Integrand size = 22, antiderivative size = 104 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^3}+\frac {d (B d-A e) (c d-b e)}{2 e^4 (d+e x)^2}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^4 (d+e x)}-\frac {(3 B c d-b B e-A c e) \log (d+e x)}{e^4} \]
B*c*x/e^3+1/2*d*(-A*e+B*d)*(-b*e+c*d)/e^4/(e*x+d)^2+(-B*d*(-2*b*e+3*c*d)+A *e*(-b*e+2*c*d))/e^4/(e*x+d)-(-A*c*e-B*b*e+3*B*c*d)*ln(e*x+d)/e^4
Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 B c e x+\frac {d (B d-A e) (c d-b e)}{(d+e x)^2}+\frac {-6 B c d^2+4 b B d e+4 A c d e-2 A b e^2}{d+e x}+2 (-3 B c d+b B e+A c e) \log (d+e x)}{2 e^4} \]
(2*B*c*e*x + (d*(B*d - A*e)*(c*d - b*e))/(d + e*x)^2 + (-6*B*c*d^2 + 4*b*B *d*e + 4*A*c*d*e - 2*A*b*e^2)/(d + e*x) + 2*(-3*B*c*d + b*B*e + A*c*e)*Log [d + e*x])/(2*e^4)
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {A c e+b B e-3 B c d}{e^3 (d+e x)}+\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^3 (d+e x)^2}-\frac {d (B d-A e) (c d-b e)}{e^3 (d+e x)^3}+\frac {B c}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d (B d-A e) (c d-b e)}{2 e^4 (d+e x)^2}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^4 (d+e x)}-\frac {\log (d+e x) (-A c e-b B e+3 B c d)}{e^4}+\frac {B c x}{e^3}\) |
(B*c*x)/e^3 + (d*(B*d - A*e)*(c*d - b*e))/(2*e^4*(d + e*x)^2) - (B*d*(3*c* d - 2*b*e) - A*e*(2*c*d - b*e))/(e^4*(d + e*x)) - ((3*B*c*d - b*B*e - A*c* e)*Log[d + e*x])/e^4
3.12.10.3.1 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 = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {\frac {B c \,x^{3}}{e}-\frac {d \left (A b \,e^{2}-3 A c d e -3 B b d e +9 B c \,d^{2}\right )}{2 e^{4}}-\frac {\left (A b \,e^{2}-2 A c d e -2 B b d e +6 B c \,d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {\left (A c e +B b e -3 B c d \right ) \ln \left (e x +d \right )}{e^{4}}\) | \(108\) |
default | \(\frac {B c x}{e^{3}}-\frac {A b \,e^{2}-2 A c d e -2 B b d e +3 B c \,d^{2}}{e^{4} \left (e x +d \right )}+\frac {\left (A c e +B b e -3 B c d \right ) \ln \left (e x +d \right )}{e^{4}}+\frac {d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{2 e^{4} \left (e x +d \right )^{2}}\) | \(109\) |
risch | \(\frac {B c x}{e^{3}}+\frac {\left (-A b \,e^{2}+2 A c d e +2 B b d e -3 B c \,d^{2}\right ) x -\frac {d \left (A b \,e^{2}-3 A c d e -3 B b d e +5 B c \,d^{2}\right )}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {\ln \left (e x +d \right ) A c}{e^{3}}+\frac {\ln \left (e x +d \right ) B b}{e^{3}}-\frac {3 \ln \left (e x +d \right ) B c d}{e^{4}}\) | \(120\) |
parallelrisch | \(\frac {2 A \ln \left (e x +d \right ) x^{2} c \,e^{3}+2 B \ln \left (e x +d \right ) x^{2} b \,e^{3}-6 B \ln \left (e x +d \right ) x^{2} c d \,e^{2}+2 B c \,x^{3} e^{3}+4 A \ln \left (e x +d \right ) x c d \,e^{2}+4 B \ln \left (e x +d \right ) x b d \,e^{2}-12 B \ln \left (e x +d \right ) x c \,d^{2} e +2 A \ln \left (e x +d \right ) c \,d^{2} e -2 A b \,e^{3} x +4 A c d \,e^{2} x +2 B \ln \left (e x +d \right ) b \,d^{2} e -6 B \ln \left (e x +d \right ) c \,d^{3}+4 B x b d \,e^{2}-12 B c \,d^{2} e x -A b d \,e^{2}+3 A c \,d^{2} e +3 B b \,d^{2} e -9 B c \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(225\) |
(B*c*x^3/e-1/2*d*(A*b*e^2-3*A*c*d*e-3*B*b*d*e+9*B*c*d^2)/e^4-(A*b*e^2-2*A* c*d*e-2*B*b*d*e+6*B*c*d^2)/e^3*x)/(e*x+d)^2+1/e^4*(A*c*e+B*b*e-3*B*c*d)*ln (e*x+d)
Time = 0.33 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} - A b d e^{2} + 3 \, {\left (B b + A c\right )} d^{2} e - 2 \, {\left (2 \, B c d^{2} e + A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x - 2 \, {\left (3 \, B c d^{3} - {\left (B b + A c\right )} d^{2} e + {\left (3 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 2 \, {\left (3 \, B c d^{2} e - {\left (B b + A c\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]
1/2*(2*B*c*e^3*x^3 + 4*B*c*d*e^2*x^2 - 5*B*c*d^3 - A*b*d*e^2 + 3*(B*b + A* c)*d^2*e - 2*(2*B*c*d^2*e + A*b*e^3 - 2*(B*b + A*c)*d*e^2)*x - 2*(3*B*c*d^ 3 - (B*b + A*c)*d^2*e + (3*B*c*d*e^2 - (B*b + A*c)*e^3)*x^2 + 2*(3*B*c*d^2 *e - (B*b + A*c)*d*e^2)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Time = 0.67 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} + \frac {- A b d e^{2} + 3 A c d^{2} e + 3 B b d^{2} e - 5 B c d^{3} + x \left (- 2 A b e^{3} + 4 A c d e^{2} + 4 B b d e^{2} - 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} + \frac {\left (A c e + B b e - 3 B c d\right ) \log {\left (d + e x \right )}}{e^{4}} \]
B*c*x/e**3 + (-A*b*d*e**2 + 3*A*c*d**2*e + 3*B*b*d**2*e - 5*B*c*d**3 + x*( -2*A*b*e**3 + 4*A*c*d*e**2 + 4*B*b*d*e**2 - 6*B*c*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2) + (A*c*e + B*b*e - 3*B*c*d)*log(d + e*x)/e**4
Time = 0.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=-\frac {5 \, B c d^{3} + A b d e^{2} - 3 \, {\left (B b + A c\right )} d^{2} e + 2 \, {\left (3 \, B c d^{2} e + A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - {\left (B b + A c\right )} e\right )} \log \left (e x + d\right )}{e^{4}} \]
-1/2*(5*B*c*d^3 + A*b*d*e^2 - 3*(B*b + A*c)*d^2*e + 2*(3*B*c*d^2*e + A*b*e ^3 - 2*(B*b + A*c)*d*e^2)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + B*c*x/e^3 - (3*B*c*d - (B*b + A*c)*e)*log(e*x + d)/e^4
Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - B b e - A c e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {5 \, B c d^{3} - 3 \, B b d^{2} e - 3 \, A c d^{2} e + A b d e^{2} + 2 \, {\left (3 \, B c d^{2} e - 2 \, B b d e^{2} - 2 \, A c d e^{2} + A b e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \]
B*c*x/e^3 - (3*B*c*d - B*b*e - A*c*e)*log(abs(e*x + d))/e^4 - 1/2*(5*B*c*d ^3 - 3*B*b*d^2*e - 3*A*c*d^2*e + A*b*d*e^2 + 2*(3*B*c*d^2*e - 2*B*b*d*e^2 - 2*A*c*d*e^2 + A*b*e^3)*x)/((e*x + d)^2*e^4)
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e+B\,b\,e-3\,B\,c\,d\right )}{e^4}-\frac {x\,\left (A\,b\,e^2+3\,B\,c\,d^2-2\,A\,c\,d\,e-2\,B\,b\,d\,e\right )+\frac {5\,B\,c\,d^3+A\,b\,d\,e^2-3\,A\,c\,d^2\,e-3\,B\,b\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,c\,x}{e^3} \]