Integrand size = 36, antiderivative size = 228 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=\frac {\left (100 d^4+68 d^3 e+51 d^2 e^2+8 d e^3+21 e^4\right ) x}{e^6}-\frac {\left (80 d^3+51 d^2 e+34 d e^2+4 e^3\right ) x^2}{2 e^5}+\frac {\left (60 d^2+34 d e+17 e^2\right ) x^3}{3 e^4}-\frac {(40 d+17 e) x^4}{4 e^3}+\frac {4 x^5}{e^2}-\frac {\left (5 d^2-2 d e+3 e^2\right ) \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right )}{e^7 (d+e x)}-\frac {\left (120 d^5+85 d^4 e+68 d^3 e^2+12 d^2 e^3+42 d e^4-7 e^5\right ) \log (d+e x)}{e^7} \]
(100*d^4+68*d^3*e+51*d^2*e^2+8*d*e^3+21*e^4)*x/e^6-1/2*(80*d^3+51*d^2*e+34 *d*e^2+4*e^3)*x^2/e^5+1/3*(60*d^2+34*d*e+17*e^2)*x^3/e^4-1/4*(40*d+17*e)*x ^4/e^3+4*x^5/e^2-(5*d^2-2*d*e+3*e^2)*(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4) /e^7/(e*x+d)-(120*d^5+85*d^4*e+68*d^3*e^2+12*d^2*e^3+42*d*e^4-7*e^5)*ln(e* x+d)/e^7
Time = 0.05 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.98 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=\frac {12 e \left (100 d^4+68 d^3 e+51 d^2 e^2+8 d e^3+21 e^4\right ) x-6 e^2 \left (80 d^3+51 d^2 e+34 d e^2+4 e^3\right ) x^2+4 e^3 \left (60 d^2+34 d e+17 e^2\right ) x^3-3 e^4 (40 d+17 e) x^4+48 e^5 x^5-\frac {12 \left (20 d^6+17 d^5 e+17 d^4 e^2+4 d^3 e^3+21 d^2 e^4-7 d e^5+6 e^6\right )}{d+e x}-12 \left (120 d^5+85 d^4 e+68 d^3 e^2+12 d^2 e^3+42 d e^4-7 e^5\right ) \log (d+e x)}{12 e^7} \]
(12*e*(100*d^4 + 68*d^3*e + 51*d^2*e^2 + 8*d*e^3 + 21*e^4)*x - 6*e^2*(80*d ^3 + 51*d^2*e + 34*d*e^2 + 4*e^3)*x^2 + 4*e^3*(60*d^2 + 34*d*e + 17*e^2)*x ^3 - 3*e^4*(40*d + 17*e)*x^4 + 48*e^5*x^5 - (12*(20*d^6 + 17*d^5*e + 17*d^ 4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d*e^5 + 6*e^6))/(d + e*x) - 12*(120*d^5 + 85*d^4*e + 68*d^3*e^2 + 12*d^2*e^3 + 42*d*e^4 - 7*e^5)*Log[d + e*x])/(1 2*e^7)
Time = 0.45 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, 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.
| \(\displaystyle \int \frac {\left (5 x^2+2 x+3\right ) \left (4 x^4-5 x^3+3 x^2+x+2\right )}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {x^2 \left (60 d^2+34 d e+17 e^2\right )}{e^4}-\frac {x \left (80 d^3+51 d^2 e+34 d e^2+4 e^3\right )}{e^5}+\frac {100 d^4+68 d^3 e+51 d^2 e^2+8 d e^3+21 e^4}{e^6}+\frac {-120 d^5-85 d^4 e-68 d^3 e^2-12 d^2 e^3-42 d e^4+7 e^5}{e^6 (d+e x)}+\frac {20 d^6+17 d^5 e+17 d^4 e^2+4 d^3 e^3+21 d^2 e^4-7 d e^5+6 e^6}{e^6 (d+e x)^2}-\frac {x^3 (40 d+17 e)}{e^3}+\frac {20 x^4}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^3 \left (60 d^2+34 d e+17 e^2\right )}{3 e^4}-\frac {x^2 \left (80 d^3+51 d^2 e+34 d e^2+4 e^3\right )}{2 e^5}-\frac {\left (5 d^2-2 d e+3 e^2\right ) \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right )}{e^7 (d+e x)}+\frac {x \left (100 d^4+68 d^3 e+51 d^2 e^2+8 d e^3+21 e^4\right )}{e^6}-\frac {\left (120 d^5+85 d^4 e+68 d^3 e^2+12 d^2 e^3+42 d e^4-7 e^5\right ) \log (d+e x)}{e^7}-\frac {x^4 (40 d+17 e)}{4 e^3}+\frac {4 x^5}{e^2}\) |
((100*d^4 + 68*d^3*e + 51*d^2*e^2 + 8*d*e^3 + 21*e^4)*x)/e^6 - ((80*d^3 + 51*d^2*e + 34*d*e^2 + 4*e^3)*x^2)/(2*e^5) + ((60*d^2 + 34*d*e + 17*e^2)*x^ 3)/(3*e^4) - ((40*d + 17*e)*x^4)/(4*e^3) + (4*x^5)/e^2 - ((5*d^2 - 2*d*e + 3*e^2)*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4))/(e^7*(d + e*x)) - ( (120*d^5 + 85*d^4*e + 68*d^3*e^2 + 12*d^2*e^3 + 42*d*e^4 - 7*e^5)*Log[d + e*x])/e^7
3.3.94.3.1 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 = 0.46 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.02
| method | result | size |
| norman | \(\frac {\frac {\left (120 d^{6}+85 d^{5} e +68 d^{4} e^{2}+12 d^{3} e^{3}+42 d^{2} e^{4}-7 d \,e^{5}+6 e^{6}\right ) x}{e^{6} d}+\frac {4 x^{6}}{e}-\frac {\left (24 d +17 e \right ) x^{5}}{4 e^{2}}+\frac {\left (120 d^{2}+85 d e +68 e^{2}\right ) x^{4}}{12 e^{3}}-\frac {\left (120 d^{3}+85 d^{2} e +68 d \,e^{2}+12 e^{3}\right ) x^{3}}{6 e^{4}}+\frac {\left (120 d^{4}+85 d^{3} e +68 d^{2} e^{2}+12 d \,e^{3}+42 e^{4}\right ) x^{2}}{2 e^{5}}}{e x +d}-\frac {\left (120 d^{5}+85 d^{4} e +68 d^{3} e^{2}+12 d^{2} e^{3}+42 d \,e^{4}-7 e^{5}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(232\) |
| default | \(\frac {4 e^{4} x^{5}-10 d \,e^{3} x^{4}-\frac {17}{4} e^{4} x^{4}+20 d^{2} e^{2} x^{3}+\frac {34}{3} d \,e^{3} x^{3}+\frac {17}{3} e^{4} x^{3}-40 d^{3} e \,x^{2}-\frac {51}{2} d^{2} e^{2} x^{2}-17 d \,e^{3} x^{2}-2 e^{4} x^{2}+100 d^{4} x +68 d^{3} e x +51 d^{2} e^{2} x +8 d \,e^{3} x +21 e^{4} x}{e^{6}}-\frac {20 d^{6}+17 d^{5} e +17 d^{4} e^{2}+4 d^{3} e^{3}+21 d^{2} e^{4}-7 d \,e^{5}+6 e^{6}}{e^{7} \left (e x +d \right )}+\frac {\left (-120 d^{5}-85 d^{4} e -68 d^{3} e^{2}-12 d^{2} e^{3}-42 d \,e^{4}+7 e^{5}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(240\) |
| risch | \(\frac {7 \ln \left (e x +d \right )}{e^{2}}-\frac {17 x^{4}}{4 e^{2}}-\frac {120 \ln \left (e x +d \right ) d^{5}}{e^{7}}-\frac {85 \ln \left (e x +d \right ) d^{4}}{e^{6}}-\frac {68 \ln \left (e x +d \right ) d^{3}}{e^{5}}-\frac {12 \ln \left (e x +d \right ) d^{2}}{e^{4}}-\frac {42 \ln \left (e x +d \right ) d}{e^{3}}-\frac {10 d \,x^{4}}{e^{3}}+\frac {20 d^{2} x^{3}}{e^{4}}+\frac {34 d \,x^{3}}{3 e^{3}}-\frac {40 d^{3} x^{2}}{e^{5}}-\frac {51 d^{2} x^{2}}{2 e^{4}}-\frac {17 d \,x^{2}}{e^{3}}+\frac {100 d^{4} x}{e^{6}}+\frac {68 d^{3} x}{e^{5}}+\frac {51 d^{2} x}{e^{4}}+\frac {8 d x}{e^{3}}-\frac {20 d^{6}}{e^{7} \left (e x +d \right )}-\frac {17 d^{5}}{e^{6} \left (e x +d \right )}-\frac {17 d^{4}}{e^{5} \left (e x +d \right )}-\frac {4 d^{3}}{e^{4} \left (e x +d \right )}-\frac {21 d^{2}}{e^{3} \left (e x +d \right )}+\frac {7 d}{e^{2} \left (e x +d \right )}+\frac {17 x^{3}}{3 e^{2}}-\frac {2 x^{2}}{e^{2}}-\frac {6}{e \left (e x +d \right )}+\frac {4 x^{5}}{e^{2}}+\frac {21 x}{e^{2}}\) | \(313\) |
| parallelrisch | \(-\frac {51 e^{6} x^{5}+72 e^{6}-48 e^{6} x^{6}+1020 d^{5} e +504 \ln \left (e x +d \right ) x d \,e^{5}+1440 \ln \left (e x +d \right ) x \,d^{5} e +1020 \ln \left (e x +d \right ) x \,d^{4} e^{2}+816 \ln \left (e x +d \right ) x \,d^{3} e^{3}+144 \ln \left (e x +d \right ) x \,d^{2} e^{4}-84 d \,e^{5}+816 d^{4} e^{2}+144 d^{3} e^{3}+504 d^{2} e^{4}+240 d^{3} x^{3} e^{3}-720 d^{4} e^{2} x^{2}-120 d^{2} e^{4} x^{4}+72 d \,e^{5} x^{5}-68 x^{4} e^{6}+24 x^{3} e^{6}-252 x^{2} e^{6}+1440 \ln \left (e x +d \right ) d^{6}-85 d \,e^{5} x^{4}+170 d^{2} e^{4} x^{3}-510 d^{3} e^{3} x^{2}+136 x^{3} d \,e^{5}-408 x^{2} d^{2} e^{4}-72 x^{2} d \,e^{5}+1020 \ln \left (e x +d \right ) d^{5} e +816 \ln \left (e x +d \right ) d^{4} e^{2}+144 \ln \left (e x +d \right ) d^{3} e^{3}+504 \ln \left (e x +d \right ) d^{2} e^{4}-84 \ln \left (e x +d \right ) d \,e^{5}+1440 d^{6}-84 \ln \left (e x +d \right ) x \,e^{6}}{12 e^{7} \left (e x +d \right )}\) | \(362\) |
((120*d^6+85*d^5*e+68*d^4*e^2+12*d^3*e^3+42*d^2*e^4-7*d*e^5+6*e^6)/e^6/d*x +4*x^6/e-1/4*(24*d+17*e)/e^2*x^5+1/12*(120*d^2+85*d*e+68*e^2)/e^3*x^4-1/6* (120*d^3+85*d^2*e+68*d*e^2+12*e^3)/e^4*x^3+1/2*(120*d^4+85*d^3*e+68*d^2*e^ 2+12*d*e^3+42*e^4)/e^5*x^2)/(e*x+d)-(120*d^5+85*d^4*e+68*d^3*e^2+12*d^2*e^ 3+42*d*e^4-7*e^5)*ln(e*x+d)/e^7
Time = 0.24 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.40 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=\frac {48 \, e^{6} x^{6} - 240 \, d^{6} - 204 \, d^{5} e - 204 \, d^{4} e^{2} - 48 \, d^{3} e^{3} - 252 \, d^{2} e^{4} + 84 \, d e^{5} - 72 \, e^{6} - 3 \, {\left (24 \, d e^{5} + 17 \, e^{6}\right )} x^{5} + {\left (120 \, d^{2} e^{4} + 85 \, d e^{5} + 68 \, e^{6}\right )} x^{4} - 2 \, {\left (120 \, d^{3} e^{3} + 85 \, d^{2} e^{4} + 68 \, d e^{5} + 12 \, e^{6}\right )} x^{3} + 6 \, {\left (120 \, d^{4} e^{2} + 85 \, d^{3} e^{3} + 68 \, d^{2} e^{4} + 12 \, d e^{5} + 42 \, e^{6}\right )} x^{2} + 12 \, {\left (100 \, d^{5} e + 68 \, d^{4} e^{2} + 51 \, d^{3} e^{3} + 8 \, d^{2} e^{4} + 21 \, d e^{5}\right )} x - 12 \, {\left (120 \, d^{6} + 85 \, d^{5} e + 68 \, d^{4} e^{2} + 12 \, d^{3} e^{3} + 42 \, d^{2} e^{4} - 7 \, d e^{5} + {\left (120 \, d^{5} e + 85 \, d^{4} e^{2} + 68 \, d^{3} e^{3} + 12 \, d^{2} e^{4} + 42 \, d e^{5} - 7 \, e^{6}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{8} x + d e^{7}\right )}} \]
1/12*(48*e^6*x^6 - 240*d^6 - 204*d^5*e - 204*d^4*e^2 - 48*d^3*e^3 - 252*d^ 2*e^4 + 84*d*e^5 - 72*e^6 - 3*(24*d*e^5 + 17*e^6)*x^5 + (120*d^2*e^4 + 85* d*e^5 + 68*e^6)*x^4 - 2*(120*d^3*e^3 + 85*d^2*e^4 + 68*d*e^5 + 12*e^6)*x^3 + 6*(120*d^4*e^2 + 85*d^3*e^3 + 68*d^2*e^4 + 12*d*e^5 + 42*e^6)*x^2 + 12* (100*d^5*e + 68*d^4*e^2 + 51*d^3*e^3 + 8*d^2*e^4 + 21*d*e^5)*x - 12*(120*d ^6 + 85*d^5*e + 68*d^4*e^2 + 12*d^3*e^3 + 42*d^2*e^4 - 7*d*e^5 + (120*d^5* e + 85*d^4*e^2 + 68*d^3*e^3 + 12*d^2*e^4 + 42*d*e^5 - 7*e^6)*x)*log(e*x + d))/(e^8*x + d*e^7)
Time = 0.45 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.04 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=x^{4} \left (- \frac {10 d}{e^{3}} - \frac {17}{4 e^{2}}\right ) + x^{3} \cdot \left (\frac {20 d^{2}}{e^{4}} + \frac {34 d}{3 e^{3}} + \frac {17}{3 e^{2}}\right ) + x^{2} \left (- \frac {40 d^{3}}{e^{5}} - \frac {51 d^{2}}{2 e^{4}} - \frac {17 d}{e^{3}} - \frac {2}{e^{2}}\right ) + x \left (\frac {100 d^{4}}{e^{6}} + \frac {68 d^{3}}{e^{5}} + \frac {51 d^{2}}{e^{4}} + \frac {8 d}{e^{3}} + \frac {21}{e^{2}}\right ) + \frac {- 20 d^{6} - 17 d^{5} e - 17 d^{4} e^{2} - 4 d^{3} e^{3} - 21 d^{2} e^{4} + 7 d e^{5} - 6 e^{6}}{d e^{7} + e^{8} x} + \frac {4 x^{5}}{e^{2}} - \frac {\left (120 d^{5} + 85 d^{4} e + 68 d^{3} e^{2} + 12 d^{2} e^{3} + 42 d e^{4} - 7 e^{5}\right ) \log {\left (d + e x \right )}}{e^{7}} \]
x**4*(-10*d/e**3 - 17/(4*e**2)) + x**3*(20*d**2/e**4 + 34*d/(3*e**3) + 17/ (3*e**2)) + x**2*(-40*d**3/e**5 - 51*d**2/(2*e**4) - 17*d/e**3 - 2/e**2) + x*(100*d**4/e**6 + 68*d**3/e**5 + 51*d**2/e**4 + 8*d/e**3 + 21/e**2) + (- 20*d**6 - 17*d**5*e - 17*d**4*e**2 - 4*d**3*e**3 - 21*d**2*e**4 + 7*d*e**5 - 6*e**6)/(d*e**7 + e**8*x) + 4*x**5/e**2 - (120*d**5 + 85*d**4*e + 68*d* *3*e**2 + 12*d**2*e**3 + 42*d*e**4 - 7*e**5)*log(d + e*x)/e**7
Time = 0.20 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.03 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=-\frac {20 \, d^{6} + 17 \, d^{5} e + 17 \, d^{4} e^{2} + 4 \, d^{3} e^{3} + 21 \, d^{2} e^{4} - 7 \, d e^{5} + 6 \, e^{6}}{e^{8} x + d e^{7}} + \frac {48 \, e^{4} x^{5} - 3 \, {\left (40 \, d e^{3} + 17 \, e^{4}\right )} x^{4} + 4 \, {\left (60 \, d^{2} e^{2} + 34 \, d e^{3} + 17 \, e^{4}\right )} x^{3} - 6 \, {\left (80 \, d^{3} e + 51 \, d^{2} e^{2} + 34 \, d e^{3} + 4 \, e^{4}\right )} x^{2} + 12 \, {\left (100 \, d^{4} + 68 \, d^{3} e + 51 \, d^{2} e^{2} + 8 \, d e^{3} + 21 \, e^{4}\right )} x}{12 \, e^{6}} - \frac {{\left (120 \, d^{5} + 85 \, d^{4} e + 68 \, d^{3} e^{2} + 12 \, d^{2} e^{3} + 42 \, d e^{4} - 7 \, e^{5}\right )} \log \left (e x + d\right )}{e^{7}} \]
-(20*d^6 + 17*d^5*e + 17*d^4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d*e^5 + 6*e^ 6)/(e^8*x + d*e^7) + 1/12*(48*e^4*x^5 - 3*(40*d*e^3 + 17*e^4)*x^4 + 4*(60* d^2*e^2 + 34*d*e^3 + 17*e^4)*x^3 - 6*(80*d^3*e + 51*d^2*e^2 + 34*d*e^3 + 4 *e^4)*x^2 + 12*(100*d^4 + 68*d^3*e + 51*d^2*e^2 + 8*d*e^3 + 21*e^4)*x)/e^6 - (120*d^5 + 85*d^4*e + 68*d^3*e^2 + 12*d^2*e^3 + 42*d*e^4 - 7*e^5)*log(e *x + d)/e^7
Time = 0.43 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.43 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=-\frac {{\left (e x + d\right )}^{5} {\left (\frac {3 \, {\left (120 \, d e + 17 \, e^{2}\right )}}{{\left (e x + d\right )} e} - \frac {4 \, {\left (300 \, d^{2} e^{2} + 85 \, d e^{3} + 17 \, e^{4}\right )}}{{\left (e x + d\right )}^{2} e^{2}} + \frac {12 \, {\left (200 \, d^{3} e^{3} + 85 \, d^{2} e^{4} + 34 \, d e^{5} + 2 \, e^{6}\right )}}{{\left (e x + d\right )}^{3} e^{3}} - \frac {12 \, {\left (300 \, d^{4} e^{4} + 170 \, d^{3} e^{5} + 102 \, d^{2} e^{6} + 12 \, d e^{7} + 21 \, e^{8}\right )}}{{\left (e x + d\right )}^{4} e^{4}} - 48\right )}}{12 \, e^{7}} + \frac {{\left (120 \, d^{5} + 85 \, d^{4} e + 68 \, d^{3} e^{2} + 12 \, d^{2} e^{3} + 42 \, d e^{4} - 7 \, e^{5}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{7}} - \frac {\frac {20 \, d^{6} e^{5}}{e x + d} + \frac {17 \, d^{5} e^{6}}{e x + d} + \frac {17 \, d^{4} e^{7}}{e x + d} + \frac {4 \, d^{3} e^{8}}{e x + d} + \frac {21 \, d^{2} e^{9}}{e x + d} - \frac {7 \, d e^{10}}{e x + d} + \frac {6 \, e^{11}}{e x + d}}{e^{12}} \]
-1/12*(e*x + d)^5*(3*(120*d*e + 17*e^2)/((e*x + d)*e) - 4*(300*d^2*e^2 + 8 5*d*e^3 + 17*e^4)/((e*x + d)^2*e^2) + 12*(200*d^3*e^3 + 85*d^2*e^4 + 34*d* e^5 + 2*e^6)/((e*x + d)^3*e^3) - 12*(300*d^4*e^4 + 170*d^3*e^5 + 102*d^2*e ^6 + 12*d*e^7 + 21*e^8)/((e*x + d)^4*e^4) - 48)/e^7 + (120*d^5 + 85*d^4*e + 68*d^3*e^2 + 12*d^2*e^3 + 42*d*e^4 - 7*e^5)*log(abs(e*x + d)/((e*x + d)^ 2*abs(e)))/e^7 - (20*d^6*e^5/(e*x + d) + 17*d^5*e^6/(e*x + d) + 17*d^4*e^7 /(e*x + d) + 4*d^3*e^8/(e*x + d) + 21*d^2*e^9/(e*x + d) - 7*d*e^10/(e*x + d) + 6*e^11/(e*x + d))/e^12
Time = 13.31 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.59 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{(d+e x)^2} \, dx=x^3\,\left (\frac {17}{3\,e^2}-\frac {20\,d^2}{3\,e^4}+\frac {2\,d\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{3\,e}\right )-x^2\,\left (\frac {2}{e^2}+\frac {d\,\left (\frac {17}{e^2}-\frac {20\,d^2}{e^4}+\frac {2\,d\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{e}\right )}{e}-\frac {d^2\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{2\,e^2}\right )-x^4\,\left (\frac {10\,d}{e^3}+\frac {17}{4\,e^2}\right )+x\,\left (\frac {21}{e^2}+\frac {2\,d\,\left (\frac {4}{e^2}+\frac {2\,d\,\left (\frac {17}{e^2}-\frac {20\,d^2}{e^4}+\frac {2\,d\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{e}\right )}{e}-\frac {d^2\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{e^2}\right )}{e}-\frac {d^2\,\left (\frac {17}{e^2}-\frac {20\,d^2}{e^4}+\frac {2\,d\,\left (\frac {40\,d}{e^3}+\frac {17}{e^2}\right )}{e}\right )}{e^2}\right )+\frac {4\,x^5}{e^2}-\frac {\ln \left (d+e\,x\right )\,\left (120\,d^5+85\,d^4\,e+68\,d^3\,e^2+12\,d^2\,e^3+42\,d\,e^4-7\,e^5\right )}{e^7}-\frac {20\,d^6+17\,d^5\,e+17\,d^4\,e^2+4\,d^3\,e^3+21\,d^2\,e^4-7\,d\,e^5+6\,e^6}{e\,\left (x\,e^7+d\,e^6\right )} \]
x^3*(17/(3*e^2) - (20*d^2)/(3*e^4) + (2*d*((40*d)/e^3 + 17/e^2))/(3*e)) - x^2*(2/e^2 + (d*(17/e^2 - (20*d^2)/e^4 + (2*d*((40*d)/e^3 + 17/e^2))/e))/e - (d^2*((40*d)/e^3 + 17/e^2))/(2*e^2)) - x^4*((10*d)/e^3 + 17/(4*e^2)) + x*(21/e^2 + (2*d*(4/e^2 + (2*d*(17/e^2 - (20*d^2)/e^4 + (2*d*((40*d)/e^3 + 17/e^2))/e))/e - (d^2*((40*d)/e^3 + 17/e^2))/e^2))/e - (d^2*(17/e^2 - (20 *d^2)/e^4 + (2*d*((40*d)/e^3 + 17/e^2))/e))/e^2) + (4*x^5)/e^2 - (log(d + e*x)*(42*d*e^4 + 85*d^4*e + 120*d^5 - 7*e^5 + 12*d^2*e^3 + 68*d^3*e^2))/e^ 7 - (17*d^5*e - 7*d*e^5 + 20*d^6 + 6*e^6 + 21*d^2*e^4 + 4*d^3*e^3 + 17*d^4 *e^2)/(e*(d*e^6 + e^7*x))