Integrand size = 25, antiderivative size = 153 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=\frac {\left (a C e^2+c \left (3 C d^2-e (2 B d-A e)\right )\right ) x}{e^4}-\frac {c (2 C d-B e) x^2}{2 e^3}+\frac {c C x^3}{3 e^2}-\frac {\left (c d^2+a e^2\right ) \left (C d^2-B d e+A e^2\right )}{e^5 (d+e x)}-\frac {\left (a e^2 (2 C d-B e)+c d \left (4 C d^2-e (3 B d-2 A e)\right )\right ) \log (d+e x)}{e^5} \]
(a*C*e^2+c*(3*C*d^2-e*(-A*e+2*B*d)))*x/e^4-1/2*c*(-B*e+2*C*d)*x^2/e^3+1/3* c*C*x^3/e^2-(a*e^2+c*d^2)*(A*e^2-B*d*e+C*d^2)/e^5/(e*x+d)-(a*e^2*(-B*e+2*C *d)+c*d*(4*C*d^2-e*(-2*A*e+3*B*d)))*ln(e*x+d)/e^5
Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=\frac {6 e \left (3 c C d^2+a C e^2+c e (-2 B d+A e)\right ) x+3 c e^2 (-2 C d+B e) x^2+2 c C e^3 x^3-\frac {6 \left (c d^2+a e^2\right ) \left (C d^2+e (-B d+A e)\right )}{d+e x}+6 \left (-4 c C d^3+c d e (3 B d-2 A e)+a e^2 (-2 C d+B e)\right ) \log (d+e x)}{6 e^5} \]
(6*e*(3*c*C*d^2 + a*C*e^2 + c*e*(-2*B*d + A*e))*x + 3*c*e^2*(-2*C*d + B*e) *x^2 + 2*c*C*e^3*x^3 - (6*(c*d^2 + a*e^2)*(C*d^2 + e*(-(B*d) + A*e)))/(d + e*x) + 6*(-4*c*C*d^3 + c*d*e*(3*B*d - 2*A*e) + a*e^2*(-2*C*d + B*e))*Log[ d + e*x])/(6*e^5)
Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2159, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {-a e^2 (2 C d-B e)+c d e (3 B d-2 A e)-4 c C d^3}{e^4 (d+e x)}+\frac {\left (a e^2+c d^2\right ) \left (A e^2-B d e+C d^2\right )}{e^4 (d+e x)^2}+\frac {a C e^2-c e (2 B d-A e)+3 c C d^2}{e^4}+\frac {c x (B e-2 C d)}{e^3}+\frac {c C x^2}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\log (d+e x) \left (a e^2 (2 C d-B e)-c d e (3 B d-2 A e)+4 c C d^3\right )}{e^5}-\frac {\left (a e^2+c d^2\right ) \left (A e^2-B d e+C d^2\right )}{e^5 (d+e x)}+\frac {x \left (a C e^2-c e (2 B d-A e)+3 c C d^2\right )}{e^4}-\frac {c x^2 (2 C d-B e)}{2 e^3}+\frac {c C x^3}{3 e^2}\) |
((3*c*C*d^2 + a*C*e^2 - c*e*(2*B*d - A*e))*x)/e^4 - (c*(2*C*d - B*e)*x^2)/ (2*e^3) + (c*C*x^3)/(3*e^2) - ((c*d^2 + a*e^2)*(C*d^2 - B*d*e + A*e^2))/(e ^5*(d + e*x)) - ((4*c*C*d^3 - c*d*e*(3*B*d - 2*A*e) + a*e^2*(2*C*d - B*e)) *Log[d + e*x])/e^5
3.1.23.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.45 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12
| method | result | size |
| default | \(\frac {\frac {1}{3} c C \,x^{3} e^{2}+\frac {1}{2} B c \,e^{2} x^{2}-C c d e \,x^{2}+A c \,e^{2} x -2 B c d e x +a C \,e^{2} x +3 C c \,d^{2} x}{e^{4}}-\frac {A a \,e^{4}+A c \,d^{2} e^{2}-B a d \,e^{3}-B c \,d^{3} e +C a \,d^{2} e^{2}+C c \,d^{4}}{e^{5} \left (e x +d \right )}+\frac {\left (-2 A c d \,e^{2}+B \,e^{3} a +3 B c \,d^{2} e -2 C a d \,e^{2}-4 C c \,d^{3}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(172\) |
| norman | \(\frac {\frac {\left (A a \,e^{4}+2 A c \,d^{2} e^{2}-B a d \,e^{3}-3 B c \,d^{3} e +2 C a \,d^{2} e^{2}+4 C c \,d^{4}\right ) x}{e^{4} d}+\frac {\left (2 A c \,e^{2}-3 B c d e +2 a C \,e^{2}+4 C c \,d^{2}\right ) x^{2}}{2 e^{3}}+\frac {c C \,x^{4}}{3 e}+\frac {c \left (3 B e -4 C d \right ) x^{3}}{6 e^{2}}}{e x +d}-\frac {\left (2 A c d \,e^{2}-B \,e^{3} a -3 B c \,d^{2} e +2 C a d \,e^{2}+4 C c \,d^{3}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(183\) |
| risch | \(\frac {c C \,x^{3}}{3 e^{2}}+\frac {B c \,x^{2}}{2 e^{2}}-\frac {C c d \,x^{2}}{e^{3}}+\frac {A c x}{e^{2}}-\frac {2 B c d x}{e^{3}}+\frac {a C x}{e^{2}}+\frac {3 C c \,d^{2} x}{e^{4}}-\frac {A a}{e \left (e x +d \right )}-\frac {A c \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {B a d}{e^{2} \left (e x +d \right )}+\frac {B c \,d^{3}}{e^{4} \left (e x +d \right )}-\frac {C a \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {C c \,d^{4}}{e^{5} \left (e x +d \right )}-\frac {2 \ln \left (e x +d \right ) A c d}{e^{3}}+\frac {\ln \left (e x +d \right ) B a}{e^{2}}+\frac {3 \ln \left (e x +d \right ) B c \,d^{2}}{e^{4}}-\frac {2 \ln \left (e x +d \right ) C a d}{e^{3}}-\frac {4 \ln \left (e x +d \right ) C c \,d^{3}}{e^{5}}\) | \(234\) |
| parallelrisch | \(-\frac {6 A a \,e^{4}+24 C c \,d^{4}+12 A \ln \left (e x +d \right ) x c d \,e^{3}+24 C \ln \left (e x +d \right ) x c \,d^{3} e +24 C \ln \left (e x +d \right ) c \,d^{4}-3 B \,x^{3} c \,e^{4}-6 A \,x^{2} c \,e^{4}-6 C \,x^{2} a \,e^{4}-2 c C \,x^{4} e^{4}+12 A c \,d^{2} e^{2}-18 B c \,d^{3} e -6 B a d \,e^{3}+12 C a \,d^{2} e^{2}+9 B \,x^{2} c d \,e^{3}+12 C \ln \left (e x +d \right ) x a d \,e^{3}-12 C \,x^{2} c \,d^{2} e^{2}+12 A \ln \left (e x +d \right ) c \,d^{2} e^{2}-6 B \ln \left (e x +d \right ) a d \,e^{3}-18 B \ln \left (e x +d \right ) c \,d^{3} e +12 C \ln \left (e x +d \right ) a \,d^{2} e^{2}+4 C \,x^{3} c d \,e^{3}-18 B \ln \left (e x +d \right ) x c \,d^{2} e^{2}-6 B \ln \left (e x +d \right ) x a \,e^{4}}{6 e^{5} \left (e x +d \right )}\) | \(288\) |
1/e^4*(1/3*c*C*x^3*e^2+1/2*B*c*e^2*x^2-C*c*d*e*x^2+A*c*e^2*x-2*B*c*d*e*x+a *C*e^2*x+3*C*c*d^2*x)-(A*a*e^4+A*c*d^2*e^2-B*a*d*e^3-B*c*d^3*e+C*a*d^2*e^2 +C*c*d^4)/e^5/(e*x+d)+(-2*A*c*d*e^2+B*a*e^3+3*B*c*d^2*e-2*C*a*d*e^2-4*C*c* d^3)/e^5*ln(e*x+d)
Time = 0.29 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=\frac {2 \, C c e^{4} x^{4} - 6 \, C c d^{4} + 6 \, B c d^{3} e + 6 \, B a d e^{3} - 6 \, A a e^{4} - 6 \, {\left (C a + A c\right )} d^{2} e^{2} - {\left (4 \, C c d e^{3} - 3 \, B c e^{4}\right )} x^{3} + 3 \, {\left (4 \, C c d^{2} e^{2} - 3 \, B c d e^{3} + 2 \, {\left (C a + A c\right )} e^{4}\right )} x^{2} + 6 \, {\left (3 \, C c d^{3} e - 2 \, B c d^{2} e^{2} + {\left (C a + A c\right )} d e^{3}\right )} x - 6 \, {\left (4 \, C c d^{4} - 3 \, B c d^{3} e - B a d e^{3} + 2 \, {\left (C a + A c\right )} d^{2} e^{2} + {\left (4 \, C c d^{3} e - 3 \, B c d^{2} e^{2} - B a e^{4} + 2 \, {\left (C a + A c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{6} x + d e^{5}\right )}} \]
1/6*(2*C*c*e^4*x^4 - 6*C*c*d^4 + 6*B*c*d^3*e + 6*B*a*d*e^3 - 6*A*a*e^4 - 6 *(C*a + A*c)*d^2*e^2 - (4*C*c*d*e^3 - 3*B*c*e^4)*x^3 + 3*(4*C*c*d^2*e^2 - 3*B*c*d*e^3 + 2*(C*a + A*c)*e^4)*x^2 + 6*(3*C*c*d^3*e - 2*B*c*d^2*e^2 + (C *a + A*c)*d*e^3)*x - 6*(4*C*c*d^4 - 3*B*c*d^3*e - B*a*d*e^3 + 2*(C*a + A*c )*d^2*e^2 + (4*C*c*d^3*e - 3*B*c*d^2*e^2 - B*a*e^4 + 2*(C*a + A*c)*d*e^3)* x)*log(e*x + d))/(e^6*x + d*e^5)
Time = 0.53 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=\frac {C c x^{3}}{3 e^{2}} + x^{2} \left (\frac {B c}{2 e^{2}} - \frac {C c d}{e^{3}}\right ) + x \left (\frac {A c}{e^{2}} - \frac {2 B c d}{e^{3}} + \frac {C a}{e^{2}} + \frac {3 C c d^{2}}{e^{4}}\right ) + \frac {- A a e^{4} - A c d^{2} e^{2} + B a d e^{3} + B c d^{3} e - C a d^{2} e^{2} - C c d^{4}}{d e^{5} + e^{6} x} - \frac {\left (2 A c d e^{2} - B a e^{3} - 3 B c d^{2} e + 2 C a d e^{2} + 4 C c d^{3}\right ) \log {\left (d + e x \right )}}{e^{5}} \]
C*c*x**3/(3*e**2) + x**2*(B*c/(2*e**2) - C*c*d/e**3) + x*(A*c/e**2 - 2*B*c *d/e**3 + C*a/e**2 + 3*C*c*d**2/e**4) + (-A*a*e**4 - A*c*d**2*e**2 + B*a*d *e**3 + B*c*d**3*e - C*a*d**2*e**2 - C*c*d**4)/(d*e**5 + e**6*x) - (2*A*c* d*e**2 - B*a*e**3 - 3*B*c*d**2*e + 2*C*a*d*e**2 + 4*C*c*d**3)*log(d + e*x) /e**5
Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=-\frac {C c d^{4} - B c d^{3} e - B a d e^{3} + A a e^{4} + {\left (C a + A c\right )} d^{2} e^{2}}{e^{6} x + d e^{5}} + \frac {2 \, C c e^{2} x^{3} - 3 \, {\left (2 \, C c d e - B c e^{2}\right )} x^{2} + 6 \, {\left (3 \, C c d^{2} - 2 \, B c d e + {\left (C a + A c\right )} e^{2}\right )} x}{6 \, e^{4}} - \frac {{\left (4 \, C c d^{3} - 3 \, B c d^{2} e - B a e^{3} + 2 \, {\left (C a + A c\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \]
-(C*c*d^4 - B*c*d^3*e - B*a*d*e^3 + A*a*e^4 + (C*a + A*c)*d^2*e^2)/(e^6*x + d*e^5) + 1/6*(2*C*c*e^2*x^3 - 3*(2*C*c*d*e - B*c*e^2)*x^2 + 6*(3*C*c*d^2 - 2*B*c*d*e + (C*a + A*c)*e^2)*x)/e^4 - (4*C*c*d^3 - 3*B*c*d^2*e - B*a*e^ 3 + 2*(C*a + A*c)*d*e^2)*log(e*x + d)/e^5
Time = 0.30 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=\frac {{\left (2 \, C c - \frac {3 \, {\left (4 \, C c d e - B c e^{2}\right )}}{{\left (e x + d\right )} e} + \frac {6 \, {\left (6 \, C c d^{2} e^{2} - 3 \, B c d e^{3} + C a e^{4} + A c e^{4}\right )}}{{\left (e x + d\right )}^{2} e^{2}}\right )} {\left (e x + d\right )}^{3}}{6 \, e^{5}} + \frac {{\left (4 \, C c d^{3} - 3 \, B c d^{2} e + 2 \, C a d e^{2} + 2 \, A c d e^{2} - B a e^{3}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} - \frac {\frac {C c d^{4} e^{3}}{e x + d} - \frac {B c d^{3} e^{4}}{e x + d} + \frac {C a d^{2} e^{5}}{e x + d} + \frac {A c d^{2} e^{5}}{e x + d} - \frac {B a d e^{6}}{e x + d} + \frac {A a e^{7}}{e x + d}}{e^{8}} \]
1/6*(2*C*c - 3*(4*C*c*d*e - B*c*e^2)/((e*x + d)*e) + 6*(6*C*c*d^2*e^2 - 3* B*c*d*e^3 + C*a*e^4 + A*c*e^4)/((e*x + d)^2*e^2))*(e*x + d)^3/e^5 + (4*C*c *d^3 - 3*B*c*d^2*e + 2*C*a*d*e^2 + 2*A*c*d*e^2 - B*a*e^3)*log(abs(e*x + d) /((e*x + d)^2*abs(e)))/e^5 - (C*c*d^4*e^3/(e*x + d) - B*c*d^3*e^4/(e*x + d ) + C*a*d^2*e^5/(e*x + d) + A*c*d^2*e^5/(e*x + d) - B*a*d*e^6/(e*x + d) + A*a*e^7/(e*x + d))/e^8
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^2} \, dx=x^2\,\left (\frac {B\,c}{2\,e^2}-\frac {C\,c\,d}{e^3}\right )-x\,\left (\frac {2\,d\,\left (\frac {B\,c}{e^2}-\frac {2\,C\,c\,d}{e^3}\right )}{e}-\frac {A\,c+C\,a}{e^2}+\frac {C\,c\,d^2}{e^4}\right )-\frac {\ln \left (d+e\,x\right )\,\left (4\,C\,c\,d^3-B\,a\,e^3+2\,A\,c\,d\,e^2+2\,C\,a\,d\,e^2-3\,B\,c\,d^2\,e\right )}{e^5}-\frac {A\,a\,e^4+C\,c\,d^4-B\,a\,d\,e^3-B\,c\,d^3\,e+A\,c\,d^2\,e^2+C\,a\,d^2\,e^2}{e\,\left (x\,e^5+d\,e^4\right )}+\frac {C\,c\,x^3}{3\,e^2} \]
x^2*((B*c)/(2*e^2) - (C*c*d)/e^3) - x*((2*d*((B*c)/e^2 - (2*C*c*d)/e^3))/e - (A*c + C*a)/e^2 + (C*c*d^2)/e^4) - (log(d + e*x)*(4*C*c*d^3 - B*a*e^3 + 2*A*c*d*e^2 + 2*C*a*d*e^2 - 3*B*c*d^2*e))/e^5 - (A*a*e^4 + C*c*d^4 - B*a* d*e^3 - B*c*d^3*e + A*c*d^2*e^2 + C*a*d^2*e^2)/(e*(d*e^4 + e^5*x)) + (C*c* x^3)/(3*e^2)