Integrand size = 25, antiderivative size = 357 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}-\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^4} \]
e*(A*c*e*(-b*e+3*c*d)+B*(3*c^2*d^2+b^2*e^2-c*e*(a*e+3*b*d)))*x/c^3+1/2*e^2 *(A*c*e-B*b*e+3*B*c*d)*x^2/c^2+1/3*B*e^3*x^3/c+1/2*(A*c*e*(3*c^2*d^2+b^2*e ^2-c*e*(a*e+3*b*d))+B*(c^3*d^3-b^3*e^3-3*c^2*d*e*(a*e+b*d)+b*c*e^2*(2*a*e+ 3*b*d)))*ln(c*x^2+b*x+a)/c^4-(b^4*B*e^3-b^3*c*e^2*(A*e+3*B*d)+b^2*c*e*(3*A *c*d*e-4*B*a*e^2+3*B*c*d^2)-b*c^2*(-3*A*a*e^3+3*A*c*d^2*e-9*B*a*d*e^2+B*c* d^3)+2*c^2*(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))/c^4/(-4*a*c+b^2)^(1/2)
Time = 0.19 (sec) , antiderivative size = 352, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\frac {6 c e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x+3 c^2 e^2 (3 B c d-b B e+A c e) x^2+2 B c^3 e^3 x^3+\frac {6 \left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )+b c^2 \left (-B c d^3-3 A c d^2 e+9 a B d e^2+3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (-3 c d^2+a e^2\right )\right )\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+3 \left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log (a+x (b+c x))}{6 c^4} \]
(6*c*e*(A*c*e*(3*c*d - b*e) + B*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))) *x + 3*c^2*e^2*(3*B*c*d - b*B*e + A*c*e)*x^2 + 2*B*c^3*e^3*x^3 + (6*(b^4*B *e^3 - b^3*c*e^2*(3*B*d + A*e) + b^2*c*e*(3*B*c*d^2 + 3*A*c*d*e - 4*a*B*e^ 2) + b*c^2*(-(B*c*d^3) - 3*A*c*d^2*e + 9*a*B*d*e^2 + 3*a*A*e^3) + 2*c^2*(A *c*d*(c*d^2 - 3*a*e^2) + a*B*e*(-3*c*d^2 + a*e^2)))*ArcTan[(b + 2*c*x)/Sqr t[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + 3*(A*c*e*(3*c^2*d^2 + b^2*e^2 - c*e *(3*b*d + a*e)) + B*(c^3*d^3 - b^3*e^3 - 3*c^2*d*e*(b*d + a*e) + b*c*e^2*( 3*b*d + 2*a*e)))*Log[a + x*(b + c*x)])/(6*c^4)
Time = 0.75 (sec) , antiderivative size = 357, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \int \left (\frac {e \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}+\frac {x \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )+A c \left (a b e^3-3 a c d e^2+c^2 d^3\right )-a B e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{c^3 \left (a+b x+c x^2\right )}+\frac {e^2 x (A c e-b B e+3 B c d)}{c^2}+\frac {B e^3 x^2}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )-b c^2 \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}+\frac {\log \left (a+b x+c x^2\right ) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )}{2 c^4}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {B e^3 x^3}{3 c}\) |
(e*(A*c*e*(3*c*d - b*e) + B*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e)))*x)/ c^3 + (e^2*(3*B*c*d - b*B*e + A*c*e)*x^2)/(2*c^2) + (B*e^3*x^3)/(3*c) - (( b^4*B*e^3 - b^3*c*e^2*(3*B*d + A*e) + b^2*c*e*(3*B*c*d^2 + 3*A*c*d*e - 4*a *B*e^2) - b*c^2*(B*c*d^3 + 3*A*c*d^2*e - 9*a*B*d*e^2 - 3*a*A*e^3) + 2*c^2* (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)/S qrt[b^2 - 4*a*c]])/(c^4*Sqrt[b^2 - 4*a*c]) + ((A*c*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e)) + B*(c^3*d^3 - b^3*e^3 - 3*c^2*d*e*(b*d + a*e) + b*c* e^2*(3*b*d + 2*a*e)))*Log[a + b*x + c*x^2])/(2*c^4)
3.24.64.3.1 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 = 0.55 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(-\frac {e \left (-\frac {1}{3} B \,c^{2} x^{3} e^{2}-\frac {1}{2} A \,c^{2} e^{2} x^{2}+\frac {1}{2} B b c \,e^{2} x^{2}-\frac {3}{2} B \,c^{2} d e \,x^{2}+A b c \,e^{2} x -3 A \,c^{2} d e x +B a c \,e^{2} x -B \,b^{2} e^{2} x +3 B b c d e x -3 B \,c^{2} d^{2} x \right )}{c^{3}}+\frac {\frac {\left (-A a \,c^{2} e^{3}+A \,b^{2} c \,e^{3}-3 A b \,c^{2} d \,e^{2}+3 A \,c^{3} d^{2} e +2 B a b c \,e^{3}-3 B a \,c^{2} d \,e^{2}-B \,b^{3} e^{3}+3 B \,b^{2} c d \,e^{2}-3 B b \,c^{2} d^{2} e +B \,c^{3} d^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (A a b c \,e^{3}-3 A a \,c^{2} d \,e^{2}+A \,c^{3} d^{3}+B \,e^{3} c \,a^{2}-B a \,b^{2} e^{3}+3 B a b c d \,e^{2}-3 B a \,c^{2} d^{2} e -\frac {\left (-A a \,c^{2} e^{3}+A \,b^{2} c \,e^{3}-3 A b \,c^{2} d \,e^{2}+3 A \,c^{3} d^{2} e +2 B a b c \,e^{3}-3 B a \,c^{2} d \,e^{2}-B \,b^{3} e^{3}+3 B \,b^{2} c d \,e^{2}-3 B b \,c^{2} d^{2} e +B \,c^{3} d^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\) | \(437\) |
| risch | \(\text {Expression too large to display}\) | \(23578\) |
-e/c^3*(-1/3*B*c^2*x^3*e^2-1/2*A*c^2*e^2*x^2+1/2*B*b*c*e^2*x^2-3/2*B*c^2*d *e*x^2+A*b*c*e^2*x-3*A*c^2*d*e*x+B*a*c*e^2*x-B*b^2*e^2*x+3*B*b*c*d*e*x-3*B *c^2*d^2*x)+1/c^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+2*B*a*b*c*e^3-3*B*a*c^2*d*e^2-B*b^3*e^3+3*B*b^2*c*d*e^2-3*B*b*c^2*d^2 *e+B*c^3*d^3)/c*ln(c*x^2+b*x+a)+2*(A*a*b*c*e^3-3*A*a*c^2*d*e^2+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+2*B*a*b*c*e^3-3*B*a*c^2*d*e^2-B*b ^3*e^3+3*B*b^2*c*d*e^2-3*B*b*c^2*d^2*e+B*c^3*d^3)*b/c)/(4*a*c-b^2)^(1/2)*a rctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Time = 0.56 (sec) , antiderivative size = 1139, normalized size of antiderivative = 3.19 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\text {Too large to display} \]
[1/6*(2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 + 3*(3*(B*b^2*c^3 - 4*B*a*c^4)*d*e ^2 - (B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*e^3)*x^2 - 3*((B*b*c^ 3 - 2*A*c^4)*d^3 - 3*(B*b^2*c^2 - (2*B*a + A*b)*c^3)*d^2*e + 3*(B*b^3*c + 2*A*a*c^3 - (3*B*a*b + A*b^2)*c^2)*d*e^2 - (B*b^4 + (2*B*a^2 + 3*A*a*b)*c^ 2 - (4*B*a*b^2 + A*b^3)*c)*e^3)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 6*(3* (B*b^2*c^3 - 4*B*a*c^4)*d^2*e - 3*(B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^ 2)*c^3)*d*e^2 + (B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^2 + A*b^3)*c^2 )*e^3)*x + 3*((B*b^2*c^3 - 4*B*a*c^4)*d^3 - 3*(B*b^3*c^2 + 4*A*a*c^4 - (4* B*a*b + A*b^2)*c^3)*d^2*e + 3*(B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^ 2 + A*b^3)*c^2)*d*e^2 - (B*b^5 - 4*A*a^2*c^3 + (8*B*a^2*b + 5*A*a*b^2)*c^2 - (6*B*a*b^3 + A*b^4)*c)*e^3)*log(c*x^2 + b*x + a))/(b^2*c^4 - 4*a*c^5), 1/6*(2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 + 3*(3*(B*b^2*c^3 - 4*B*a*c^4)*d*e^ 2 - (B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*e^3)*x^2 + 6*((B*b*c^3 - 2*A*c^4)*d^3 - 3*(B*b^2*c^2 - (2*B*a + A*b)*c^3)*d^2*e + 3*(B*b^3*c + 2 *A*a*c^3 - (3*B*a*b + A*b^2)*c^2)*d*e^2 - (B*b^4 + (2*B*a^2 + 3*A*a*b)*c^2 - (4*B*a*b^2 + A*b^3)*c)*e^3)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a* c)*(2*c*x + b)/(b^2 - 4*a*c)) + 6*(3*(B*b^2*c^3 - 4*B*a*c^4)*d^2*e - 3*(B* b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*d*e^2 + (B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^2 + A*b^3)*c^2)*e^3)*x + 3*((B*b^2*c^3 - 4*B*a*c...
Leaf count of result is larger than twice the leaf count of optimal. 2759 vs. \(2 (360) = 720\).
Time = 13.66 (sec) , antiderivative size = 2759, normalized size of antiderivative = 7.73 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\text {Too large to display} \]
B*e**3*x**3/(3*c) + x**2*(A*e**3/(2*c) - B*b*e**3/(2*c**2) + 3*B*d*e**2/(2 *c)) + x*(-A*b*e**3/c**2 + 3*A*d*e**2/c - B*a*e**3/c**2 + B*b**2*e**3/c**3 - 3*B*b*d*e**2/c**2 + 3*B*d**2*e/c) + (-sqrt(-4*a*c + b**2)*(3*A*a*b*c**2 *e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c **3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B* a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3* B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**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 + 2*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2*e + B*c**3*d**3)/(2*c**4))*log(x + (2*A*a**2*c**2*e**3 - A*a*b**2*c*e**3 + 3*A*a*b*c**2*d*e**2 - 6*A*a*c**3*d**2*e + A*b*c**3*d**3 - 3*B*a**2*b*c*e** 3 + 6*B*a**2*c**2*d*e**2 + B*a*b**3*e**3 - 3*B*a*b**2*c*d*e**2 + 3*B*a*b*c **2*d**2*e - 2*B*a*c**3*d**3 + 4*a*c**4*(-sqrt(-4*a*c + b**2)*(3*A*a*b*c** 2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b* c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B *a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3 *B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**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 + 2*B*a*b*c*e** 3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2* e + B*c**3*d**3)/(2*c**4)) - b**2*c**3*(-sqrt(-4*a*c + b**2)*(3*A*a*b*c...
Exception generated. \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.27 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=\frac {2 \, B c^{2} e^{3} x^{3} + 9 \, B c^{2} d e^{2} x^{2} - 3 \, B b c e^{3} x^{2} + 3 \, A c^{2} e^{3} x^{2} + 18 \, B c^{2} d^{2} e x - 18 \, B b c d e^{2} x + 18 \, A c^{2} d e^{2} x + 6 \, B b^{2} e^{3} x - 6 \, B a c e^{3} x - 6 \, A b c e^{3} x}{6 \, c^{3}} + \frac {{\left (B c^{3} d^{3} - 3 \, B b c^{2} d^{2} e + 3 \, A c^{3} d^{2} e + 3 \, B b^{2} c d e^{2} - 3 \, B a c^{2} d e^{2} - 3 \, A b c^{2} d e^{2} - B b^{3} e^{3} + 2 \, B a b c e^{3} + A b^{2} c e^{3} - A a c^{2} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} - \frac {{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 6 \, B a c^{3} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 9 \, B a b c^{2} d e^{2} - 3 \, A b^{2} c^{2} d e^{2} + 6 \, A a c^{3} d e^{2} - B b^{4} e^{3} + 4 \, B a b^{2} c e^{3} + A b^{3} c e^{3} - 2 \, B a^{2} c^{2} e^{3} - 3 \, A a b c^{2} e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{4}} \]
1/6*(2*B*c^2*e^3*x^3 + 9*B*c^2*d*e^2*x^2 - 3*B*b*c*e^3*x^2 + 3*A*c^2*e^3*x ^2 + 18*B*c^2*d^2*e*x - 18*B*b*c*d*e^2*x + 18*A*c^2*d*e^2*x + 6*B*b^2*e^3* x - 6*B*a*c*e^3*x - 6*A*b*c*e^3*x)/c^3 + 1/2*(B*c^3*d^3 - 3*B*b*c^2*d^2*e + 3*A*c^3*d^2*e + 3*B*b^2*c*d*e^2 - 3*B*a*c^2*d*e^2 - 3*A*b*c^2*d*e^2 - B* b^3*e^3 + 2*B*a*b*c*e^3 + A*b^2*c*e^3 - A*a*c^2*e^3)*log(c*x^2 + b*x + a)/ c^4 - (B*b*c^3*d^3 - 2*A*c^4*d^3 - 3*B*b^2*c^2*d^2*e + 6*B*a*c^3*d^2*e + 3 *A*b*c^3*d^2*e + 3*B*b^3*c*d*e^2 - 9*B*a*b*c^2*d*e^2 - 3*A*b^2*c^2*d*e^2 + 6*A*a*c^3*d*e^2 - B*b^4*e^3 + 4*B*a*b^2*c*e^3 + A*b^3*c*e^3 - 2*B*a^2*c^2 *e^3 - 3*A*a*b*c^2*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^4)
Time = 11.90 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx=x^2\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{2\,c}-\frac {B\,b\,e^3}{2\,c^2}\right )-x\,\left (\frac {b\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{c}-\frac {B\,b\,e^3}{c^2}\right )}{c}-\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}+\frac {B\,a\,e^3}{c^2}\right )-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (-8\,B\,a^2\,b\,c^2\,e^3+12\,B\,a^2\,c^3\,d\,e^2+4\,A\,a^2\,c^3\,e^3+6\,B\,a\,b^3\,c\,e^3-15\,B\,a\,b^2\,c^2\,d\,e^2-5\,A\,a\,b^2\,c^2\,e^3+12\,B\,a\,b\,c^3\,d^2\,e+12\,A\,a\,b\,c^3\,d\,e^2-4\,B\,a\,c^4\,d^3-12\,A\,a\,c^4\,d^2\,e-B\,b^5\,e^3+3\,B\,b^4\,c\,d\,e^2+A\,b^4\,c\,e^3-3\,B\,b^3\,c^2\,d^2\,e-3\,A\,b^3\,c^2\,d\,e^2+B\,b^2\,c^3\,d^3+3\,A\,b^2\,c^3\,d^2\,e\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\right )}+\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (2\,B\,a^2\,c^2\,e^3-4\,B\,a\,b^2\,c\,e^3+9\,B\,a\,b\,c^2\,d\,e^2+3\,A\,a\,b\,c^2\,e^3-6\,B\,a\,c^3\,d^2\,e-6\,A\,a\,c^3\,d\,e^2+B\,b^4\,e^3-3\,B\,b^3\,c\,d\,e^2-A\,b^3\,c\,e^3+3\,B\,b^2\,c^2\,d^2\,e+3\,A\,b^2\,c^2\,d\,e^2-B\,b\,c^3\,d^3-3\,A\,b\,c^3\,d^2\,e+2\,A\,c^4\,d^3\right )}{c^4\,\sqrt {4\,a\,c-b^2}}+\frac {B\,e^3\,x^3}{3\,c} \]
x^2*((A*e^3 + 3*B*d*e^2)/(2*c) - (B*b*e^3)/(2*c^2)) - x*((b*((A*e^3 + 3*B* d*e^2)/c - (B*b*e^3)/c^2))/c - (3*d*e*(A*e + B*d))/c + (B*a*e^3)/c^2) - (l og(a + b*x + c*x^2)*(A*b^4*c*e^3 - 4*B*a*c^4*d^3 - B*b^5*e^3 + 4*A*a^2*c^3 *e^3 + B*b^2*c^3*d^3 - 5*A*a*b^2*c^2*e^3 - 8*B*a^2*b*c^2*e^3 + 3*A*b^2*c^3 *d^2*e - 3*A*b^3*c^2*d*e^2 + 12*B*a^2*c^3*d*e^2 - 3*B*b^3*c^2*d^2*e + 6*B* a*b^3*c*e^3 - 12*A*a*c^4*d^2*e + 3*B*b^4*c*d*e^2 + 12*A*a*b*c^3*d*e^2 + 12 *B*a*b*c^3*d^2*e - 15*B*a*b^2*c^2*d*e^2))/(2*(4*a*c^5 - b^2*c^4)) + (atan( b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2))*(2*A*c^4*d^3 + B*b^4* e^3 - A*b^3*c*e^3 - B*b*c^3*d^3 + 2*B*a^2*c^2*e^3 + 3*A*b^2*c^2*d*e^2 + 3* B*b^2*c^2*d^2*e + 3*A*a*b*c^2*e^3 - 4*B*a*b^2*c*e^3 - 6*A*a*c^3*d*e^2 - 3* A*b*c^3*d^2*e - 6*B*a*c^3*d^2*e - 3*B*b^3*c*d*e^2 + 9*B*a*b*c^2*d*e^2))/(c ^4*(4*a*c - b^2)^(1/2)) + (B*e^3*x^3)/(3*c)