sqrt {-4 a c +b^{2}}\, b c +\sqrt {-4 a c +b^{2}}\, c^{2}+3 c a b +2 a \,c^{2}-b^{3}-b^{2} c +b \,c^{2}+2 c^{3}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (\sqrt {-4 a c +b^{2}}\, a c -\sqrt {-4 a c +b^{2}}\, b^{2}-\sqrt {-4 a c +b^{2}}\, b c +\sqrt {-4 a c +b^{2}}\, c^{2}-3 c a b -2 a \,c^{2}+b^{3}+b^{2} c -b \,c^{2}-2 c^{3}\right ) \arctan \left (\frac {\left (a -b +c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{c^{3}}-\frac {2 \left (\frac {\left (c b +\frac {1}{2} c^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (c b -\frac {1}{2} c^{2}\right ) \tan \left (\frac {x}{2}\right )}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}+\frac {\left (2 a c -2 b^{2}+3 c^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{c^{3}}\) | \(406\) |
| risch | \(\text {Expression too large to display}\) | \(5253\) |
2/c^3*(a-b+c)*(1/2*((-4*a*c+b^2)^(1/2)*a*c-(-4*a*c+b^2)^(1/2)*b^2-(-4*a*c+ b^2)^(1/2)*b*c+(-4*a*c+b^2)^(1/2)*c^2+3*c*a*b+2*a*c^2-b^3-b^2*c+b*c^2+2*c^ 3)/(-4*a*c+b^2)^(1/2)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2)*arctanh((-a +b-c)*tan(1/2*x)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2))+1/2*((-4*a*c+b^ 2)^(1/2)*a*c-(-4*a*c+b^2)^(1/2)*b^2-(-4*a*c+b^2)^(1/2)*b*c+(-4*a*c+b^2)^(1 /2)*c^2-3*c*a*b-2*a*c^2+b^3+b^2*c-b*c^2-2*c^3)/(-4*a*c+b^2)^(1/2)/(((-4*a* c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)*arctan((a-b+c)*tan(1/2*x)/(((-4*a*c+b^2)^ (1/2)+a-c)*(a-b+c))^(1/2)))-2/c^3*(((c*b+1/2*c^2)*tan(1/2*x)^3+(c*b-1/2*c^ 2)*tan(1/2*x))/(1+tan(1/2*x)^2)^2+1/2*(2*a*c-2*b^2+3*c^2)*arctan(tan(1/2*x )))
Leaf count of result is larger than twice the leaf count of optimal. 5045 vs. \(2 (323) = 646\).
Time = 2.40 (sec) , antiderivative size = 5045, normalized size of antiderivative = 13.00 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int { \frac {\sin \left (x\right )^{4}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a} \,d x } \]
1/4*(4*c^3*integrate(-2*(2*(b^4 - 2*a*b^2*c - 2*b^2*c^2)*cos(3*x)^2 + 4*(2 *a^2*b^2 - 5*a^2*c^2 - 4*a*c^3 - c^4 - (2*a^3 - a*b^2)*c)*cos(2*x)^2 + 2*( b^4 - 2*a*b^2*c - 2*b^2*c^2)*cos(x)^2 + 2*(b^4 - 2*a*b^2*c - 2*b^2*c^2)*si n(3*x)^2 + 4*(2*a^2*b^2 - 5*a^2*c^2 - 4*a*c^3 - c^4 - (2*a^3 - a*b^2)*c)*s in(2*x)^2 + 2*(4*a*b^3 - 10*a*b*c^2 - 4*b*c^3 - (6*a^2*b - b^3)*c)*sin(2*x )*sin(x) + 2*(b^4 - 2*a*b^2*c - 2*b^2*c^2)*sin(x)^2 + ((b^3*c - 2*a*b*c^2 - 2*b*c^3)*cos(3*x) + 2*(a*b^2*c - a^2*c^2 - 2*a*c^3 - c^4)*cos(2*x) + (b^ 3*c - 2*a*b*c^2 - 2*b*c^3)*cos(x))*cos(4*x) + (b^3*c - 2*a*b*c^2 - 2*b*c^3 + 2*(4*a*b^3 - 10*a*b*c^2 - 4*b*c^3 - (6*a^2*b - b^3)*c)*cos(2*x) + 4*(b^ 4 - 2*a*b^2*c - 2*b^2*c^2)*cos(x))*cos(3*x) + 2*(a*b^2*c - a^2*c^2 - 2*a*c ^3 - c^4 + (4*a*b^3 - 10*a*b*c^2 - 4*b*c^3 - (6*a^2*b - b^3)*c)*cos(x))*co s(2*x) + (b^3*c - 2*a*b*c^2 - 2*b*c^3)*cos(x) + ((b^3*c - 2*a*b*c^2 - 2*b* c^3)*sin(3*x) + 2*(a*b^2*c - a^2*c^2 - 2*a*c^3 - c^4)*sin(2*x) + (b^3*c - 2*a*b*c^2 - 2*b*c^3)*sin(x))*sin(4*x) + 2*((4*a*b^3 - 10*a*b*c^2 - 4*b*c^3 - (6*a^2*b - b^3)*c)*sin(2*x) + 2*(b^4 - 2*a*b^2*c - 2*b^2*c^2)*sin(x))*s in(3*x))/(c^5*cos(4*x)^2 + 4*b^2*c^3*cos(3*x)^2 + 4*b^2*c^3*cos(x)^2 + c^5 *sin(4*x)^2 + 4*b^2*c^3*sin(3*x)^2 + 4*b^2*c^3*sin(x)^2 + 4*b*c^4*cos(x) + c^5 + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*cos(2*x)^2 + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*sin(2*x)^2 + 8*(2*a*b*c^3 + b*c^4)*sin(2*x)*sin(x) + 2*(2*b*c^4*cos( 3*x) + 2*b*c^4*cos(x) + c^5 + 2*(2*a*c^4 + c^5)*cos(2*x))*cos(4*x) + 4*...
Leaf count of result is larger than twice the leaf count of optimal. 11375 vs. \(2 (323) = 646\).
Time = 2.80 (sec) , antiderivative size = 11375, normalized size of antiderivative = 29.32 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
((2*a^2*b^6 - 4*a*b^7 + 2*b^8 - 18*a^3*b^4*c + 38*a^2*b^5*c - 18*a*b^6*c - 2*b^7*c + 48*a^4*b^2*c^2 - 112*a^3*b^3*c^2 + 42*a^2*b^4*c^2 + 28*a*b^5*c^ 2 - 4*b^6*c^2 - 32*a^5*c^3 + 96*a^4*b*c^3 + 16*a^3*b^2*c^3 - 128*a^2*b^3*c ^3 + 26*a*b^4*c^3 + 6*b^5*c^3 - 96*a^4*c^4 + 192*a^3*b*c^4 - 16*a^2*b^2*c^ 4 - 48*a*b^3*c^4 - 2*b^4*c^4 - 96*a^3*c^5 + 96*a^2*b*c^5 + 16*a*b^2*c^5 - 32*a^2*c^6 + 3*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c)) *sqrt(b^2 - 4*a*c)*a^2*b^4 - 2*(b^2 - 4*a*c)*a^2*b^4 - 2*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a*b^5 + 4*(b^ 2 - 4*a*c)*a*b^5 - 5*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*b^6 - 2*(b^2 - 4*a*c)*b^6 - 15*sqrt(a^2 - a*b + b *c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^3*b^2*c + 10 *(b^2 - 4*a*c)*a^3*b^2*c + 13*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a* c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^2*b^3*c - 22*(b^2 - 4*a*c)*a^2*b^3*c + 37*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a*b^4*c + 10*(b^2 - 4*a*c)*a*b^4*c + sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*b^5*c + 2*(b^2 - 4*a*c)*b ^5*c + 12*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt (b^2 - 4*a*c)*a^4*c^2 - 8*(b^2 - 4*a*c)*a^4*c^2 - 20*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^3*b*c^2 + 24*(b ^2 - 4*a*c)*a^3*b*c^2 - 85*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*...
Time = 15.41 (sec) , antiderivative size = 46613, normalized size of antiderivative = 120.14 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
atan(((((((2048*(48*a*c^15 + 272*a^2*c^14 + 576*a^3*c^13 + 576*a^4*c^12 + 272*a^5*c^11 + 48*a^6*c^10 - 12*b^2*c^14 + 20*b^3*c^13 + 18*b^4*c^12 - 46* b^5*c^11 + 6*b^6*c^10 + 26*b^7*c^9 - 12*b^8*c^8 - 140*a*b^2*c^13 + 288*a*b ^3*c^12 + 30*a*b^4*c^11 - 240*a*b^5*c^10 + 74*a*b^6*c^9 + 20*a*b^7*c^8 - 4 16*a^2*b*c^13 - 736*a^3*b*c^12 - 544*a^4*b*c^11 - 144*a^5*b*c^10 - 360*a^2 *b^2*c^12 + 728*a^2*b^3*c^11 - 50*a^2*b^4*c^10 - 182*a^2*b^5*c^9 + 4*a^2*b ^6*c^8 - 360*a^3*b^2*c^11 + 544*a^3*b^3*c^10 + 10*a^3*b^4*c^9 - 20*a^3*b^5 *c^8 - 172*a^4*b^2*c^10 + 116*a^4*b^3*c^9 + 8*a^4*b^4*c^8 - 44*a^5*b^2*c^9 - 80*a*b*c^14))/c^8 - (2048*tan(x/2)*(-(8*a*c^7 + b^8 + 24*a^2*c^6 + 24*a ^3*c^5 + 8*a^4*c^4 + b^5*(-(4*a*c - b^2)^3)^(1/2) - 2*b^2*c^6 + 3*b^4*c^4 - 3*b^6*c^2 - 18*a*b^2*c^5 + 24*a*b^4*c^3 + 3*b*c^4*(-(4*a*c - b^2)^3)^(1/ 2) - 54*a^2*b^2*c^4 + 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - 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) + 6*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*( 16*a^2*c^8 + b^4*c^6 - 8*a*b^2*c^7)))^(1/2)*(32*a*c^16 - 64*a^2*c^15 - 128 *a^3*c^14 + 64*a^4*c^13 + 96*a^5*c^12 - 8*b^2*c^15 + 24*b^3*c^14 - 32*b^4* c^13 + 32*b^5*c^12 - 24*b^6*c^11 + 8*b^7*c^10 + 144*a*b^2*c^14 - 200*a*b^3 *c^13 + 184*a*b^4*c^12 - 56*a*b^5*c^11 - 8*a*b^6*c^10 + 288*a^2*b*c^14 + 3 52*a^3*b*c^13 - 32*a^4*b*c^12 - 320*a^2*b^2*c^13 + 8*a^2*b^3*c^12 + 96*a^2 *b^4*c^11 - 8*a^2*b^5*c^10 - 272*a^3*b^2*c^12 + 40*a^3*b^3*c^11 + 8*a^3...