Integrand size = 35, antiderivative size = 185 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {e^3 \left (10 c^2 d^4-15 a c d^2 e^2+6 a^2 e^4\right ) x}{c^5 d^5}+\frac {e^4 \left (5 c d^2-3 a e^2\right ) x^2}{2 c^4 d^4}+\frac {e^5 x^3}{3 c^3 d^3}-\frac {\left (c d^2-a e^2\right )^5}{2 c^6 d^6 (a e+c d x)^2}-\frac {5 e \left (c d^2-a e^2\right )^4}{c^6 d^6 (a e+c d x)}+\frac {10 e^2 \left (c d^2-a e^2\right )^3 \log (a e+c d x)}{c^6 d^6} \]
e^3*(6*a^2*e^4-15*a*c*d^2*e^2+10*c^2*d^4)*x/c^5/d^5+1/2*e^4*(-3*a*e^2+5*c* d^2)*x^2/c^4/d^4+1/3*e^5*x^3/c^3/d^3-1/2*(-a*e^2+c*d^2)^5/c^6/d^6/(c*d*x+a *e)^2-5*e*(-a*e^2+c*d^2)^4/c^6/d^6/(c*d*x+a*e)+10*e^2*(-a*e^2+c*d^2)^3*ln( c*d*x+a*e)/c^6/d^6
Time = 0.06 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.42 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {-27 a^5 e^{10}+3 a^4 c d e^8 (35 d+2 e x)+3 a^3 c^2 d^2 e^6 \left (-50 d^2+10 d e x+21 e^2 x^2\right )+5 a^2 c^3 d^3 e^4 \left (18 d^3-24 d^2 e x-33 d e^2 x^2+4 e^3 x^3\right )-5 a c^4 d^4 e^2 \left (3 d^4-24 d^3 e x-24 d^2 e^2 x^2+12 d e^3 x^3+e^4 x^4\right )+c^5 d^5 \left (-3 d^5-30 d^4 e x+60 d^2 e^3 x^3+15 d e^4 x^4+2 e^5 x^5\right )-60 e^2 \left (-c d^2+a e^2\right )^3 (a e+c d x)^2 \log (a e+c d x)}{6 c^6 d^6 (a e+c d x)^2} \]
(-27*a^5*e^10 + 3*a^4*c*d*e^8*(35*d + 2*e*x) + 3*a^3*c^2*d^2*e^6*(-50*d^2 + 10*d*e*x + 21*e^2*x^2) + 5*a^2*c^3*d^3*e^4*(18*d^3 - 24*d^2*e*x - 33*d*e ^2*x^2 + 4*e^3*x^3) - 5*a*c^4*d^4*e^2*(3*d^4 - 24*d^3*e*x - 24*d^2*e^2*x^2 + 12*d*e^3*x^3 + e^4*x^4) + c^5*d^5*(-3*d^5 - 30*d^4*e*x + 60*d^2*e^3*x^3 + 15*d*e^4*x^4 + 2*e^5*x^5) - 60*e^2*(-(c*d^2) + a*e^2)^3*(a*e + c*d*x)^2 *Log[a*e + c*d*x])/(6*c^6*d^6*(a*e + c*d*x)^2)
Time = 0.40 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {1121, 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.
| \(\displaystyle \int \frac {(d+e x)^8}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1121 |
| \(\displaystyle \int \left (\frac {6 a^2 e^7-15 a c d^2 e^5+10 c^2 d^4 e^3}{c^5 d^5}+\frac {10 e^2 \left (c d^2-a e^2\right )^3}{c^5 d^5 (a e+c d x)}+\frac {5 e \left (c d^2-a e^2\right )^4}{c^5 d^5 (a e+c d x)^2}+\frac {\left (c d^2-a e^2\right )^5}{c^5 d^5 (a e+c d x)^3}+\frac {e^4 x \left (5 c d^2-3 a e^2\right )}{c^4 d^4}+\frac {e^5 x^2}{c^3 d^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^3 x \left (6 a^2 e^4-15 a c d^2 e^2+10 c^2 d^4\right )}{c^5 d^5}-\frac {5 e \left (c d^2-a e^2\right )^4}{c^6 d^6 (a e+c d x)}-\frac {\left (c d^2-a e^2\right )^5}{2 c^6 d^6 (a e+c d x)^2}+\frac {10 e^2 \left (c d^2-a e^2\right )^3 \log (a e+c d x)}{c^6 d^6}+\frac {e^4 x^2 \left (5 c d^2-3 a e^2\right )}{2 c^4 d^4}+\frac {e^5 x^3}{3 c^3 d^3}\) |
(e^3*(10*c^2*d^4 - 15*a*c*d^2*e^2 + 6*a^2*e^4)*x)/(c^5*d^5) + (e^4*(5*c*d^ 2 - 3*a*e^2)*x^2)/(2*c^4*d^4) + (e^5*x^3)/(3*c^3*d^3) - (c*d^2 - a*e^2)^5/ (2*c^6*d^6*(a*e + c*d*x)^2) - (5*e*(c*d^2 - a*e^2)^4)/(c^6*d^6*(a*e + c*d* x)) + (10*e^2*(c*d^2 - a*e^2)^3*Log[a*e + c*d*x])/(c^6*d^6)
3.19.86.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
Time = 2.47 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.61
| method | result | size |
| default | \(\frac {e^{3} \left (\frac {1}{3} x^{3} c^{2} d^{2} e^{2}-\frac {3}{2} x^{2} a c d \,e^{3}+\frac {5}{2} x^{2} c^{2} d^{3} e +6 a^{2} e^{4} x -15 a c \,d^{2} e^{2} x +10 c^{2} d^{4} x \right )}{c^{5} d^{5}}-\frac {-a^{5} e^{10}+5 a^{4} c \,d^{2} e^{8}-10 a^{3} c^{2} d^{4} e^{6}+10 a^{2} c^{3} d^{6} e^{4}-5 a \,c^{4} d^{8} e^{2}+c^{5} d^{10}}{2 c^{6} d^{6} \left (c d x +a e \right )^{2}}-\frac {10 e^{2} \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \ln \left (c d x +a e \right )}{c^{6} d^{6}}-\frac {5 e \left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right )}{c^{6} d^{6} \left (c d x +a e \right )}\) | \(297\) |
| risch | \(\frac {e^{5} x^{3}}{3 c^{3} d^{3}}-\frac {3 e^{6} x^{2} a}{2 c^{4} d^{4}}+\frac {5 e^{4} x^{2}}{2 c^{3} d^{2}}+\frac {6 e^{7} a^{2} x}{c^{5} d^{5}}-\frac {15 e^{5} a x}{c^{4} d^{3}}+\frac {10 e^{3} x}{c^{3} d}+\frac {\left (-5 a^{4} e^{9}+20 a^{3} c \,d^{2} e^{7}-30 a^{2} e^{5} d^{4} c^{2}+20 a \,e^{3} d^{6} c^{3}-5 c^{4} d^{8} e \right ) x -\frac {9 a^{5} e^{10}-35 a^{4} c \,d^{2} e^{8}+50 a^{3} c^{2} d^{4} e^{6}-30 a^{2} c^{3} d^{6} e^{4}+5 a \,c^{4} d^{8} e^{2}+c^{5} d^{10}}{2 c d}}{c^{5} d^{5} \left (c d x +a e \right )^{2}}-\frac {10 e^{8} \ln \left (c d x +a e \right ) a^{3}}{c^{6} d^{6}}+\frac {30 e^{6} \ln \left (c d x +a e \right ) a^{2}}{c^{5} d^{4}}-\frac {30 e^{4} \ln \left (c d x +a e \right ) a}{c^{4} d^{2}}+\frac {10 e^{2} \ln \left (c d x +a e \right )}{c^{3}}\) | \(321\) |
| norman | \(\frac {-\frac {90 a^{5} e^{10}-230 a^{4} c \,d^{2} e^{8}+145 a^{3} c^{2} d^{4} e^{6}+45 a^{2} c^{3} d^{6} e^{4}+15 a \,c^{4} d^{8} e^{2}+3 c^{5} d^{10}}{6 d^{4} c^{6}}+\frac {e^{7} x^{7}}{3 c d}-\frac {\left (90 a^{5} e^{14}+10 c \,d^{2} a^{4} e^{12}-415 a^{3} c^{2} d^{4} e^{10}+305 a^{2} c^{3} d^{6} e^{8}+190 a \,c^{4} d^{8} e^{6}+198 c^{5} d^{10} e^{4}\right ) x^{2}}{6 c^{6} d^{6} e^{2}}-\frac {\left (90 a^{5} e^{12}-170 a^{4} c \,d^{2} e^{10}+5 a^{3} c^{2} d^{4} e^{8}+100 a^{2} c^{3} d^{6} e^{6}+90 a \,c^{4} d^{8} e^{4}+18 c^{5} d^{10} e^{2}\right ) x}{3 c^{6} d^{5} e}-\frac {\left (60 a^{4} e^{12}-140 a^{3} c \,d^{2} e^{10}+85 d^{4} a^{2} c^{2} e^{8}-20 a \,c^{3} d^{6} e^{6}+120 c^{4} d^{8} e^{4}\right ) x^{3}}{3 c^{5} d^{5} e}+\frac {e^{5} \left (10 a^{2} e^{4}-35 a c \,d^{2} e^{2}+46 c^{2} d^{4}\right ) x^{5}}{3 c^{3} d^{3}}-\frac {e^{6} \left (5 e^{2} a -19 c \,d^{2}\right ) x^{6}}{6 c^{2} d^{2}}}{\left (c d x +a e \right )^{2} \left (e x +d \right )^{2}}-\frac {10 e^{2} \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \ln \left (c d x +a e \right )}{c^{6} d^{6}}\) | \(482\) |
| parallelrisch | \(-\frac {-360 a^{3} c^{2} d^{3} e^{7} x +360 a^{2} c^{3} d^{5} e^{5} x +120 a^{4} c d \,e^{9} x -270 a^{4} c \,d^{2} e^{8}+270 a^{3} c^{2} d^{4} e^{6}-90 a^{2} c^{3} d^{6} e^{4}+15 a \,c^{4} d^{8} e^{2}+30 c^{5} d^{9} e x -60 \ln \left (c d x +a e \right ) x^{2} c^{5} d^{8} e^{2}+60 \ln \left (c d x +a e \right ) x^{2} a^{3} c^{2} d^{2} e^{8}-180 \ln \left (c d x +a e \right ) x^{2} a^{2} c^{3} d^{4} e^{6}+120 \ln \left (c d x +a e \right ) x \,a^{4} c d \,e^{9}-360 \ln \left (c d x +a e \right ) x \,a^{3} c^{2} d^{3} e^{7}+360 \ln \left (c d x +a e \right ) x \,a^{2} c^{3} d^{5} e^{5}-120 \ln \left (c d x +a e \right ) x a \,c^{4} d^{7} e^{3}+90 a^{5} e^{10}+3 c^{5} d^{10}+60 \ln \left (c d x +a e \right ) a^{5} e^{10}-15 x^{4} c^{5} d^{6} e^{4}-2 x^{5} e^{5} c^{5} d^{5}-60 x^{3} c^{5} d^{7} e^{3}+180 \ln \left (c d x +a e \right ) x^{2} a \,c^{4} d^{6} e^{4}-120 x a \,c^{4} d^{7} e^{3}-180 \ln \left (c d x +a e \right ) a^{4} c \,d^{2} e^{8}+180 \ln \left (c d x +a e \right ) a^{3} c^{2} d^{4} e^{6}-60 \ln \left (c d x +a e \right ) a^{2} c^{3} d^{6} e^{4}-20 x^{3} a^{2} c^{3} d^{3} e^{7}+60 x^{3} a \,c^{4} d^{5} e^{5}+5 x^{4} a \,c^{4} d^{4} e^{6}}{6 c^{6} d^{6} \left (c d x +a e \right )^{2}}\) | \(514\) |
e^3/c^5/d^5*(1/3*x^3*c^2*d^2*e^2-3/2*x^2*a*c*d*e^3+5/2*x^2*c^2*d^3*e+6*a^2 *e^4*x-15*a*c*d^2*e^2*x+10*c^2*d^4*x)-1/2/c^6/d^6*(-a^5*e^10+5*a^4*c*d^2*e ^8-10*a^3*c^2*d^4*e^6+10*a^2*c^3*d^6*e^4-5*a*c^4*d^8*e^2+c^5*d^10)/(c*d*x+ a*e)^2-10/c^6/d^6*e^2*(a^3*e^6-3*a^2*c*d^2*e^4+3*a*c^2*d^4*e^2-c^3*d^6)*ln (c*d*x+a*e)-5/c^6/d^6*e*(a^4*e^8-4*a^3*c*d^2*e^6+6*a^2*c^2*d^4*e^4-4*a*c^3 *d^6*e^2+c^4*d^8)/(c*d*x+a*e)
Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (179) = 358\).
Time = 0.26 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.54 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2 \, c^{5} d^{5} e^{5} x^{5} - 3 \, c^{5} d^{10} - 15 \, a c^{4} d^{8} e^{2} + 90 \, a^{2} c^{3} d^{6} e^{4} - 150 \, a^{3} c^{2} d^{4} e^{6} + 105 \, a^{4} c d^{2} e^{8} - 27 \, a^{5} e^{10} + 5 \, {\left (3 \, c^{5} d^{6} e^{4} - a c^{4} d^{4} e^{6}\right )} x^{4} + 20 \, {\left (3 \, c^{5} d^{7} e^{3} - 3 \, a c^{4} d^{5} e^{5} + a^{2} c^{3} d^{3} e^{7}\right )} x^{3} + 3 \, {\left (40 \, a c^{4} d^{6} e^{4} - 55 \, a^{2} c^{3} d^{4} e^{6} + 21 \, a^{3} c^{2} d^{2} e^{8}\right )} x^{2} - 6 \, {\left (5 \, c^{5} d^{9} e - 20 \, a c^{4} d^{7} e^{3} + 20 \, a^{2} c^{3} d^{5} e^{5} - 5 \, a^{3} c^{2} d^{3} e^{7} - a^{4} c d e^{9}\right )} x + 60 \, {\left (a^{2} c^{3} d^{6} e^{4} - 3 \, a^{3} c^{2} d^{4} e^{6} + 3 \, a^{4} c d^{2} e^{8} - a^{5} e^{10} + {\left (c^{5} d^{8} e^{2} - 3 \, a c^{4} d^{6} e^{4} + 3 \, a^{2} c^{3} d^{4} e^{6} - a^{3} c^{2} d^{2} e^{8}\right )} x^{2} + 2 \, {\left (a c^{4} d^{7} e^{3} - 3 \, a^{2} c^{3} d^{5} e^{5} + 3 \, a^{3} c^{2} d^{3} e^{7} - a^{4} c d e^{9}\right )} x\right )} \log \left (c d x + a e\right )}{6 \, {\left (c^{8} d^{8} x^{2} + 2 \, a c^{7} d^{7} e x + a^{2} c^{6} d^{6} e^{2}\right )}} \]
1/6*(2*c^5*d^5*e^5*x^5 - 3*c^5*d^10 - 15*a*c^4*d^8*e^2 + 90*a^2*c^3*d^6*e^ 4 - 150*a^3*c^2*d^4*e^6 + 105*a^4*c*d^2*e^8 - 27*a^5*e^10 + 5*(3*c^5*d^6*e ^4 - a*c^4*d^4*e^6)*x^4 + 20*(3*c^5*d^7*e^3 - 3*a*c^4*d^5*e^5 + a^2*c^3*d^ 3*e^7)*x^3 + 3*(40*a*c^4*d^6*e^4 - 55*a^2*c^3*d^4*e^6 + 21*a^3*c^2*d^2*e^8 )*x^2 - 6*(5*c^5*d^9*e - 20*a*c^4*d^7*e^3 + 20*a^2*c^3*d^5*e^5 - 5*a^3*c^2 *d^3*e^7 - a^4*c*d*e^9)*x + 60*(a^2*c^3*d^6*e^4 - 3*a^3*c^2*d^4*e^6 + 3*a^ 4*c*d^2*e^8 - a^5*e^10 + (c^5*d^8*e^2 - 3*a*c^4*d^6*e^4 + 3*a^2*c^3*d^4*e^ 6 - a^3*c^2*d^2*e^8)*x^2 + 2*(a*c^4*d^7*e^3 - 3*a^2*c^3*d^5*e^5 + 3*a^3*c^ 2*d^3*e^7 - a^4*c*d*e^9)*x)*log(c*d*x + a*e))/(c^8*d^8*x^2 + 2*a*c^7*d^7*e *x + a^2*c^6*d^6*e^2)
Time = 47.70 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.64 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=x^{2} \left (- \frac {3 a e^{6}}{2 c^{4} d^{4}} + \frac {5 e^{4}}{2 c^{3} d^{2}}\right ) + x \left (\frac {6 a^{2} e^{7}}{c^{5} d^{5}} - \frac {15 a e^{5}}{c^{4} d^{3}} + \frac {10 e^{3}}{c^{3} d}\right ) + \frac {- 9 a^{5} e^{10} + 35 a^{4} c d^{2} e^{8} - 50 a^{3} c^{2} d^{4} e^{6} + 30 a^{2} c^{3} d^{6} e^{4} - 5 a c^{4} d^{8} e^{2} - c^{5} d^{10} + x \left (- 10 a^{4} c d e^{9} + 40 a^{3} c^{2} d^{3} e^{7} - 60 a^{2} c^{3} d^{5} e^{5} + 40 a c^{4} d^{7} e^{3} - 10 c^{5} d^{9} e\right )}{2 a^{2} c^{6} d^{6} e^{2} + 4 a c^{7} d^{7} e x + 2 c^{8} d^{8} x^{2}} + \frac {e^{5} x^{3}}{3 c^{3} d^{3}} - \frac {10 e^{2} \left (a e^{2} - c d^{2}\right )^{3} \log {\left (a e + c d x \right )}}{c^{6} d^{6}} \]
x**2*(-3*a*e**6/(2*c**4*d**4) + 5*e**4/(2*c**3*d**2)) + x*(6*a**2*e**7/(c* *5*d**5) - 15*a*e**5/(c**4*d**3) + 10*e**3/(c**3*d)) + (-9*a**5*e**10 + 35 *a**4*c*d**2*e**8 - 50*a**3*c**2*d**4*e**6 + 30*a**2*c**3*d**6*e**4 - 5*a* c**4*d**8*e**2 - c**5*d**10 + x*(-10*a**4*c*d*e**9 + 40*a**3*c**2*d**3*e** 7 - 60*a**2*c**3*d**5*e**5 + 40*a*c**4*d**7*e**3 - 10*c**5*d**9*e))/(2*a** 2*c**6*d**6*e**2 + 4*a*c**7*d**7*e*x + 2*c**8*d**8*x**2) + e**5*x**3/(3*c* *3*d**3) - 10*e**2*(a*e**2 - c*d**2)**3*log(a*e + c*d*x)/(c**6*d**6)
Time = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.68 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {c^{5} d^{10} + 5 \, a c^{4} d^{8} e^{2} - 30 \, a^{2} c^{3} d^{6} e^{4} + 50 \, a^{3} c^{2} d^{4} e^{6} - 35 \, a^{4} c d^{2} e^{8} + 9 \, a^{5} e^{10} + 10 \, {\left (c^{5} d^{9} e - 4 \, a c^{4} d^{7} e^{3} + 6 \, a^{2} c^{3} d^{5} e^{5} - 4 \, a^{3} c^{2} d^{3} e^{7} + a^{4} c d e^{9}\right )} x}{2 \, {\left (c^{8} d^{8} x^{2} + 2 \, a c^{7} d^{7} e x + a^{2} c^{6} d^{6} e^{2}\right )}} + \frac {2 \, c^{2} d^{2} e^{5} x^{3} + 3 \, {\left (5 \, c^{2} d^{3} e^{4} - 3 \, a c d e^{6}\right )} x^{2} + 6 \, {\left (10 \, c^{2} d^{4} e^{3} - 15 \, a c d^{2} e^{5} + 6 \, a^{2} e^{7}\right )} x}{6 \, c^{5} d^{5}} + \frac {10 \, {\left (c^{3} d^{6} e^{2} - 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} - a^{3} e^{8}\right )} \log \left (c d x + a e\right )}{c^{6} d^{6}} \]
-1/2*(c^5*d^10 + 5*a*c^4*d^8*e^2 - 30*a^2*c^3*d^6*e^4 + 50*a^3*c^2*d^4*e^6 - 35*a^4*c*d^2*e^8 + 9*a^5*e^10 + 10*(c^5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2 *c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)*x)/(c^8*d^8*x^2 + 2*a*c^7* d^7*e*x + a^2*c^6*d^6*e^2) + 1/6*(2*c^2*d^2*e^5*x^3 + 3*(5*c^2*d^3*e^4 - 3 *a*c*d*e^6)*x^2 + 6*(10*c^2*d^4*e^3 - 15*a*c*d^2*e^5 + 6*a^2*e^7)*x)/(c^5* d^5) + 10*(c^3*d^6*e^2 - 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 - a^3*e^8)*log( c*d*x + a*e)/(c^6*d^6)
Time = 0.27 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {10 \, {\left (c^{3} d^{6} e^{2} - 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} - a^{3} e^{8}\right )} \log \left ({\left | c d x + a e \right |}\right )}{c^{6} d^{6}} - \frac {c^{5} d^{10} + 5 \, a c^{4} d^{8} e^{2} - 30 \, a^{2} c^{3} d^{6} e^{4} + 50 \, a^{3} c^{2} d^{4} e^{6} - 35 \, a^{4} c d^{2} e^{8} + 9 \, a^{5} e^{10} + 10 \, {\left (c^{5} d^{9} e - 4 \, a c^{4} d^{7} e^{3} + 6 \, a^{2} c^{3} d^{5} e^{5} - 4 \, a^{3} c^{2} d^{3} e^{7} + a^{4} c d e^{9}\right )} x}{2 \, {\left (c d x + a e\right )}^{2} c^{6} d^{6}} + \frac {2 \, c^{6} d^{6} e^{5} x^{3} + 15 \, c^{6} d^{7} e^{4} x^{2} - 9 \, a c^{5} d^{5} e^{6} x^{2} + 60 \, c^{6} d^{8} e^{3} x - 90 \, a c^{5} d^{6} e^{5} x + 36 \, a^{2} c^{4} d^{4} e^{7} x}{6 \, c^{9} d^{9}} \]
10*(c^3*d^6*e^2 - 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 - a^3*e^8)*log(abs(c*d *x + a*e))/(c^6*d^6) - 1/2*(c^5*d^10 + 5*a*c^4*d^8*e^2 - 30*a^2*c^3*d^6*e^ 4 + 50*a^3*c^2*d^4*e^6 - 35*a^4*c*d^2*e^8 + 9*a^5*e^10 + 10*(c^5*d^9*e - 4 *a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)*x)/( (c*d*x + a*e)^2*c^6*d^6) + 1/6*(2*c^6*d^6*e^5*x^3 + 15*c^6*d^7*e^4*x^2 - 9 *a*c^5*d^5*e^6*x^2 + 60*c^6*d^8*e^3*x - 90*a*c^5*d^6*e^5*x + 36*a^2*c^4*d^ 4*e^7*x)/(c^9*d^9)
Time = 9.87 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.84 \[ \int \frac {(d+e x)^8}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=x^2\,\left (\frac {5\,e^4}{2\,c^3\,d^2}-\frac {3\,a\,e^6}{2\,c^4\,d^4}\right )-x\,\left (\frac {3\,a^2\,e^7}{c^5\,d^5}-\frac {10\,e^3}{c^3\,d}+\frac {3\,a\,e\,\left (\frac {5\,e^4}{c^3\,d^2}-\frac {3\,a\,e^6}{c^4\,d^4}\right )}{c\,d}\right )-\frac {x\,\left (5\,a^4\,e^9-20\,a^3\,c\,d^2\,e^7+30\,a^2\,c^2\,d^4\,e^5-20\,a\,c^3\,d^6\,e^3+5\,c^4\,d^8\,e\right )+\frac {9\,a^5\,e^{10}-35\,a^4\,c\,d^2\,e^8+50\,a^3\,c^2\,d^4\,e^6-30\,a^2\,c^3\,d^6\,e^4+5\,a\,c^4\,d^8\,e^2+c^5\,d^{10}}{2\,c\,d}}{a^2\,c^5\,d^5\,e^2+2\,a\,c^6\,d^6\,e\,x+c^7\,d^7\,x^2}-\frac {\ln \left (a\,e+c\,d\,x\right )\,\left (10\,a^3\,e^8-30\,a^2\,c\,d^2\,e^6+30\,a\,c^2\,d^4\,e^4-10\,c^3\,d^6\,e^2\right )}{c^6\,d^6}+\frac {e^5\,x^3}{3\,c^3\,d^3} \]
x^2*((5*e^4)/(2*c^3*d^2) - (3*a*e^6)/(2*c^4*d^4)) - x*((3*a^2*e^7)/(c^5*d^ 5) - (10*e^3)/(c^3*d) + (3*a*e*((5*e^4)/(c^3*d^2) - (3*a*e^6)/(c^4*d^4)))/ (c*d)) - (x*(5*a^4*e^9 + 5*c^4*d^8*e - 20*a*c^3*d^6*e^3 - 20*a^3*c*d^2*e^7 + 30*a^2*c^2*d^4*e^5) + (9*a^5*e^10 + c^5*d^10 + 5*a*c^4*d^8*e^2 - 35*a^4 *c*d^2*e^8 - 30*a^2*c^3*d^6*e^4 + 50*a^3*c^2*d^4*e^6)/(2*c*d))/(c^7*d^7*x^ 2 + a^2*c^5*d^5*e^2 + 2*a*c^6*d^6*e*x) - (log(a*e + c*d*x)*(10*a^3*e^8 - 1 0*c^3*d^6*e^2 + 30*a*c^2*d^4*e^4 - 30*a^2*c*d^2*e^6))/(c^6*d^6) + (e^5*x^3 )/(3*c^3*d^3)