Integrand size = 27, antiderivative size = 240 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=-\frac {\left (a e^2 (3 C d+B e)-c d \left (C d^2+3 e (B d+A e)\right )\right ) x}{c^2}-\frac {e \left (a C e^2-c \left (3 C d^2+e (3 B d+A e)\right )\right ) x^2}{2 c^2}+\frac {e^2 (3 C d+B e) x^3}{3 c}+\frac {C e^3 x^4}{4 c}+\frac {\left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (3 C d+B e)-c d^2 (C d+3 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {\left (B c d \left (c d^2-3 a e^2\right )+(A c-a C) e \left (3 c d^2-a e^2\right )\right ) \log \left (a+c x^2\right )}{2 c^3} \]
-(a*e^2*(B*e+3*C*d)-c*d*(C*d^2+3*e*(A*e+B*d)))*x/c^2-1/2*e*(a*C*e^2-c*(3*C *d^2+e*(A*e+3*B*d)))*x^2/c^2+1/3*e^2*(B*e+3*C*d)*x^3/c+1/4*C*e^3*x^4/c+1/2 *(B*c*d*(-3*a*e^2+c*d^2)+(A*c-C*a)*e*(-a*e^2+3*c*d^2))*ln(c*x^2+a)/c^3+(A* c*d*(-3*a*e^2+c*d^2)+a*(a*e^2*(B*e+3*C*d)-c*d^2*(3*B*e+C*d)))*arctan(x*c^( 1/2)/a^(1/2))/c^(5/2)/a^(1/2)
Time = 0.13 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {\left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (3 C d+B e)-c d^2 (C d+3 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {c x \left (-6 a e^2 (6 C d+2 B e+C e x)+3 c C \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+2 c e \left (3 A e (6 d+e x)+B \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )\right )+6 \left (B c d \left (c d^2-3 a e^2\right )+(A c-a C) e \left (3 c d^2-a e^2\right )\right ) \log \left (a+c x^2\right )}{12 c^3} \]
((A*c*d*(c*d^2 - 3*a*e^2) + a*(a*e^2*(3*C*d + B*e) - c*d^2*(C*d + 3*B*e))) *ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + (c*x*(-6*a*e^2*(6*C*d + 2*B*e + C*e*x) + 3*c*C*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 2*c*e *(3*A*e*(6*d + e*x) + B*(18*d^2 + 9*d*e*x + 2*e^2*x^2))) + 6*(B*c*d*(c*d^2 - 3*a*e^2) + (A*c - a*C)*e*(3*c*d^2 - a*e^2))*Log[a + c*x^2])/(12*c^3)
Time = 0.53 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2160, 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.
| \(\displaystyle \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \int \left (\frac {-a e^2 (B e+3 C d)+3 c d e (A e+B d)+c C d^3}{c^2}+\frac {x \left (e (A c-a C) \left (3 c d^2-a e^2\right )+B c d \left (c d^2-3 a e^2\right )\right )+A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (B e+3 C d)-c d^2 (3 B e+C d)\right )}{c^2 \left (a+c x^2\right )}+\frac {e x \left (-a C e^2+c e (A e+3 B d)+3 c C d^2\right )}{c^2}+\frac {e^2 x^2 (B e+3 C d)}{c}+\frac {C e^3 x^3}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (B e+3 C d)-c d^2 (3 B e+C d)\right )\right )}{\sqrt {a} c^{5/2}}+\frac {\log \left (a+c x^2\right ) \left (e (A c-a C) \left (3 c d^2-a e^2\right )+B c d \left (c d^2-3 a e^2\right )\right )}{2 c^3}+\frac {x \left (-a e^2 (B e+3 C d)+3 c d e (A e+B d)+c C d^3\right )}{c^2}+\frac {e x^2 \left (-a C e^2+c e (A e+3 B d)+3 c C d^2\right )}{2 c^2}+\frac {e^2 x^3 (B e+3 C d)}{3 c}+\frac {C e^3 x^4}{4 c}\) |
((c*C*d^3 + 3*c*d*e*(B*d + A*e) - a*e^2*(3*C*d + B*e))*x)/c^2 + (e*(3*c*C* d^2 - a*C*e^2 + c*e*(3*B*d + A*e))*x^2)/(2*c^2) + (e^2*(3*C*d + B*e)*x^3)/ (3*c) + (C*e^3*x^4)/(4*c) + ((A*c*d*(c*d^2 - 3*a*e^2) + a*(a*e^2*(3*C*d + B*e) - c*d^2*(C*d + 3*B*e)))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2) ) + ((B*c*d*(c*d^2 - 3*a*e^2) + (A*c - a*C)*e*(3*c*d^2 - a*e^2))*Log[a + c *x^2])/(2*c^3)
3.1.43.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.68 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.08
| method | result | size |
| default | \(\frac {\frac {1}{4} c C \,x^{4} e^{3}+\frac {1}{3} B c \,x^{3} e^{3}+C c d \,e^{2} x^{3}+\frac {1}{2} A c \,e^{3} x^{2}+\frac {3}{2} B \,x^{2} c d \,e^{2}-\frac {1}{2} C a \,e^{3} x^{2}+\frac {3}{2} C c \,d^{2} e \,x^{2}+3 A c d \,e^{2} x -B x a \,e^{3}+3 B c \,d^{2} e x -3 C a d \,e^{2} x +C c \,d^{3} x}{c^{2}}+\frac {\frac {\left (-A a c \,e^{3}+3 A \,c^{2} d^{2} e -3 B a c d \,e^{2}+B \,c^{2} d^{3}+C \,a^{2} e^{3}-3 C a c \,d^{2} e \right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (-3 A a c d \,e^{2}+A \,d^{3} c^{2}+a^{2} B \,e^{3}-3 B a c \,d^{2} e +3 C \,a^{2} d \,e^{2}-C a c \,d^{3}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{c^{2}}\) | \(260\) |
| risch | \(\text {Expression too large to display}\) | \(2166\) |
1/c^2*(1/4*c*C*x^4*e^3+1/3*B*c*x^3*e^3+C*c*d*e^2*x^3+1/2*A*c*e^3*x^2+3/2*B *x^2*c*d*e^2-1/2*C*a*e^3*x^2+3/2*C*c*d^2*e*x^2+3*A*c*d*e^2*x-B*x*a*e^3+3*B *c*d^2*e*x-3*C*a*d*e^2*x+C*c*d^3*x)+1/c^2*(1/2*(-A*a*c*e^3+3*A*c^2*d^2*e-3 *B*a*c*d*e^2+B*c^2*d^3+C*a^2*e^3-3*C*a*c*d^2*e)/c*ln(c*x^2+a)+(-3*A*a*c*d* e^2+A*c^2*d^3+B*a^2*e^3-3*B*a*c*d^2*e+3*C*a^2*d*e^2-C*a*c*d^3)/(a*c)^(1/2) *arctan(c*x/(a*c)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 592, normalized size of antiderivative = 2.47 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\left [\frac {3 \, C a c^{2} e^{3} x^{4} + 4 \, {\left (3 \, C a c^{2} d e^{2} + B a c^{2} e^{3}\right )} x^{3} + 6 \, {\left (3 \, C a c^{2} d^{2} e + 3 \, B a c^{2} d e^{2} - {\left (C a^{2} c - A a c^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 12 \, {\left (C a c^{2} d^{3} + 3 \, B a c^{2} d^{2} e - B a^{2} c e^{3} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d e^{2}\right )} x + 6 \, {\left (B a c^{2} d^{3} - 3 \, B a^{2} c d e^{2} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d^{2} e + {\left (C a^{3} - A a^{2} c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}, \frac {3 \, C a c^{2} e^{3} x^{4} + 4 \, {\left (3 \, C a c^{2} d e^{2} + B a c^{2} e^{3}\right )} x^{3} + 6 \, {\left (3 \, C a c^{2} d^{2} e + 3 \, B a c^{2} d e^{2} - {\left (C a^{2} c - A a c^{2}\right )} e^{3}\right )} x^{2} - 12 \, {\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 12 \, {\left (C a c^{2} d^{3} + 3 \, B a c^{2} d^{2} e - B a^{2} c e^{3} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d e^{2}\right )} x + 6 \, {\left (B a c^{2} d^{3} - 3 \, B a^{2} c d e^{2} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d^{2} e + {\left (C a^{3} - A a^{2} c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}\right ] \]
[1/12*(3*C*a*c^2*e^3*x^4 + 4*(3*C*a*c^2*d*e^2 + B*a*c^2*e^3)*x^3 + 6*(3*C* a*c^2*d^2*e + 3*B*a*c^2*d*e^2 - (C*a^2*c - A*a*c^2)*e^3)*x^2 + 6*(3*B*a*c* d^2*e - B*a^2*e^3 + (C*a*c - A*c^2)*d^3 - 3*(C*a^2 - A*a*c)*d*e^2)*sqrt(-a *c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 12*(C*a*c^2*d^3 + 3*B* a*c^2*d^2*e - B*a^2*c*e^3 - 3*(C*a^2*c - A*a*c^2)*d*e^2)*x + 6*(B*a*c^2*d^ 3 - 3*B*a^2*c*d*e^2 - 3*(C*a^2*c - A*a*c^2)*d^2*e + (C*a^3 - A*a^2*c)*e^3) *log(c*x^2 + a))/(a*c^3), 1/12*(3*C*a*c^2*e^3*x^4 + 4*(3*C*a*c^2*d*e^2 + B *a*c^2*e^3)*x^3 + 6*(3*C*a*c^2*d^2*e + 3*B*a*c^2*d*e^2 - (C*a^2*c - A*a*c^ 2)*e^3)*x^2 - 12*(3*B*a*c*d^2*e - B*a^2*e^3 + (C*a*c - A*c^2)*d^3 - 3*(C*a ^2 - A*a*c)*d*e^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 12*(C*a*c^2*d^3 + 3*B *a*c^2*d^2*e - B*a^2*c*e^3 - 3*(C*a^2*c - A*a*c^2)*d*e^2)*x + 6*(B*a*c^2*d ^3 - 3*B*a^2*c*d*e^2 - 3*(C*a^2*c - A*a*c^2)*d^2*e + (C*a^3 - A*a^2*c)*e^3 )*log(c*x^2 + a))/(a*c^3)]
Leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (224) = 448\).
Time = 5.52 (sec) , antiderivative size = 1008, normalized size of antiderivative = 4.20 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {C e^{3} x^{4}}{4 c} + x^{3} \left (\frac {B e^{3}}{3 c} + \frac {C d e^{2}}{c}\right ) + x^{2} \left (\frac {A e^{3}}{2 c} + \frac {3 B d e^{2}}{2 c} - \frac {C a e^{3}}{2 c^{2}} + \frac {3 C d^{2} e}{2 c}\right ) + x \left (\frac {3 A d e^{2}}{c} - \frac {B a e^{3}}{c^{2}} + \frac {3 B d^{2} e}{c} - \frac {3 C a d e^{2}}{c^{2}} + \frac {C d^{3}}{c}\right ) + \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {A a^{2} c e^{3} - 3 A a c^{2} d^{2} e + 3 B a^{2} c d e^{2} - B a c^{2} d^{3} - C a^{3} e^{3} + 3 C a^{2} c d^{2} e + 2 a c^{3} \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right )}{- 3 A a c^{2} d e^{2} + A c^{3} d^{3} + B a^{2} c e^{3} - 3 B a c^{2} d^{2} e + 3 C a^{2} c d e^{2} - C a c^{2} d^{3}} \right )} + \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {A a^{2} c e^{3} - 3 A a c^{2} d^{2} e + 3 B a^{2} c d e^{2} - B a c^{2} d^{3} - C a^{3} e^{3} + 3 C a^{2} c d^{2} e + 2 a c^{3} \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right )}{- 3 A a c^{2} d e^{2} + A c^{3} d^{3} + B a^{2} c e^{3} - 3 B a c^{2} d^{2} e + 3 C a^{2} c d e^{2} - C a c^{2} d^{3}} \right )} \]
C*e**3*x**4/(4*c) + x**3*(B*e**3/(3*c) + C*d*e**2/c) + x**2*(A*e**3/(2*c) + 3*B*d*e**2/(2*c) - C*a*e**3/(2*c**2) + 3*C*d**2*e/(2*c)) + x*(3*A*d*e**2 /c - B*a*e**3/c**2 + 3*B*d**2*e/c - 3*C*a*d*e**2/c**2 + C*d**3/c) + ((-A*a *c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3 *C*a*c*d**2*e)/(2*c**3) - sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B *a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e**2 - C*a*c*d**3)/(2*a*c**6))*lo g(x + (A*a**2*c*e**3 - 3*A*a*c**2*d**2*e + 3*B*a**2*c*d*e**2 - B*a*c**2*d* *3 - C*a**3*e**3 + 3*C*a**2*c*d**2*e + 2*a*c**3*((-A*a*c*e**3 + 3*A*c**2*d **2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3*C*a*c*d**2*e)/(2*c* *3) - sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c *d**2*e + 3*C*a**2*d*e**2 - C*a*c*d**3)/(2*a*c**6)))/(-3*A*a*c**2*d*e**2 + A*c**3*d**3 + B*a**2*c*e**3 - 3*B*a*c**2*d**2*e + 3*C*a**2*c*d*e**2 - C*a *c**2*d**3)) + ((-A*a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d **3 + C*a**2*e**3 - 3*C*a*c*d**2*e)/(2*c**3) + sqrt(-a*c**7)*(-3*A*a*c*d*e **2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e**2 - C*a*c *d**3)/(2*a*c**6))*log(x + (A*a**2*c*e**3 - 3*A*a*c**2*d**2*e + 3*B*a**2*c *d*e**2 - B*a*c**2*d**3 - C*a**3*e**3 + 3*C*a**2*c*d**2*e + 2*a*c**3*((-A* a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3*C*a*c*d**2*e)/(2*c**3) + sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e**2 - C*a*c*d**3)/(2*a*c**6)...
Time = 0.27 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=-\frac {{\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {3 \, C c e^{3} x^{4} + 4 \, {\left (3 \, C c d e^{2} + B c e^{3}\right )} x^{3} + 6 \, {\left (3 \, C c d^{2} e + 3 \, B c d e^{2} - {\left (C a - A c\right )} e^{3}\right )} x^{2} + 12 \, {\left (C c d^{3} + 3 \, B c d^{2} e - B a e^{3} - 3 \, {\left (C a - A c\right )} d e^{2}\right )} x}{12 \, c^{2}} + \frac {{\left (B c^{2} d^{3} - 3 \, B a c d e^{2} - 3 \, {\left (C a c - A c^{2}\right )} d^{2} e + {\left (C a^{2} - A a c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} \]
-(3*B*a*c*d^2*e - B*a^2*e^3 + (C*a*c - A*c^2)*d^3 - 3*(C*a^2 - A*a*c)*d*e^ 2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/12*(3*C*c*e^3*x^4 + 4*(3*C*c* d*e^2 + B*c*e^3)*x^3 + 6*(3*C*c*d^2*e + 3*B*c*d*e^2 - (C*a - A*c)*e^3)*x^2 + 12*(C*c*d^3 + 3*B*c*d^2*e - B*a*e^3 - 3*(C*a - A*c)*d*e^2)*x)/c^2 + 1/2 *(B*c^2*d^3 - 3*B*a*c*d*e^2 - 3*(C*a*c - A*c^2)*d^2*e + (C*a^2 - A*a*c)*e^ 3)*log(c*x^2 + a)/c^3
Time = 0.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=-\frac {{\left (C a c d^{3} - A c^{2} d^{3} + 3 \, B a c d^{2} e - 3 \, C a^{2} d e^{2} + 3 \, A a c d e^{2} - B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {{\left (B c^{2} d^{3} - 3 \, C a c d^{2} e + 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + C a^{2} e^{3} - A a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {3 \, C c^{3} e^{3} x^{4} + 12 \, C c^{3} d e^{2} x^{3} + 4 \, B c^{3} e^{3} x^{3} + 18 \, C c^{3} d^{2} e x^{2} + 18 \, B c^{3} d e^{2} x^{2} - 6 \, C a c^{2} e^{3} x^{2} + 6 \, A c^{3} e^{3} x^{2} + 12 \, C c^{3} d^{3} x + 36 \, B c^{3} d^{2} e x - 36 \, C a c^{2} d e^{2} x + 36 \, A c^{3} d e^{2} x - 12 \, B a c^{2} e^{3} x}{12 \, c^{4}} \]
-(C*a*c*d^3 - A*c^2*d^3 + 3*B*a*c*d^2*e - 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 - B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/2*(B*c^2*d^3 - 3*C*a* c*d^2*e + 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + C*a^2*e^3 - A*a*c*e^3)*log(c*x^2 + a)/c^3 + 1/12*(3*C*c^3*e^3*x^4 + 12*C*c^3*d*e^2*x^3 + 4*B*c^3*e^3*x^3 + 18*C*c^3*d^2*e*x^2 + 18*B*c^3*d*e^2*x^2 - 6*C*a*c^2*e^3*x^2 + 6*A*c^3*e^3 *x^2 + 12*C*c^3*d^3*x + 36*B*c^3*d^2*e*x - 36*C*a*c^2*d*e^2*x + 36*A*c^3*d *e^2*x - 12*B*a*c^2*e^3*x)/c^4
Time = 12.87 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=x^2\,\left (\frac {3\,C\,d^2\,e+3\,B\,d\,e^2+A\,e^3}{2\,c}-\frac {C\,a\,e^3}{2\,c^2}\right )+x\,\left (\frac {C\,d^3+3\,B\,d^2\,e+3\,A\,d\,e^2}{c}-\frac {a\,\left (B\,e^3+3\,C\,d\,e^2\right )}{c^2}\right )+\frac {x^3\,\left (B\,e^3+3\,C\,d\,e^2\right )}{3\,c}+\frac {C\,e^3\,x^4}{4\,c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,C\,a^2\,d\,e^2+B\,a^2\,e^3-C\,a\,c\,d^3-3\,B\,a\,c\,d^2\,e-3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (4\,C\,a^3\,c^3\,e^3-12\,C\,a^2\,c^4\,d^2\,e-12\,B\,a^2\,c^4\,d\,e^2-4\,A\,a^2\,c^4\,e^3+4\,B\,a\,c^5\,d^3+12\,A\,a\,c^5\,d^2\,e\right )}{8\,a\,c^6} \]
x^2*((A*e^3 + 3*B*d*e^2 + 3*C*d^2*e)/(2*c) - (C*a*e^3)/(2*c^2)) + x*((C*d^ 3 + 3*A*d*e^2 + 3*B*d^2*e)/c - (a*(B*e^3 + 3*C*d*e^2))/c^2) + (x^3*(B*e^3 + 3*C*d*e^2))/(3*c) + (C*e^3*x^4)/(4*c) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^ 2*d^3 + B*a^2*e^3 - C*a*c*d^3 + 3*C*a^2*d*e^2 - 3*A*a*c*d*e^2 - 3*B*a*c*d^ 2*e))/(a^(1/2)*c^(5/2)) + (log(a + c*x^2)*(4*B*a*c^5*d^3 - 4*A*a^2*c^4*e^3 + 4*C*a^3*c^3*e^3 - 12*B*a^2*c^4*d*e^2 - 12*C*a^2*c^4*d^2*e + 12*A*a*c^5* d^2*e))/(8*a*c^6)