Integrand size = 22, antiderivative size = 116 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=\frac {d (B d-A e) (c d-b e)}{4 e^4 (d+e x)^4}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{3 e^4 (d+e x)^3}+\frac {3 B c d-b B e-A c e}{2 e^4 (d+e x)^2}-\frac {B c}{e^4 (d+e x)} \]
1/4*d*(-A*e+B*d)*(-b*e+c*d)/e^4/(e*x+d)^4+1/3*(-B*d*(-2*b*e+3*c*d)+A*e*(-b *e+2*c*d))/e^4/(e*x+d)^3+1/2*(-A*c*e-B*b*e+3*B*c*d)/e^4/(e*x+d)^2-B*c/e^4/ (e*x+d)
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=-\frac {A e \left (b e (d+4 e x)+c \left (d^2+4 d e x+6 e^2 x^2\right )\right )+B \left (b e \left (d^2+4 d e x+6 e^2 x^2\right )+3 c \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )}{12 e^4 (d+e x)^4} \]
-1/12*(A*e*(b*e*(d + 4*e*x) + c*(d^2 + 4*d*e*x + 6*e^2*x^2)) + B*(b*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3*c*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) )/(e^4*(d + e*x)^4)
Time = 0.29 (sec) , antiderivative size = 116, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {A c e+b B e-3 B c d}{e^3 (d+e x)^3}+\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^3 (d+e x)^4}-\frac {d (B d-A e) (c d-b e)}{e^3 (d+e x)^5}+\frac {B c}{e^3 (d+e x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-A c e-b B e+3 B c d}{2 e^4 (d+e x)^2}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{3 e^4 (d+e x)^3}+\frac {d (B d-A e) (c d-b e)}{4 e^4 (d+e x)^4}-\frac {B c}{e^4 (d+e x)}\) |
(d*(B*d - A*e)*(c*d - b*e))/(4*e^4*(d + e*x)^4) - (B*d*(3*c*d - 2*b*e) - A *e*(2*c*d - b*e))/(3*e^4*(d + e*x)^3) + (3*B*c*d - b*B*e - A*c*e)/(2*e^4*( d + e*x)^2) - (B*c)/(e^4*(d + e*x))
3.12.12.3.1 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 = 102, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {-\frac {B c \,x^{3}}{e}-\frac {\left (A c e +B b e +3 B c d \right ) x^{2}}{2 e^{2}}-\frac {\left (A b \,e^{2}+A c d e +B b d e +3 B c \,d^{2}\right ) x}{3 e^{3}}-\frac {d \left (A b \,e^{2}+A c d e +B b d e +3 B c \,d^{2}\right )}{12 e^{4}}}{\left (e x +d \right )^{4}}\) | \(102\) |
risch | \(\frac {-\frac {B c \,x^{3}}{e}-\frac {\left (A c e +B b e +3 B c d \right ) x^{2}}{2 e^{2}}-\frac {\left (A b \,e^{2}+A c d e +B b d e +3 B c \,d^{2}\right ) x}{3 e^{3}}-\frac {d \left (A b \,e^{2}+A c d e +B b d e +3 B c \,d^{2}\right )}{12 e^{4}}}{\left (e x +d \right )^{4}}\) | \(102\) |
gosper | \(-\frac {12 B c \,x^{3} e^{3}+6 A c \,e^{3} x^{2}+6 B \,x^{2} b \,e^{3}+18 B \,x^{2} c d \,e^{2}+4 A b \,e^{3} x +4 A c d \,e^{2} x +4 B x b d \,e^{2}+12 B c \,d^{2} e x +A b d \,e^{2}+A c \,d^{2} e +B b \,d^{2} e +3 B c \,d^{3}}{12 e^{4} \left (e x +d \right )^{4}}\) | \(118\) |
default | \(-\frac {A b \,e^{2}-2 A c d e -2 B b d e +3 B c \,d^{2}}{3 e^{4} \left (e x +d \right )^{3}}-\frac {B c}{e^{4} \left (e x +d \right )}+\frac {d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{4 e^{4} \left (e x +d \right )^{4}}-\frac {A c e +B b e -3 B c d}{2 e^{4} \left (e x +d \right )^{2}}\) | \(118\) |
parallelrisch | \(-\frac {12 B c \,x^{3} e^{3}+6 A c \,e^{3} x^{2}+6 B \,x^{2} b \,e^{3}+18 B \,x^{2} c d \,e^{2}+4 A b \,e^{3} x +4 A c d \,e^{2} x +4 B x b d \,e^{2}+12 B c \,d^{2} e x +A b d \,e^{2}+A c \,d^{2} e +B b \,d^{2} e +3 B c \,d^{3}}{12 e^{4} \left (e x +d \right )^{4}}\) | \(118\) |
(-B*c*x^3/e-1/2*(A*c*e+B*b*e+3*B*c*d)/e^2*x^2-1/3*(A*b*e^2+A*c*d*e+B*b*d*e +3*B*c*d^2)/e^3*x-1/12*d*(A*b*e^2+A*c*d*e+B*b*d*e+3*B*c*d^2)/e^4)/(e*x+d)^ 4
Time = 0.34 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=-\frac {12 \, B c e^{3} x^{3} + 3 \, B c d^{3} + A b d e^{2} + {\left (B b + A c\right )} d^{2} e + 6 \, {\left (3 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} x^{2} + 4 \, {\left (3 \, B c d^{2} e + A b e^{3} + {\left (B b + A c\right )} d e^{2}\right )} x}{12 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]
-1/12*(12*B*c*e^3*x^3 + 3*B*c*d^3 + A*b*d*e^2 + (B*b + A*c)*d^2*e + 6*(3*B *c*d*e^2 + (B*b + A*c)*e^3)*x^2 + 4*(3*B*c*d^2*e + A*b*e^3 + (B*b + A*c)*d *e^2)*x)/(e^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4*d^3*e^5*x + d^4*e^4)
Time = 2.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=\frac {- A b d e^{2} - A c d^{2} e - B b d^{2} e - 3 B c d^{3} - 12 B c e^{3} x^{3} + x^{2} \left (- 6 A c e^{3} - 6 B b e^{3} - 18 B c d e^{2}\right ) + x \left (- 4 A b e^{3} - 4 A c d e^{2} - 4 B b d e^{2} - 12 B c d^{2} e\right )}{12 d^{4} e^{4} + 48 d^{3} e^{5} x + 72 d^{2} e^{6} x^{2} + 48 d e^{7} x^{3} + 12 e^{8} x^{4}} \]
(-A*b*d*e**2 - A*c*d**2*e - B*b*d**2*e - 3*B*c*d**3 - 12*B*c*e**3*x**3 + x **2*(-6*A*c*e**3 - 6*B*b*e**3 - 18*B*c*d*e**2) + x*(-4*A*b*e**3 - 4*A*c*d* e**2 - 4*B*b*d*e**2 - 12*B*c*d**2*e))/(12*d**4*e**4 + 48*d**3*e**5*x + 72* d**2*e**6*x**2 + 48*d*e**7*x**3 + 12*e**8*x**4)
Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=-\frac {12 \, B c e^{3} x^{3} + 3 \, B c d^{3} + A b d e^{2} + {\left (B b + A c\right )} d^{2} e + 6 \, {\left (3 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} x^{2} + 4 \, {\left (3 \, B c d^{2} e + A b e^{3} + {\left (B b + A c\right )} d e^{2}\right )} x}{12 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]
-1/12*(12*B*c*e^3*x^3 + 3*B*c*d^3 + A*b*d*e^2 + (B*b + A*c)*d^2*e + 6*(3*B *c*d*e^2 + (B*b + A*c)*e^3)*x^2 + 4*(3*B*c*d^2*e + A*b*e^3 + (B*b + A*c)*d *e^2)*x)/(e^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4*d^3*e^5*x + d^4*e^4)
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=-\frac {\frac {6 \, B b}{{\left (e x + d\right )}^{2}} + \frac {6 \, A c}{{\left (e x + d\right )}^{2}} - \frac {8 \, B b d}{{\left (e x + d\right )}^{3}} - \frac {8 \, A c d}{{\left (e x + d\right )}^{3}} + \frac {3 \, B b d^{2}}{{\left (e x + d\right )}^{4}} + \frac {3 \, A c d^{2}}{{\left (e x + d\right )}^{4}} + \frac {12 \, B c}{{\left (e x + d\right )} e} - \frac {18 \, B c d}{{\left (e x + d\right )}^{2} e} + \frac {12 \, B c d^{2}}{{\left (e x + d\right )}^{3} e} - \frac {3 \, B c d^{3}}{{\left (e x + d\right )}^{4} e} + \frac {4 \, A b e}{{\left (e x + d\right )}^{3}} - \frac {3 \, A b d e}{{\left (e x + d\right )}^{4}}}{12 \, e^{3}} \]
-1/12*(6*B*b/(e*x + d)^2 + 6*A*c/(e*x + d)^2 - 8*B*b*d/(e*x + d)^3 - 8*A*c *d/(e*x + d)^3 + 3*B*b*d^2/(e*x + d)^4 + 3*A*c*d^2/(e*x + d)^4 + 12*B*c/(( e*x + d)*e) - 18*B*c*d/((e*x + d)^2*e) + 12*B*c*d^2/((e*x + d)^3*e) - 3*B* c*d^3/((e*x + d)^4*e) + 4*A*b*e/(e*x + d)^3 - 3*A*b*d*e/(e*x + d)^4)/e^3
Time = 0.07 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^5} \, dx=-\frac {\frac {d\,\left (A\,b\,e^2+3\,B\,c\,d^2+A\,c\,d\,e+B\,b\,d\,e\right )}{12\,e^4}+\frac {x\,\left (A\,b\,e^2+3\,B\,c\,d^2+A\,c\,d\,e+B\,b\,d\,e\right )}{3\,e^3}+\frac {x^2\,\left (A\,c\,e+B\,b\,e+3\,B\,c\,d\right )}{2\,e^2}+\frac {B\,c\,x^3}{e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]