Integrand size = 24, antiderivative size = 111 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 c^2 x}{e^3}+\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right )}{2 e^4 (d+e x)^2}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)}-\frac {3 c (2 c d-b e) \log (d+e x)}{e^4} \]
2*c^2*x/e^3+1/2*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)/e^4/(e*x+d)^2+(-6*c^2*d^2 -b^2*e^2+2*c*e*(-a*e+3*b*d))/e^4/(e*x+d)-3*c*(-b*e+2*c*d)*ln(e*x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.06 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {c^2 \left (-10 d^3-8 d^2 e x+8 d e^2 x^2+4 e^3 x^3\right )-b e^2 (a e+b (d+2 e x))+c e (-2 a e (d+2 e x)+3 b d (3 d+4 e x))-6 c (2 c d-b e) (d+e x)^2 \log (d+e x)}{2 e^4 (d+e x)^2} \]
(c^2*(-10*d^3 - 8*d^2*e*x + 8*d*e^2*x^2 + 4*e^3*x^3) - b*e^2*(a*e + b*(d + 2*e*x)) + c*e*(-2*a*e*(d + 2*e*x) + 3*b*d*(3*d + 4*e*x)) - 6*c*(2*c*d - b *e)*(d + e*x)^2*Log[d + e*x])/(2*e^4*(d + e*x)^2)
Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1195, 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.
\(\displaystyle \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^3 (d+e x)^2}+\frac {(b e-2 c d) \left (a e^2-b d e+c d^2\right )}{e^3 (d+e x)^3}-\frac {3 c (2 c d-b e)}{e^3 (d+e x)}+\frac {2 c^2}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^4 (d+e x)}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{2 e^4 (d+e x)^2}-\frac {3 c (2 c d-b e) \log (d+e x)}{e^4}+\frac {2 c^2 x}{e^3}\) |
(2*c^2*x)/e^3 + ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(2*e^4*(d + e*x)^2 ) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(e^4*(d + e*x)) - (3*c*(2* c*d - b*e)*Log[d + e*x])/e^4
3.16.1.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.09
method | result | size |
norman | \(\frac {-\frac {a b \,e^{3}+2 a c d \,e^{2}+b^{2} d \,e^{2}-9 b c \,d^{2} e +18 c^{2} d^{3}}{2 e^{4}}+\frac {2 c^{2} x^{3}}{e}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +12 c^{2} d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {3 c \left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{4}}\) | \(121\) |
default | \(\frac {2 c^{2} x}{e^{3}}-\frac {2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{e^{4} \left (e x +d \right )}+\frac {3 c \left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{4}}-\frac {a b \,e^{3}-2 a c d \,e^{2}-b^{2} d \,e^{2}+3 b c \,d^{2} e -2 c^{2} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(124\) |
risch | \(\frac {2 c^{2} x}{e^{3}}+\frac {\left (-2 a c \,e^{2}-b^{2} e^{2}+6 b c d e -6 c^{2} d^{2}\right ) x -\frac {a b \,e^{3}+2 a c d \,e^{2}+b^{2} d \,e^{2}-9 b c \,d^{2} e +10 c^{2} d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {3 c \ln \left (e x +d \right ) b}{e^{3}}-\frac {6 c^{2} \ln \left (e x +d \right ) d}{e^{4}}\) | \(127\) |
parallelrisch | \(\frac {6 \ln \left (e x +d \right ) x^{2} b c \,e^{3}-12 \ln \left (e x +d \right ) x^{2} c^{2} d \,e^{2}+4 c^{2} x^{3} e^{3}+12 \ln \left (e x +d \right ) x b c d \,e^{2}-24 \ln \left (e x +d \right ) x \,c^{2} d^{2} e +6 \ln \left (e x +d \right ) b c \,d^{2} e -12 \ln \left (e x +d \right ) c^{2} d^{3}-4 x a c \,e^{3}-2 x \,b^{2} e^{3}+12 x b c d \,e^{2}-24 x \,c^{2} d^{2} e -a b \,e^{3}-2 a c d \,e^{2}-b^{2} d \,e^{2}+9 b c \,d^{2} e -18 c^{2} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(194\) |
(-1/2*(a*b*e^3+2*a*c*d*e^2+b^2*d*e^2-9*b*c*d^2*e+18*c^2*d^3)/e^4+2*c^2*x^3 /e-(2*a*c*e^2+b^2*e^2-6*b*c*d*e+12*c^2*d^2)/e^3*x)/(e*x+d)^2+3*c/e^4*(b*e- 2*c*d)*ln(e*x+d)
Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.68 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {4 \, c^{2} e^{3} x^{3} + 8 \, c^{2} d e^{2} x^{2} - 10 \, c^{2} d^{3} + 9 \, b c d^{2} e - a b e^{3} - {\left (b^{2} + 2 \, a c\right )} d e^{2} - 2 \, {\left (4 \, c^{2} d^{2} e - 6 \, b c d e^{2} + {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x - 6 \, {\left (2 \, c^{2} d^{3} - b c d^{2} e + {\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} x^{2} + 2 \, {\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]
1/2*(4*c^2*e^3*x^3 + 8*c^2*d*e^2*x^2 - 10*c^2*d^3 + 9*b*c*d^2*e - a*b*e^3 - (b^2 + 2*a*c)*d*e^2 - 2*(4*c^2*d^2*e - 6*b*c*d*e^2 + (b^2 + 2*a*c)*e^3)* x - 6*(2*c^2*d^3 - b*c*d^2*e + (2*c^2*d*e^2 - b*c*e^3)*x^2 + 2*(2*c^2*d^2* e - b*c*d*e^2)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Time = 0.73 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.25 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 c^{2} x}{e^{3}} + \frac {3 c \left (b e - 2 c d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- a b e^{3} - 2 a c d e^{2} - b^{2} d e^{2} + 9 b c d^{2} e - 10 c^{2} d^{3} + x \left (- 4 a c e^{3} - 2 b^{2} e^{3} + 12 b c d e^{2} - 12 c^{2} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \]
2*c**2*x/e**3 + 3*c*(b*e - 2*c*d)*log(d + e*x)/e**4 + (-a*b*e**3 - 2*a*c*d *e**2 - b**2*d*e**2 + 9*b*c*d**2*e - 10*c**2*d**3 + x*(-4*a*c*e**3 - 2*b** 2*e**3 + 12*b*c*d*e**2 - 12*c**2*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e* *6*x**2)
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=-\frac {10 \, c^{2} d^{3} - 9 \, b c d^{2} e + a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2} + 2 \, {\left (6 \, c^{2} d^{2} e - 6 \, b c d e^{2} + {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {2 \, c^{2} x}{e^{3}} - \frac {3 \, {\left (2 \, c^{2} d - b c e\right )} \log \left (e x + d\right )}{e^{4}} \]
-1/2*(10*c^2*d^3 - 9*b*c*d^2*e + a*b*e^3 + (b^2 + 2*a*c)*d*e^2 + 2*(6*c^2* d^2*e - 6*b*c*d*e^2 + (b^2 + 2*a*c)*e^3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4 ) + 2*c^2*x/e^3 - 3*(2*c^2*d - b*c*e)*log(e*x + d)/e^4
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, c^{2} x}{e^{3}} - \frac {3 \, {\left (2 \, c^{2} d - b c e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {10 \, c^{2} d^{3} - 9 \, b c d^{2} e + b^{2} d e^{2} + 2 \, a c d e^{2} + a b e^{3} + 2 \, {\left (6 \, c^{2} d^{2} e - 6 \, b c d e^{2} + b^{2} e^{3} + 2 \, a c e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \]
2*c^2*x/e^3 - 3*(2*c^2*d - b*c*e)*log(abs(e*x + d))/e^4 - 1/2*(10*c^2*d^3 - 9*b*c*d^2*e + b^2*d*e^2 + 2*a*c*d*e^2 + a*b*e^3 + 2*(6*c^2*d^2*e - 6*b*c *d*e^2 + b^2*e^3 + 2*a*c*e^3)*x)/((e*x + d)^2*e^4)
Time = 10.73 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.22 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2\,c^2\,x}{e^3}-\frac {\frac {b^2\,d\,e^2-9\,b\,c\,d^2\,e+a\,b\,e^3+10\,c^2\,d^3+2\,a\,c\,d\,e^2}{2\,e}+x\,\left (b^2\,e^2-6\,b\,c\,d\,e+6\,c^2\,d^2+2\,a\,c\,e^2\right )}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}-\frac {\ln \left (d+e\,x\right )\,\left (6\,c^2\,d-3\,b\,c\,e\right )}{e^4} \]