Integrand size = 38, antiderivative size = 317 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{2 e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^2}+\frac {40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2 (d+e x)}-\frac {\left (423 d^3-4101 d^2 e+879 d e^2+703 e^3\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{5 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^3}+\frac {\left (100 d^6-120 d^5 e+228 d^4 e^2-242 d^3 e^3+141 d^2 e^4+120 d e^5-e^6\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^3}+\frac {\left (458 d^3-21 d^2 e-816 d e^2+113 e^3\right ) \log \left (3+2 x+5 x^2\right )}{10 \left (5 d^2-2 d e+3 e^2\right )^3} \]
1/2*(-4*d^4-5*d^3*e-3*d^2*e^2+d*e^3-2*e^4)/e^3/(5*d^2-2*d*e+3*e^2)/(e*x+d) ^2+(40*d^5+d^4*e+28*d^3*e^2+44*d^2*e^3-2*d*e^4+e^5)/e^3/(5*d^2-2*d*e+3*e^2 )^2/(e*x+d)+(100*d^6-120*d^5*e+228*d^4*e^2-242*d^3*e^3+141*d^2*e^4+120*d*e ^5-e^6)*ln(e*x+d)/e^3/(5*d^2-2*d*e+3*e^2)^3+1/10*(458*d^3-21*d^2*e-816*d*e ^2+113*e^3)*ln(5*x^2+2*x+3)/(5*d^2-2*d*e+3*e^2)^3-1/70*(423*d^3-4101*d^2*e +879*d*e^2+703*e^3)*arctan(1/14*(1+5*x)*14^(1/2))/(5*d^2-2*d*e+3*e^2)^3*14 ^(1/2)
Time = 0.26 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.88 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=-\frac {\frac {35 \left (5 d^2-2 d e+3 e^2\right )^2 \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right )}{e^3 (d+e x)^2}-\frac {70 \left (5 d^2-2 d e+3 e^2\right ) \left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right )}{e^3 (d+e x)}+\sqrt {14} \left (423 d^3-4101 d^2 e+879 d e^2+703 e^3\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )+\frac {70 \left (-100 d^6+120 d^5 e-228 d^4 e^2+242 d^3 e^3-141 d^2 e^4-120 d e^5+e^6\right ) \log (d+e x)}{e^3}-7 \left (458 d^3-21 d^2 e-816 d e^2+113 e^3\right ) \log \left (3+2 x+5 x^2\right )}{70 \left (5 d^2-2 d e+3 e^2\right )^3} \]
-1/70*((35*(5*d^2 - 2*d*e + 3*e^2)^2*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4))/(e^3*(d + e*x)^2) - (70*(5*d^2 - 2*d*e + 3*e^2)*(40*d^5 + d^4*e + 28*d^3*e^2 + 44*d^2*e^3 - 2*d*e^4 + e^5))/(e^3*(d + e*x)) + Sqrt[14]*(42 3*d^3 - 4101*d^2*e + 879*d*e^2 + 703*e^3)*ArcTan[(1 + 5*x)/Sqrt[14]] + (70 *(-100*d^6 + 120*d^5*e - 228*d^4*e^2 + 242*d^3*e^3 - 141*d^2*e^4 - 120*d*e ^5 + e^6)*Log[d + e*x])/e^3 - 7*(458*d^3 - 21*d^2*e - 816*d*e^2 + 113*e^3) *Log[3 + 2*x + 5*x^2])/(5*d^2 - 2*d*e + 3*e^2)^3
Time = 0.55 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, 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.
| \(\displaystyle \int \frac {4 x^4-5 x^3+3 x^2+x+2}{\left (5 x^2+2 x+3\right ) (d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {7 d^3+816 d^2 e+x \left (458 d^3-21 d^2 e-816 d e^2+113 e^3\right )-339 d e^2-118 e^3}{\left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^3}+\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^2 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^3}+\frac {-40 d^5-d^4 e-28 d^3 e^2-44 d^2 e^3+2 d e^4-e^5}{e^2 \left (5 d^2-2 d e+3 e^2\right )^2 (d+e x)^2}+\frac {100 d^6-120 d^5 e+228 d^4 e^2-242 d^3 e^3+141 d^2 e^4+120 d e^5-e^6}{e^2 \left (5 d^2-2 d e+3 e^2\right )^3 (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {5 x+1}{\sqrt {14}}\right ) \left (423 d^3-4101 d^2 e+879 d e^2+703 e^3\right )}{5 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^3}+\frac {\left (458 d^3-21 d^2 e-816 d e^2+113 e^3\right ) \log \left (5 x^2+2 x+3\right )}{10 \left (5 d^2-2 d e+3 e^2\right )^3}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{2 e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^2}+\frac {40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2 (d+e x)}+\frac {\left (100 d^6-120 d^5 e+228 d^4 e^2-242 d^3 e^3+141 d^2 e^4+120 d e^5-e^6\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^3}\) |
-1/2*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)/(e^3*(5*d^2 - 2*d*e + 3 *e^2)*(d + e*x)^2) + (40*d^5 + d^4*e + 28*d^3*e^2 + 44*d^2*e^3 - 2*d*e^4 + e^5)/(e^3*(5*d^2 - 2*d*e + 3*e^2)^2*(d + e*x)) - ((423*d^3 - 4101*d^2*e + 879*d*e^2 + 703*e^3)*ArcTan[(1 + 5*x)/Sqrt[14]])/(5*Sqrt[14]*(5*d^2 - 2*d *e + 3*e^2)^3) + ((100*d^6 - 120*d^5*e + 228*d^4*e^2 - 242*d^3*e^3 + 141*d ^2*e^4 + 120*d*e^5 - e^6)*Log[d + e*x])/(e^3*(5*d^2 - 2*d*e + 3*e^2)^3) + ((458*d^3 - 21*d^2*e - 816*d*e^2 + 113*e^3)*Log[3 + 2*x + 5*x^2])/(10*(5*d ^2 - 2*d*e + 3*e^2)^3)
3.4.10.3.1 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 = 0.84 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.94
| method | result | size |
| default | \(\frac {\frac {\left (458 d^{3}-21 d^{2} e -816 d \,e^{2}+113 e^{3}\right ) \ln \left (5 x^{2}+2 x +3\right )}{10}+\frac {\left (-\frac {423}{5} d^{3}+\frac {4101}{5} d^{2} e -\frac {879}{5} d \,e^{2}-\frac {703}{5} e^{3}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{14}}{\left (5 d^{2}-2 d e +3 e^{2}\right )^{3}}-\frac {4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}}{2 e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right ) \left (e x +d \right )^{2}}-\frac {-40 d^{5}-d^{4} e -28 d^{3} e^{2}-44 d^{2} e^{3}+2 d \,e^{4}-e^{5}}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right )^{2} \left (e x +d \right )}+\frac {\left (100 d^{6}-120 d^{5} e +228 d^{4} e^{2}-242 d^{3} e^{3}+141 d^{2} e^{4}+120 d \,e^{5}-e^{6}\right ) \ln \left (e x +d \right )}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right )^{3}}\) | \(298\) |
| risch | \(\text {Expression too large to display}\) | \(1138\) |
1/(5*d^2-2*d*e+3*e^2)^3*(1/10*(458*d^3-21*d^2*e-816*d*e^2+113*e^3)*ln(5*x^ 2+2*x+3)+1/14*(-423/5*d^3+4101/5*d^2*e-879/5*d*e^2-703/5*e^3)*14^(1/2)*arc tan(1/28*(10*x+2)*14^(1/2)))-1/2*(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)/e^3 /(5*d^2-2*d*e+3*e^2)/(e*x+d)^2-(-40*d^5-d^4*e-28*d^3*e^2-44*d^2*e^3+2*d*e^ 4-e^5)/e^3/(5*d^2-2*d*e+3*e^2)^2/(e*x+d)+(100*d^6-120*d^5*e+228*d^4*e^2-24 2*d^3*e^3+141*d^2*e^4+120*d*e^5-e^6)*ln(e*x+d)/e^3/(5*d^2-2*d*e+3*e^2)^3
Leaf count of result is larger than twice the leaf count of optimal. 698 vs. \(2 (308) = 616\).
Time = 0.40 (sec) , antiderivative size = 698, normalized size of antiderivative = 2.20 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=\frac {10500 \, d^{8} - 6825 \, d^{7} e + 14175 \, d^{6} e^{2} + 10395 \, d^{5} e^{3} - 6160 \, d^{4} e^{4} + 12145 \, d^{3} e^{5} - 4305 \, d^{2} e^{6} + 1365 \, d e^{7} - 630 \, e^{8} - \sqrt {14} {\left (423 \, d^{5} e^{3} - 4101 \, d^{4} e^{4} + 879 \, d^{3} e^{5} + 703 \, d^{2} e^{6} + {\left (423 \, d^{3} e^{5} - 4101 \, d^{2} e^{6} + 879 \, d e^{7} + 703 \, e^{8}\right )} x^{2} + 2 \, {\left (423 \, d^{4} e^{4} - 4101 \, d^{3} e^{5} + 879 \, d^{2} e^{6} + 703 \, d e^{7}\right )} x\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + 70 \, {\left (200 \, d^{7} e - 75 \, d^{6} e^{2} + 258 \, d^{5} e^{3} + 167 \, d^{4} e^{4} - 14 \, d^{3} e^{5} + 141 \, d^{2} e^{6} - 8 \, d e^{7} + 3 \, e^{8}\right )} x + 70 \, {\left (100 \, d^{8} - 120 \, d^{7} e + 228 \, d^{6} e^{2} - 242 \, d^{5} e^{3} + 141 \, d^{4} e^{4} + 120 \, d^{3} e^{5} - d^{2} e^{6} + {\left (100 \, d^{6} e^{2} - 120 \, d^{5} e^{3} + 228 \, d^{4} e^{4} - 242 \, d^{3} e^{5} + 141 \, d^{2} e^{6} + 120 \, d e^{7} - e^{8}\right )} x^{2} + 2 \, {\left (100 \, d^{7} e - 120 \, d^{6} e^{2} + 228 \, d^{5} e^{3} - 242 \, d^{4} e^{4} + 141 \, d^{3} e^{5} + 120 \, d^{2} e^{6} - d e^{7}\right )} x\right )} \log \left (e x + d\right ) + 7 \, {\left (458 \, d^{5} e^{3} - 21 \, d^{4} e^{4} - 816 \, d^{3} e^{5} + 113 \, d^{2} e^{6} + {\left (458 \, d^{3} e^{5} - 21 \, d^{2} e^{6} - 816 \, d e^{7} + 113 \, e^{8}\right )} x^{2} + 2 \, {\left (458 \, d^{4} e^{4} - 21 \, d^{3} e^{5} - 816 \, d^{2} e^{6} + 113 \, d e^{7}\right )} x\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{70 \, {\left (125 \, d^{8} e^{3} - 150 \, d^{7} e^{4} + 285 \, d^{6} e^{5} - 188 \, d^{5} e^{6} + 171 \, d^{4} e^{7} - 54 \, d^{3} e^{8} + 27 \, d^{2} e^{9} + {\left (125 \, d^{6} e^{5} - 150 \, d^{5} e^{6} + 285 \, d^{4} e^{7} - 188 \, d^{3} e^{8} + 171 \, d^{2} e^{9} - 54 \, d e^{10} + 27 \, e^{11}\right )} x^{2} + 2 \, {\left (125 \, d^{7} e^{4} - 150 \, d^{6} e^{5} + 285 \, d^{5} e^{6} - 188 \, d^{4} e^{7} + 171 \, d^{3} e^{8} - 54 \, d^{2} e^{9} + 27 \, d e^{10}\right )} x\right )}} \]
1/70*(10500*d^8 - 6825*d^7*e + 14175*d^6*e^2 + 10395*d^5*e^3 - 6160*d^4*e^ 4 + 12145*d^3*e^5 - 4305*d^2*e^6 + 1365*d*e^7 - 630*e^8 - sqrt(14)*(423*d^ 5*e^3 - 4101*d^4*e^4 + 879*d^3*e^5 + 703*d^2*e^6 + (423*d^3*e^5 - 4101*d^2 *e^6 + 879*d*e^7 + 703*e^8)*x^2 + 2*(423*d^4*e^4 - 4101*d^3*e^5 + 879*d^2* e^6 + 703*d*e^7)*x)*arctan(1/14*sqrt(14)*(5*x + 1)) + 70*(200*d^7*e - 75*d ^6*e^2 + 258*d^5*e^3 + 167*d^4*e^4 - 14*d^3*e^5 + 141*d^2*e^6 - 8*d*e^7 + 3*e^8)*x + 70*(100*d^8 - 120*d^7*e + 228*d^6*e^2 - 242*d^5*e^3 + 141*d^4*e ^4 + 120*d^3*e^5 - d^2*e^6 + (100*d^6*e^2 - 120*d^5*e^3 + 228*d^4*e^4 - 24 2*d^3*e^5 + 141*d^2*e^6 + 120*d*e^7 - e^8)*x^2 + 2*(100*d^7*e - 120*d^6*e^ 2 + 228*d^5*e^3 - 242*d^4*e^4 + 141*d^3*e^5 + 120*d^2*e^6 - d*e^7)*x)*log( e*x + d) + 7*(458*d^5*e^3 - 21*d^4*e^4 - 816*d^3*e^5 + 113*d^2*e^6 + (458* d^3*e^5 - 21*d^2*e^6 - 816*d*e^7 + 113*e^8)*x^2 + 2*(458*d^4*e^4 - 21*d^3* e^5 - 816*d^2*e^6 + 113*d*e^7)*x)*log(5*x^2 + 2*x + 3))/(125*d^8*e^3 - 150 *d^7*e^4 + 285*d^6*e^5 - 188*d^5*e^6 + 171*d^4*e^7 - 54*d^3*e^8 + 27*d^2*e ^9 + (125*d^6*e^5 - 150*d^5*e^6 + 285*d^4*e^7 - 188*d^3*e^8 + 171*d^2*e^9 - 54*d*e^10 + 27*e^11)*x^2 + 2*(125*d^7*e^4 - 150*d^6*e^5 + 285*d^5*e^6 - 188*d^4*e^7 + 171*d^3*e^8 - 54*d^2*e^9 + 27*d*e^10)*x)
Timed out. \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.57 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=-\frac {\sqrt {14} {\left (423 \, d^{3} - 4101 \, d^{2} e + 879 \, d e^{2} + 703 \, e^{3}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{70 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {{\left (100 \, d^{6} - 120 \, d^{5} e + 228 \, d^{4} e^{2} - 242 \, d^{3} e^{3} + 141 \, d^{2} e^{4} + 120 \, d e^{5} - e^{6}\right )} \log \left (e x + d\right )}{125 \, d^{6} e^{3} - 150 \, d^{5} e^{4} + 285 \, d^{4} e^{5} - 188 \, d^{3} e^{6} + 171 \, d^{2} e^{7} - 54 \, d e^{8} + 27 \, e^{9}} + \frac {{\left (458 \, d^{3} - 21 \, d^{2} e - 816 \, d e^{2} + 113 \, e^{3}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{10 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {60 \, d^{6} - 15 \, d^{5} e + 39 \, d^{4} e^{2} + 84 \, d^{3} e^{3} - 25 \, d^{2} e^{4} + 9 \, d e^{5} - 6 \, e^{6} + 2 \, {\left (40 \, d^{5} e + d^{4} e^{2} + 28 \, d^{3} e^{3} + 44 \, d^{2} e^{4} - 2 \, d e^{5} + e^{6}\right )} x}{2 \, {\left (25 \, d^{6} e^{3} - 20 \, d^{5} e^{4} + 34 \, d^{4} e^{5} - 12 \, d^{3} e^{6} + 9 \, d^{2} e^{7} + {\left (25 \, d^{4} e^{5} - 20 \, d^{3} e^{6} + 34 \, d^{2} e^{7} - 12 \, d e^{8} + 9 \, e^{9}\right )} x^{2} + 2 \, {\left (25 \, d^{5} e^{4} - 20 \, d^{4} e^{5} + 34 \, d^{3} e^{6} - 12 \, d^{2} e^{7} + 9 \, d e^{8}\right )} x\right )}} \]
-1/70*sqrt(14)*(423*d^3 - 4101*d^2*e + 879*d*e^2 + 703*e^3)*arctan(1/14*sq rt(14)*(5*x + 1))/(125*d^6 - 150*d^5*e + 285*d^4*e^2 - 188*d^3*e^3 + 171*d ^2*e^4 - 54*d*e^5 + 27*e^6) + (100*d^6 - 120*d^5*e + 228*d^4*e^2 - 242*d^3 *e^3 + 141*d^2*e^4 + 120*d*e^5 - e^6)*log(e*x + d)/(125*d^6*e^3 - 150*d^5* e^4 + 285*d^4*e^5 - 188*d^3*e^6 + 171*d^2*e^7 - 54*d*e^8 + 27*e^9) + 1/10* (458*d^3 - 21*d^2*e - 816*d*e^2 + 113*e^3)*log(5*x^2 + 2*x + 3)/(125*d^6 - 150*d^5*e + 285*d^4*e^2 - 188*d^3*e^3 + 171*d^2*e^4 - 54*d*e^5 + 27*e^6) + 1/2*(60*d^6 - 15*d^5*e + 39*d^4*e^2 + 84*d^3*e^3 - 25*d^2*e^4 + 9*d*e^5 - 6*e^6 + 2*(40*d^5*e + d^4*e^2 + 28*d^3*e^3 + 44*d^2*e^4 - 2*d*e^5 + e^6) *x)/(25*d^6*e^3 - 20*d^5*e^4 + 34*d^4*e^5 - 12*d^3*e^6 + 9*d^2*e^7 + (25*d ^4*e^5 - 20*d^3*e^6 + 34*d^2*e^7 - 12*d*e^8 + 9*e^9)*x^2 + 2*(25*d^5*e^4 - 20*d^4*e^5 + 34*d^3*e^6 - 12*d^2*e^7 + 9*d*e^8)*x)
Time = 0.29 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.38 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=-\frac {\sqrt {14} {\left (423 \, d^{3} - 4101 \, d^{2} e + 879 \, d e^{2} + 703 \, e^{3}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{70 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {{\left (458 \, d^{3} - 21 \, d^{2} e - 816 \, d e^{2} + 113 \, e^{3}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{10 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {{\left (100 \, d^{6} - 120 \, d^{5} e + 228 \, d^{4} e^{2} - 242 \, d^{3} e^{3} + 141 \, d^{2} e^{4} + 120 \, d e^{5} - e^{6}\right )} \log \left ({\left | e x + d \right |}\right )}{125 \, d^{6} e^{3} - 150 \, d^{5} e^{4} + 285 \, d^{4} e^{5} - 188 \, d^{3} e^{6} + 171 \, d^{2} e^{7} - 54 \, d e^{8} + 27 \, e^{9}} + \frac {2 \, {\left (200 \, d^{7} - 75 \, d^{6} e + 258 \, d^{5} e^{2} + 167 \, d^{4} e^{3} - 14 \, d^{3} e^{4} + 141 \, d^{2} e^{5} - 8 \, d e^{6} + 3 \, e^{7}\right )} x + \frac {300 \, d^{8} - 195 \, d^{7} e + 405 \, d^{6} e^{2} + 297 \, d^{5} e^{3} - 176 \, d^{4} e^{4} + 347 \, d^{3} e^{5} - 123 \, d^{2} e^{6} + 39 \, d e^{7} - 18 \, e^{8}}{e}}{2 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}^{3} {\left (e x + d\right )}^{2} e^{2}} \]
-1/70*sqrt(14)*(423*d^3 - 4101*d^2*e + 879*d*e^2 + 703*e^3)*arctan(1/14*sq rt(14)*(5*x + 1))/(125*d^6 - 150*d^5*e + 285*d^4*e^2 - 188*d^3*e^3 + 171*d ^2*e^4 - 54*d*e^5 + 27*e^6) + 1/10*(458*d^3 - 21*d^2*e - 816*d*e^2 + 113*e ^3)*log(5*x^2 + 2*x + 3)/(125*d^6 - 150*d^5*e + 285*d^4*e^2 - 188*d^3*e^3 + 171*d^2*e^4 - 54*d*e^5 + 27*e^6) + (100*d^6 - 120*d^5*e + 228*d^4*e^2 - 242*d^3*e^3 + 141*d^2*e^4 + 120*d*e^5 - e^6)*log(abs(e*x + d))/(125*d^6*e^ 3 - 150*d^5*e^4 + 285*d^4*e^5 - 188*d^3*e^6 + 171*d^2*e^7 - 54*d*e^8 + 27* e^9) + 1/2*(2*(200*d^7 - 75*d^6*e + 258*d^5*e^2 + 167*d^4*e^3 - 14*d^3*e^4 + 141*d^2*e^5 - 8*d*e^6 + 3*e^7)*x + (300*d^8 - 195*d^7*e + 405*d^6*e^2 + 297*d^5*e^3 - 176*d^4*e^4 + 347*d^3*e^5 - 123*d^2*e^6 + 39*d*e^7 - 18*e^8 )/e)/((5*d^2 - 2*d*e + 3*e^2)^3*(e*x + d)^2*e^2)
Time = 13.77 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.56 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^3 \left (3+2 x+5 x^2\right )} \, dx=\frac {\frac {60\,d^6-15\,d^5\,e+39\,d^4\,e^2+84\,d^3\,e^3-25\,d^2\,e^4+9\,d\,e^5-6\,e^6}{2\,e^3\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}+\frac {x\,\left (40\,d^5+d^4\,e+28\,d^3\,e^2+44\,d^2\,e^3-2\,d\,e^4+e^5\right )}{e^2\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}}{d^2+2\,d\,e\,x+e^2\,x^2}-\frac {\ln \left (x+\frac {1}{5}-\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{140}-\frac {229}{5}{}\mathrm {i}\right )\,d^3+\left (-\frac {4101\,\sqrt {14}}{140}+\frac {21}{10}{}\mathrm {i}\right )\,d^2\,e+\left (\frac {879\,\sqrt {14}}{140}+\frac {408}{5}{}\mathrm {i}\right )\,d\,e^2+\left (\frac {703\,\sqrt {14}}{140}-\frac {113}{10}{}\mathrm {i}\right )\,e^3\right )}{d^6\,125{}\mathrm {i}-d^5\,e\,150{}\mathrm {i}+d^4\,e^2\,285{}\mathrm {i}-d^3\,e^3\,188{}\mathrm {i}+d^2\,e^4\,171{}\mathrm {i}-d\,e^5\,54{}\mathrm {i}+e^6\,27{}\mathrm {i}}+\frac {\ln \left (x+\frac {1}{5}+\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{140}+\frac {229}{5}{}\mathrm {i}\right )\,d^3+\left (-\frac {4101\,\sqrt {14}}{140}-\frac {21}{10}{}\mathrm {i}\right )\,d^2\,e+\left (\frac {879\,\sqrt {14}}{140}-\frac {408}{5}{}\mathrm {i}\right )\,d\,e^2+\left (\frac {703\,\sqrt {14}}{140}+\frac {113}{10}{}\mathrm {i}\right )\,e^3\right )}{d^6\,125{}\mathrm {i}-d^5\,e\,150{}\mathrm {i}+d^4\,e^2\,285{}\mathrm {i}-d^3\,e^3\,188{}\mathrm {i}+d^2\,e^4\,171{}\mathrm {i}-d\,e^5\,54{}\mathrm {i}+e^6\,27{}\mathrm {i}}+\frac {\ln \left (d+e\,x\right )\,\left (100\,d^6-120\,d^5\,e+228\,d^4\,e^2-242\,d^3\,e^3+141\,d^2\,e^4+120\,d\,e^5-e^6\right )}{e^3\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^3} \]
((9*d*e^5 - 15*d^5*e + 60*d^6 - 6*e^6 - 25*d^2*e^4 + 84*d^3*e^3 + 39*d^4*e ^2)/(2*e^3*(25*d^4 - 20*d^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e^2)) + (x*(d^4* e - 2*d*e^4 + 40*d^5 + e^5 + 44*d^2*e^3 + 28*d^3*e^2))/(e^2*(25*d^4 - 20*d ^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e^2)))/(d^2 + e^2*x^2 + 2*d*e*x) - (log(x - (14^(1/2)*1i)/5 + 1/5)*(d^3*((423*14^(1/2))/140 - 229i/5) + e^3*((703*1 4^(1/2))/140 - 113i/10) + d*e^2*((879*14^(1/2))/140 + 408i/5) - d^2*e*((41 01*14^(1/2))/140 - 21i/10)))/(d^6*125i - d^5*e*150i - d*e^5*54i + e^6*27i + d^2*e^4*171i - d^3*e^3*188i + d^4*e^2*285i) + (log(x + (14^(1/2)*1i)/5 + 1/5)*(d^3*((423*14^(1/2))/140 + 229i/5) + e^3*((703*14^(1/2))/140 + 113i/ 10) + d*e^2*((879*14^(1/2))/140 - 408i/5) - d^2*e*((4101*14^(1/2))/140 + 2 1i/10)))/(d^6*125i - d^5*e*150i - d*e^5*54i + e^6*27i + d^2*e^4*171i - d^3 *e^3*188i + d^4*e^2*285i) + (log(d + e*x)*(120*d*e^5 - 120*d^5*e + 100*d^6 - e^6 + 141*d^2*e^4 - 242*d^3*e^3 + 228*d^4*e^2))/(e^3*(5*d^2 - 2*d*e + 3 *e^2)^3)