Integrand size = 25, antiderivative size = 414 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {B d-A e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac {A e (2 c d-b e)-B \left (c d^2-a e^2\right )}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\left (A b^3 e^3-b^2 e^2 (3 A c d+a B e)+b c \left (B c d^3+3 A c d^2 e+3 a B d e^2-3 a A e^3\right )-2 c \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}-\frac {\left (B \left (c^2 d^3-3 a c d e^2+a b e^3\right )-A e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}+\frac {\left (B \left (c^2 d^3-3 a c d e^2+a b e^3\right )-A e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^3} \]
1/2*(-A*e+B*d)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2+(-A*e*(-b*e+2*c*d)+B*(-a*e^2+ c*d^2))/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)-(B*(a*b*e^3-3*a*c*d*e^2+c^2*d^3)-A*e *(3*c^2*d^2+b^2*e^2-c*e*(a*e+3*b*d)))*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)^3+1/2* (B*(a*b*e^3-3*a*c*d*e^2+c^2*d^3)-A*e*(3*c^2*d^2+b^2*e^2-c*e*(a*e+3*b*d)))* ln(c*x^2+b*x+a)/(a*e^2-b*d*e+c*d^2)^3+(A*b^3*e^3-b^2*e^2*(3*A*c*d+B*a*e)+b *c*(-3*A*a*e^3+3*A*c*d^2*e+3*B*a*d*e^2+B*c*d^3)-2*c*(A*c*d*(-3*a*e^2+c*d^2 )+a*B*e*(-a*e^2+3*c*d^2)))*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(a*e^2-b* d*e+c*d^2)^3/(-4*a*c+b^2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {B d-A e}{2 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}+\frac {A e (-2 c d+b e)+B \left (c d^2-a e^2\right )}{\left (c d^2+e (-b d+a e)\right )^2 (d+e x)}+\frac {\left (A b^3 e^3-b^2 e^2 (3 A c d+a B e)+b c \left (B c d^3+3 A c d^2 e+3 a B d e^2-3 a A e^3\right )+2 c \left (a B e \left (-3 c d^2+a e^2\right )+A c d \left (-c d^2+3 a e^2\right )\right )\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )^3}-\frac {\left (B \left (c^2 d^3-3 a c d e^2+a b e^3\right )+A e \left (-3 c^2 d^2-b^2 e^2+c e (3 b d+a e)\right )\right ) \log (d+e x)}{\left (c d^2+e (-b d+a e)\right )^3}+\frac {\left (B \left (c^2 d^3-3 a c d e^2+a b e^3\right )+A e \left (-3 c^2 d^2-b^2 e^2+c e (3 b d+a e)\right )\right ) \log (a+x (b+c x))}{2 \left (c d^2+e (-b d+a e)\right )^3} \]
(B*d - A*e)/(2*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^2) + (A*e*(-2*c*d + b* e) + B*(c*d^2 - a*e^2))/((c*d^2 + e*(-(b*d) + a*e))^2*(d + e*x)) + ((A*b^3 *e^3 - b^2*e^2*(3*A*c*d + a*B*e) + b*c*(B*c*d^3 + 3*A*c*d^2*e + 3*a*B*d*e^ 2 - 3*a*A*e^3) + 2*c*(a*B*e*(-3*c*d^2 + a*e^2) + A*c*d*(-(c*d^2) + 3*a*e^2 )))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(Sqrt[-b^2 + 4*a*c]*(-(c*d^2) + e*(b*d - a*e))^3) - ((B*(c^2*d^3 - 3*a*c*d*e^2 + a*b*e^3) + A*e*(-3*c^2* d^2 - b^2*e^2 + c*e*(3*b*d + a*e)))*Log[d + e*x])/(c*d^2 + e*(-(b*d) + a*e ))^3 + ((B*(c^2*d^3 - 3*a*c*d*e^2 + a*b*e^3) + A*e*(-3*c^2*d^2 - b^2*e^2 + c*e*(3*b*d + a*e)))*Log[a + x*(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a*e))^3 )
Time = 0.92 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1200, 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.
| \(\displaystyle \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \int \left (\frac {e \left (A e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )-B \left (a b e^3-3 a c d e^2+c^2 d^3\right )\right )}{(d+e x) \left (a e^2-b d e+c d^2\right )^3}+\frac {A \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )+c x \left (B \left (a b e^3-3 a c d e^2+c^2 d^3\right )-A e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )\right )+a B e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac {e (A e-B d)}{(d+e x)^3 \left (a e^2-b d e+c d^2\right )}+\frac {e \left (A e (2 c d-b e)-B \left (c d^2-a e^2\right )\right )}{(d+e x)^2 \left (a e^2-b d e+c d^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-b^2 e^2 (a B e+3 A c d)+b c \left (-3 a A e^3+3 a B d e^2+3 A c d^2 e+B c d^3\right )-2 c \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )+A b^3 e^3\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^3}+\frac {\log \left (a+b x+c x^2\right ) \left (B \left (a b e^3-3 a c d e^2+c^2 d^3\right )-A e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )\right )}{2 \left (a e^2-b d e+c d^2\right )^3}-\frac {\log (d+e x) \left (B \left (a b e^3-3 a c d e^2+c^2 d^3\right )-A e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )\right )}{\left (a e^2-b d e+c d^2\right )^3}+\frac {B d-A e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac {A e (2 c d-b e)-B \left (c d^2-a e^2\right )}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}\) |
(B*d - A*e)/(2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - (A*e*(2*c*d - b*e) - B*(c*d^2 - a*e^2))/((c*d^2 - b*d*e + a*e^2)^2*(d + e*x)) + ((A*b^3*e^3 - b^2*e^2*(3*A*c*d + a*B*e) + b*c*(B*c*d^3 + 3*A*c*d^2*e + 3*a*B*d*e^2 - 3*a *A*e^3) - 2*c*(A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(3*c*d^2 - a*e^2)))*ArcTanh [(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2 )^3) - ((B*(c^2*d^3 - 3*a*c*d*e^2 + a*b*e^3) - A*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e)))*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^3 + ((B*(c^2*d^3 - 3*a*c*d*e^2 + a*b*e^3) - A*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))) *Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^3)
3.24.70.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Time = 0.45 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.18
| method | result | size |
| default | \(-\frac {A e -B d}{2 \left (e^{2} a -b d e +c \,d^{2}\right ) \left (e x +d \right )^{2}}+\frac {A b \,e^{2}-2 A c d e -B a \,e^{2}+B c \,d^{2}}{\left (e^{2} a -b d e +c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {\left (A a c \,e^{3}-A \,b^{2} e^{3}+3 A b c d \,e^{2}-3 A \,c^{2} d^{2} e +B a b \,e^{3}-3 B a c d \,e^{2}+B \,c^{2} d^{3}\right ) \ln \left (e x +d \right )}{\left (e^{2} a -b d e +c \,d^{2}\right )^{3}}+\frac {\frac {\left (A a \,c^{2} e^{3}-A \,b^{2} c \,e^{3}+3 A b \,c^{2} d \,e^{2}-3 A \,c^{3} d^{2} e +B a b c \,e^{3}-3 B a \,c^{2} d \,e^{2}+B \,c^{3} d^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (2 A a b c \,e^{3}-3 A a \,c^{2} d \,e^{2}-A \,b^{3} e^{3}+3 A \,b^{2} c d \,e^{2}-3 A b \,c^{2} d^{2} e +A \,c^{3} d^{3}-B \,e^{3} c \,a^{2}+B a \,b^{2} e^{3}-3 B a b c d \,e^{2}+3 B a \,c^{2} d^{2} e -\frac {\left (A a \,c^{2} e^{3}-A \,b^{2} c \,e^{3}+3 A b \,c^{2} d \,e^{2}-3 A \,c^{3} d^{2} e +B a b c \,e^{3}-3 B a \,c^{2} d \,e^{2}+B \,c^{3} d^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (e^{2} a -b d e +c \,d^{2}\right )^{3}}\) | \(487\) |
| risch | \(\text {Expression too large to display}\) | \(3252\) |
-1/2*(A*e-B*d)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2+(A*b*e^2-2*A*c*d*e-B*a*e^2+B* c*d^2)/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)-(A*a*c*e^3-A*b^2*e^3+3*A*b*c*d*e^2-3* A*c^2*d^2*e+B*a*b*e^3-3*B*a*c*d*e^2+B*c^2*d^3)/(a*e^2-b*d*e+c*d^2)^3*ln(e* x+d)+1/(a*e^2-b*d*e+c*d^2)^3*(1/2*(A*a*c^2*e^3-A*b^2*c*e^3+3*A*b*c^2*d*e^2 -3*A*c^3*d^2*e+B*a*b*c*e^3-3*B*a*c^2*d*e^2+B*c^3*d^3)/c*ln(c*x^2+b*x+a)+2* (2*A*a*b*c*e^3-3*A*a*c^2*d*e^2-A*b^3*e^3+3*A*b^2*c*d*e^2-3*A*b*c^2*d^2*e+A *c^3*d^3-B*e^3*c*a^2+B*a*b^2*e^3-3*B*a*b*c*d*e^2+3*B*a*c^2*d^2*e-1/2*(A*a* c^2*e^3-A*b^2*c*e^3+3*A*b*c^2*d*e^2-3*A*c^3*d^2*e+B*a*b*c*e^3-3*B*a*c^2*d* e^2+B*c^3*d^3)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Timed out. \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 844 vs. \(2 (405) = 810\).
Time = 0.29 (sec) , antiderivative size = 844, normalized size of antiderivative = 2.04 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + 3 \, A b c d e^{2} + B a b e^{3} - A b^{2} e^{3} + A a c e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )}} - \frac {{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} + 3 \, A b c d e^{3} + B a b e^{4} - A b^{2} e^{4} + A a c e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} + 3 \, a c^{2} d^{4} e^{3} - b^{3} d^{3} e^{4} - 6 \, a b c d^{3} e^{4} + 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} c d^{2} e^{5} - 3 \, a^{2} b d e^{6} + a^{3} e^{7}} - \frac {{\left (B b c^{2} d^{3} - 2 \, A c^{3} d^{3} - 6 \, B a c^{2} d^{2} e + 3 \, A b c^{2} d^{2} e + 3 \, B a b c d e^{2} - 3 \, A b^{2} c d e^{2} + 6 \, A a c^{2} d e^{2} - B a b^{2} e^{3} + A b^{3} e^{3} + 2 \, B a^{2} c e^{3} - 3 \, A a b c e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {3 \, B c^{2} d^{5} - 4 \, B b c d^{4} e - 5 \, A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 2 \, B a c d^{3} e^{2} + 8 \, A b c d^{3} e^{2} - 3 \, A b^{2} d^{2} e^{3} - 6 \, A a c d^{2} e^{3} - B a^{2} d e^{4} + 4 \, A a b d e^{4} - A a^{2} e^{5} + 2 \, {\left (B c^{2} d^{4} e - B b c d^{3} e^{2} - 2 \, A c^{2} d^{3} e^{2} + 3 \, A b c d^{2} e^{3} + B a b d e^{4} - A b^{2} d e^{4} - 2 \, A a c d e^{4} - B a^{2} e^{5} + A a b e^{5}\right )} x}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}^{3} {\left (e x + d\right )}^{2}} \]
1/2*(B*c^2*d^3 - 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + 3*A*b*c*d*e^2 + B*a*b*e^3 - A*b^2*e^3 + A*a*c*e^3)*log(c*x^2 + b*x + a)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^ 2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a^3*e^6) - (B*c^2*d^3*e - 3* A*c^2*d^2*e^2 - 3*B*a*c*d*e^3 + 3*A*b*c*d*e^3 + B*a*b*e^4 - A*b^2*e^4 + A* a*c*e^4)*log(abs(e*x + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e^3 + 3*a*c^2*d^4*e^3 - b^3*d^3*e^4 - 6*a*b*c*d^3*e^4 + 3*a*b^2*d^2*e^5 + 3*a^ 2*c*d^2*e^5 - 3*a^2*b*d*e^6 + a^3*e^7) - (B*b*c^2*d^3 - 2*A*c^3*d^3 - 6*B* a*c^2*d^2*e + 3*A*b*c^2*d^2*e + 3*B*a*b*c*d*e^2 - 3*A*b^2*c*d*e^2 + 6*A*a* c^2*d*e^2 - B*a*b^2*e^3 + A*b^3*e^3 + 2*B*a^2*c*e^3 - 3*A*a*b*c*e^3)*arcta n((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4* e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a^3*e^6)*sqrt(-b^2 + 4*a*c)) + 1/2*(3*B* c^2*d^5 - 4*B*b*c*d^4*e - 5*A*c^2*d^4*e + B*b^2*d^3*e^2 + 2*B*a*c*d^3*e^2 + 8*A*b*c*d^3*e^2 - 3*A*b^2*d^2*e^3 - 6*A*a*c*d^2*e^3 - B*a^2*d*e^4 + 4*A* a*b*d*e^4 - A*a^2*e^5 + 2*(B*c^2*d^4*e - B*b*c*d^3*e^2 - 2*A*c^2*d^3*e^2 + 3*A*b*c*d^2*e^3 + B*a*b*d*e^4 - A*b^2*d*e^4 - 2*A*a*c*d*e^4 - B*a^2*e^5 + A*a*b*e^5)*x)/((c*d^2 - b*d*e + a*e^2)^3*(e*x + d)^2)
Time = 44.00 (sec) , antiderivative size = 7042, normalized size of antiderivative = 17.01 \[ \int \frac {A+B x}{(d+e x)^3 \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \]
(log(((A*b^4*e^3 - (3*A*b*e^3*(b^2 - 4*a*c)^(3/2))/4 + (B*a*e^3*(b^2 - 4*a *c)^(3/2))/2 - B*a*b^3*e^3 + 4*B*a*c^3*d^3 - (A*b^3*e^3*(b^2 - 4*a*c)^(1/2 ))/4 + 2*A*c^3*d^3*(b^2 - 4*a*c)^(1/2) + 4*A*a^2*c^2*e^3 - B*b^2*c^2*d^3 + (B*a*b^2*e^3*(b^2 - 4*a*c)^(1/2))/2 - B*b*c^2*d^3*(b^2 - 4*a*c)^(1/2) - ( 3*B*b^3*d*e^2*(b^2 - 4*a*c)^(1/2))/4 + 3*A*b^2*c^2*d^2*e - 12*B*a^2*c^2*d* e^2 + (3*A*c*d*e^2*(b^2 - 4*a*c)^(3/2))/2 + (3*B*b*d*e^2*(b^2 - 4*a*c)^(3/ 2))/4 - (3*B*c*d^2*e*(b^2 - 4*a*c)^(3/2))/2 - 5*A*a*b^2*c*e^3 + 4*B*a^2*b* c*e^3 - 12*A*a*c^3*d^2*e - 3*A*b^3*c*d*e^2 + 12*A*a*b*c^2*d*e^2 + 3*B*a*b^ 2*c*d*e^2 - 3*A*b*c^2*d^2*e*(b^2 - 4*a*c)^(1/2) + (3*A*b^2*c*d*e^2*(b^2 - 4*a*c)^(1/2))/2 + (3*B*b^2*c*d^2*e*(b^2 - 4*a*c)^(1/2))/2)*(4*B*d*e^5*(b^2 - 4*a*c)^(7/2) + 3*B*e^6*x*(b^2 - 4*a*c)^(7/2) - 3*A*b^2*e^6*(b^2 - 4*a*c )^(5/2) + 2*A*b^4*e^6*(b^2 - 4*a*c)^(3/2) + A*b^6*e^6*(b^2 - 4*a*c)^(1/2) + 128*B*a^4*c^3*e^6 - 32*A*b^6*c*e^6*x - 2*B*a*b^5*e^6*(b^2 - 4*a*c)^(1/2) + B*b^2*d*e^5*(b^2 - 4*a*c)^(5/2) - 10*B*b^4*d*e^5*(b^2 - 4*a*c)^(3/2) + 5*B*b^6*d*e^5*(b^2 - 4*a*c)^(1/2) + B*b^2*e^6*x*(b^2 - 4*a*c)^(5/2) - 3*B* b^4*e^6*x*(b^2 - 4*a*c)^(3/2) - B*b^6*e^6*x*(b^2 - 4*a*c)^(1/2) - 320*A*a^ 3*b*c^3*e^6 + 48*B*a^2*b^4*c*e^6 + 640*A*a^3*c^4*d*e^5 - 32*A*b^2*c^5*d^5* e + 48*B*b^3*c^4*d^5*e - 48*A*c^2*d^2*e^4*(b^2 - 4*a*c)^(5/2) - 64*A*c^4*d ^4*e^2*(b^2 - 4*a*c)^(3/2) + 48*B*c^2*d^3*e^3*(b^2 - 4*a*c)^(5/2) + 272*A* a^2*b^3*c^2*e^6 - 224*B*a^3*b^2*c^2*e^6 - 1280*A*a^2*c^5*d^3*e^3 - 48*A...