Integrand size = 29, antiderivative size = 218 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^3 \left (49 e^2 f^2+160 d e f g+112 d^2 g^2\right ) x}{e^2}+\frac {d^2 \left (23 e^2 f^2+98 d e f g+80 d^2 g^2\right ) x^2}{2 e}+\frac {1}{3} d \left (7 e^2 f^2+46 d e f g+49 d^2 g^2\right ) x^3+\frac {1}{4} e \left (e^2 f^2+14 d e f g+23 d^2 g^2\right ) x^4+\frac {1}{5} e^2 g (2 e f+7 d g) x^5+\frac {1}{6} e^3 g^2 x^6+\frac {32 d^5 (e f+d g)^2}{e^3 (d-e x)}+\frac {16 d^4 (e f+d g) (5 e f+9 d g) \log (d-e x)}{e^3} \]
d^3*(112*d^2*g^2+160*d*e*f*g+49*e^2*f^2)*x/e^2+1/2*d^2*(80*d^2*g^2+98*d*e* f*g+23*e^2*f^2)*x^2/e+1/3*d*(49*d^2*g^2+46*d*e*f*g+7*e^2*f^2)*x^3+1/4*e*(2 3*d^2*g^2+14*d*e*f*g+e^2*f^2)*x^4+1/5*e^2*g*(7*d*g+2*e*f)*x^5+1/6*e^3*g^2* x^6+32*d^5*(d*g+e*f)^2/e^3/(-e*x+d)+16*d^4*(d*g+e*f)*(9*d*g+5*e*f)*ln(-e*x +d)/e^3
Time = 0.08 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^3 \left (49 e^2 f^2+160 d e f g+112 d^2 g^2\right ) x}{e^2}+\frac {d^2 \left (23 e^2 f^2+98 d e f g+80 d^2 g^2\right ) x^2}{2 e}+\frac {1}{3} d \left (7 e^2 f^2+46 d e f g+49 d^2 g^2\right ) x^3+\frac {1}{4} e \left (e^2 f^2+14 d e f g+23 d^2 g^2\right ) x^4+\frac {1}{5} e^2 g (2 e f+7 d g) x^5+\frac {1}{6} e^3 g^2 x^6-\frac {32 d^5 (e f+d g)^2}{e^3 (-d+e x)}+\frac {16 d^4 \left (5 e^2 f^2+14 d e f g+9 d^2 g^2\right ) \log (d-e x)}{e^3} \]
(d^3*(49*e^2*f^2 + 160*d*e*f*g + 112*d^2*g^2)*x)/e^2 + (d^2*(23*e^2*f^2 + 98*d*e*f*g + 80*d^2*g^2)*x^2)/(2*e) + (d*(7*e^2*f^2 + 46*d*e*f*g + 49*d^2* g^2)*x^3)/3 + (e*(e^2*f^2 + 14*d*e*f*g + 23*d^2*g^2)*x^4)/4 + (e^2*g*(2*e* f + 7*d*g)*x^5)/5 + (e^3*g^2*x^6)/6 - (32*d^5*(e*f + d*g)^2)/(e^3*(-d + e* x)) + (16*d^4*(5*e^2*f^2 + 14*d*e*f*g + 9*d^2*g^2)*Log[d - e*x])/e^3
Time = 0.49 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {639, 99, 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.
| \(\displaystyle \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 639 |
| \(\displaystyle \int \frac {(d+e x)^5 (f+g x)^2}{(d-e x)^2}dx\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \int \left (\frac {32 d^5 (d g+e f)^2}{e^2 (e x-d)^2}+\frac {16 d^4 (-9 d g-5 e f) (d g+e f)}{e^2 (d-e x)}+e x^3 \left (23 d^2 g^2+14 d e f g+e^2 f^2\right )+d x^2 \left (49 d^2 g^2+46 d e f g+7 e^2 f^2\right )+\frac {d^2 x \left (80 d^2 g^2+98 d e f g+23 e^2 f^2\right )}{e}+\frac {d^3 \left (112 d^2 g^2+160 d e f g+49 e^2 f^2\right )}{e^2}+e^2 g x^4 (7 d g+2 e f)+e^3 g^2 x^5\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {32 d^5 (d g+e f)^2}{e^3 (d-e x)}+\frac {16 d^4 (d g+e f) (9 d g+5 e f) \log (d-e x)}{e^3}+\frac {1}{4} e x^4 \left (23 d^2 g^2+14 d e f g+e^2 f^2\right )+\frac {1}{3} d x^3 \left (49 d^2 g^2+46 d e f g+7 e^2 f^2\right )+\frac {d^2 x^2 \left (80 d^2 g^2+98 d e f g+23 e^2 f^2\right )}{2 e}+\frac {d^3 x \left (112 d^2 g^2+160 d e f g+49 e^2 f^2\right )}{e^2}+\frac {1}{5} e^2 g x^5 (7 d g+2 e f)+\frac {1}{6} e^3 g^2 x^6\) |
(d^3*(49*e^2*f^2 + 160*d*e*f*g + 112*d^2*g^2)*x)/e^2 + (d^2*(23*e^2*f^2 + 98*d*e*f*g + 80*d^2*g^2)*x^2)/(2*e) + (d*(7*e^2*f^2 + 46*d*e*f*g + 49*d^2* g^2)*x^3)/3 + (e*(e^2*f^2 + 14*d*e*f*g + 23*d^2*g^2)*x^4)/4 + (e^2*g*(2*e* f + 7*d*g)*x^5)/5 + (e^3*g^2*x^6)/6 + (32*d^5*(e*f + d*g)^2)/(e^3*(d - e*x )) + (16*d^4*(e*f + d*g)*(5*e*f + 9*d*g)*Log[d - e*x])/e^3
3.6.57.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^ 2)^(p_.), x_Symbol] :> Int[(c + d*x)^(m + p)*(e + f*x)^n*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (I ntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[m]))
Time = 0.43 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(\frac {\frac {1}{6} g^{2} e^{5} x^{6}+\frac {7}{5} x^{5} d \,e^{4} g^{2}+\frac {2}{5} x^{5} e^{5} f g +\frac {23}{4} x^{4} d^{2} e^{3} g^{2}+\frac {7}{2} x^{4} d \,e^{4} f g +\frac {1}{4} x^{4} e^{5} f^{2}+\frac {49}{3} x^{3} d^{3} e^{2} g^{2}+\frac {46}{3} x^{3} d^{2} e^{3} f g +\frac {7}{3} x^{3} d \,e^{4} f^{2}+40 x^{2} d^{4} e \,g^{2}+49 x^{2} d^{3} e^{2} f g +\frac {23}{2} x^{2} d^{2} e^{3} f^{2}+112 g^{2} d^{5} x +160 f g \,d^{4} e x +49 f^{2} d^{3} e^{2} x}{e^{2}}+\frac {32 d^{5} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {16 d^{4} \left (9 d^{2} g^{2}+14 d e f g +5 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(259\) |
| risch | \(\frac {e^{3} g^{2} x^{6}}{6}+\frac {7 e^{2} x^{5} d \,g^{2}}{5}+\frac {2 e^{3} x^{5} f g}{5}+\frac {23 e \,x^{4} d^{2} g^{2}}{4}+\frac {7 e^{2} x^{4} d f g}{2}+\frac {e^{3} x^{4} f^{2}}{4}+\frac {49 x^{3} d^{3} g^{2}}{3}+\frac {46 e \,x^{3} d^{2} f g}{3}+\frac {7 e^{2} x^{3} d \,f^{2}}{3}+\frac {40 x^{2} d^{4} g^{2}}{e}+49 x^{2} d^{3} f g +\frac {23 e \,x^{2} d^{2} f^{2}}{2}+\frac {112 g^{2} d^{5} x}{e^{2}}+\frac {160 f g \,d^{4} x}{e}+49 f^{2} d^{3} x +\frac {32 d^{7} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {64 d^{6} f g}{e^{2} \left (-e x +d \right )}+\frac {32 d^{5} f^{2}}{e \left (-e x +d \right )}+\frac {144 d^{6} \ln \left (-e x +d \right ) g^{2}}{e^{3}}+\frac {224 d^{5} \ln \left (-e x +d \right ) f g}{e^{2}}+\frac {80 d^{4} \ln \left (-e x +d \right ) f^{2}}{e}\) | \(280\) |
| norman | \(\frac {\left (-\frac {287}{3} g^{2} d^{5}-\frac {434}{3} f g \,d^{4} e -\frac {140}{3} f^{2} d^{3} e^{2}\right ) x^{3}+\left (-\frac {67}{12} g^{2} d^{2} e^{3}-\frac {7}{2} f g d \,e^{4}-\frac {1}{4} f^{2} e^{5}\right ) x^{6}+\left (-\frac {224}{15} g^{2} d^{3} e^{2}-\frac {224}{15} f g \,d^{2} e^{3}-\frac {7}{3} f^{2} d \,e^{4}\right ) x^{5}+\left (-\frac {137}{4} g^{2} d^{4} e -\frac {91}{2} f g \,d^{3} e^{2}-\frac {45}{4} f^{2} d^{2} e^{3}\right ) x^{4}+\frac {d^{5} \left (144 d^{2} g^{2}+224 d e f g +81 e^{2} f^{2}\right ) x}{e^{2}}+\frac {d^{2} \left (144 g^{2} d^{6}+226 f g e \,d^{5}+87 e^{2} f^{2} d^{4}\right )}{2 e^{3}}-\frac {g^{2} e^{5} x^{8}}{6}-\frac {e^{4} g \left (7 d g +2 e f \right ) x^{7}}{5}}{-e^{2} x^{2}+d^{2}}+\frac {16 d^{4} \left (9 d^{2} g^{2}+14 d e f g +5 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(286\) |
| parallelrisch | \(\frac {-4860 d^{5} e^{2} f^{2}+8640 \ln \left (e x -d \right ) x \,d^{6} e \,g^{2}+4800 \ln \left (e x -d \right ) x \,d^{4} e^{3} f^{2}-13440 \ln \left (e x -d \right ) d^{6} e f g -8640 d^{7} g^{2}+2250 x^{2} d^{3} e^{4} f^{2}+1420 x^{3} d^{4} e^{3} g^{2}+550 x^{3} d^{2} e^{5} f^{2}+635 x^{4} d^{3} e^{4} g^{2}+125 x^{4} d \,e^{6} f^{2}+261 x^{5} d^{2} e^{5} g^{2}+74 x^{6} d \,e^{6} g^{2}+24 x^{6} e^{7} f g -4800 \ln \left (e x -d \right ) d^{5} e^{2} f^{2}+4320 x^{2} d^{5} e^{2} g^{2}+13440 \ln \left (e x -d \right ) x \,d^{5} e^{2} f g -13440 d^{6} e f g -8640 \ln \left (e x -d \right ) d^{7} g^{2}+10 g^{2} e^{7} x^{7}+15 x^{5} e^{7} f^{2}+6660 x^{2} d^{4} e^{3} f g +2020 x^{3} d^{3} e^{4} f g +710 x^{4} d^{2} e^{5} f g +186 x^{5} d \,e^{6} f g}{60 e^{3} \left (e x -d \right )}\) | \(341\) |
1/e^2*(1/6*g^2*e^5*x^6+7/5*x^5*d*e^4*g^2+2/5*x^5*e^5*f*g+23/4*x^4*d^2*e^3* g^2+7/2*x^4*d*e^4*f*g+1/4*x^4*e^5*f^2+49/3*x^3*d^3*e^2*g^2+46/3*x^3*d^2*e^ 3*f*g+7/3*x^3*d*e^4*f^2+40*x^2*d^4*e*g^2+49*x^2*d^3*e^2*f*g+23/2*x^2*d^2*e ^3*f^2+112*g^2*d^5*x+160*f*g*d^4*e*x+49*f^2*d^3*e^2*x)+32*d^5*(d^2*g^2+2*d *e*f*g+e^2*f^2)/e^3/(-e*x+d)+16*d^4/e^3*(9*d^2*g^2+14*d*e*f*g+5*e^2*f^2)*l n(-e*x+d)
Time = 0.30 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.50 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {10 \, e^{7} g^{2} x^{7} - 1920 \, d^{5} e^{2} f^{2} - 3840 \, d^{6} e f g - 1920 \, d^{7} g^{2} + 2 \, {\left (12 \, e^{7} f g + 37 \, d e^{6} g^{2}\right )} x^{6} + 3 \, {\left (5 \, e^{7} f^{2} + 62 \, d e^{6} f g + 87 \, d^{2} e^{5} g^{2}\right )} x^{5} + 5 \, {\left (25 \, d e^{6} f^{2} + 142 \, d^{2} e^{5} f g + 127 \, d^{3} e^{4} g^{2}\right )} x^{4} + 10 \, {\left (55 \, d^{2} e^{5} f^{2} + 202 \, d^{3} e^{4} f g + 142 \, d^{4} e^{3} g^{2}\right )} x^{3} + 90 \, {\left (25 \, d^{3} e^{4} f^{2} + 74 \, d^{4} e^{3} f g + 48 \, d^{5} e^{2} g^{2}\right )} x^{2} - 60 \, {\left (49 \, d^{4} e^{3} f^{2} + 160 \, d^{5} e^{2} f g + 112 \, d^{6} e g^{2}\right )} x - 960 \, {\left (5 \, d^{5} e^{2} f^{2} + 14 \, d^{6} e f g + 9 \, d^{7} g^{2} - {\left (5 \, d^{4} e^{3} f^{2} + 14 \, d^{5} e^{2} f g + 9 \, d^{6} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{60 \, {\left (e^{4} x - d e^{3}\right )}} \]
1/60*(10*e^7*g^2*x^7 - 1920*d^5*e^2*f^2 - 3840*d^6*e*f*g - 1920*d^7*g^2 + 2*(12*e^7*f*g + 37*d*e^6*g^2)*x^6 + 3*(5*e^7*f^2 + 62*d*e^6*f*g + 87*d^2*e ^5*g^2)*x^5 + 5*(25*d*e^6*f^2 + 142*d^2*e^5*f*g + 127*d^3*e^4*g^2)*x^4 + 1 0*(55*d^2*e^5*f^2 + 202*d^3*e^4*f*g + 142*d^4*e^3*g^2)*x^3 + 90*(25*d^3*e^ 4*f^2 + 74*d^4*e^3*f*g + 48*d^5*e^2*g^2)*x^2 - 60*(49*d^4*e^3*f^2 + 160*d^ 5*e^2*f*g + 112*d^6*e*g^2)*x - 960*(5*d^5*e^2*f^2 + 14*d^6*e*f*g + 9*d^7*g ^2 - (5*d^4*e^3*f^2 + 14*d^5*e^2*f*g + 9*d^6*e*g^2)*x)*log(e*x - d))/(e^4* x - d*e^3)
Time = 0.51 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {16 d^{4} \left (d g + e f\right ) \left (9 d g + 5 e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {e^{3} g^{2} x^{6}}{6} + x^{5} \cdot \left (\frac {7 d e^{2} g^{2}}{5} + \frac {2 e^{3} f g}{5}\right ) + x^{4} \cdot \left (\frac {23 d^{2} e g^{2}}{4} + \frac {7 d e^{2} f g}{2} + \frac {e^{3} f^{2}}{4}\right ) + x^{3} \cdot \left (\frac {49 d^{3} g^{2}}{3} + \frac {46 d^{2} e f g}{3} + \frac {7 d e^{2} f^{2}}{3}\right ) + x^{2} \cdot \left (\frac {40 d^{4} g^{2}}{e} + 49 d^{3} f g + \frac {23 d^{2} e f^{2}}{2}\right ) + x \left (\frac {112 d^{5} g^{2}}{e^{2}} + \frac {160 d^{4} f g}{e} + 49 d^{3} f^{2}\right ) + \frac {- 32 d^{7} g^{2} - 64 d^{6} e f g - 32 d^{5} e^{2} f^{2}}{- d e^{3} + e^{4} x} \]
16*d**4*(d*g + e*f)*(9*d*g + 5*e*f)*log(-d + e*x)/e**3 + e**3*g**2*x**6/6 + x**5*(7*d*e**2*g**2/5 + 2*e**3*f*g/5) + x**4*(23*d**2*e*g**2/4 + 7*d*e** 2*f*g/2 + e**3*f**2/4) + x**3*(49*d**3*g**2/3 + 46*d**2*e*f*g/3 + 7*d*e**2 *f**2/3) + x**2*(40*d**4*g**2/e + 49*d**3*f*g + 23*d**2*e*f**2/2) + x*(112 *d**5*g**2/e**2 + 160*d**4*f*g/e + 49*d**3*f**2) + (-32*d**7*g**2 - 64*d** 6*e*f*g - 32*d**5*e**2*f**2)/(-d*e**3 + e**4*x)
Time = 0.19 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {32 \, {\left (d^{5} e^{2} f^{2} + 2 \, d^{6} e f g + d^{7} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {10 \, e^{5} g^{2} x^{6} + 12 \, {\left (2 \, e^{5} f g + 7 \, d e^{4} g^{2}\right )} x^{5} + 15 \, {\left (e^{5} f^{2} + 14 \, d e^{4} f g + 23 \, d^{2} e^{3} g^{2}\right )} x^{4} + 20 \, {\left (7 \, d e^{4} f^{2} + 46 \, d^{2} e^{3} f g + 49 \, d^{3} e^{2} g^{2}\right )} x^{3} + 30 \, {\left (23 \, d^{2} e^{3} f^{2} + 98 \, d^{3} e^{2} f g + 80 \, d^{4} e g^{2}\right )} x^{2} + 60 \, {\left (49 \, d^{3} e^{2} f^{2} + 160 \, d^{4} e f g + 112 \, d^{5} g^{2}\right )} x}{60 \, e^{2}} + \frac {16 \, {\left (5 \, d^{4} e^{2} f^{2} + 14 \, d^{5} e f g + 9 \, d^{6} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
-32*(d^5*e^2*f^2 + 2*d^6*e*f*g + d^7*g^2)/(e^4*x - d*e^3) + 1/60*(10*e^5*g ^2*x^6 + 12*(2*e^5*f*g + 7*d*e^4*g^2)*x^5 + 15*(e^5*f^2 + 14*d*e^4*f*g + 2 3*d^2*e^3*g^2)*x^4 + 20*(7*d*e^4*f^2 + 46*d^2*e^3*f*g + 49*d^3*e^2*g^2)*x^ 3 + 30*(23*d^2*e^3*f^2 + 98*d^3*e^2*f*g + 80*d^4*e*g^2)*x^2 + 60*(49*d^3*e ^2*f^2 + 160*d^4*e*f*g + 112*d^5*g^2)*x)/e^2 + 16*(5*d^4*e^2*f^2 + 14*d^5* e*f*g + 9*d^6*g^2)*log(e*x - d)/e^3
Time = 0.33 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.25 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {16 \, {\left (5 \, d^{4} e^{2} f^{2} + 14 \, d^{5} e f g + 9 \, d^{6} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {32 \, {\left (d^{5} e^{2} f^{2} + 2 \, d^{6} e f g + d^{7} g^{2}\right )}}{{\left (e x - d\right )} e^{3}} + \frac {10 \, e^{15} g^{2} x^{6} + 24 \, e^{15} f g x^{5} + 84 \, d e^{14} g^{2} x^{5} + 15 \, e^{15} f^{2} x^{4} + 210 \, d e^{14} f g x^{4} + 345 \, d^{2} e^{13} g^{2} x^{4} + 140 \, d e^{14} f^{2} x^{3} + 920 \, d^{2} e^{13} f g x^{3} + 980 \, d^{3} e^{12} g^{2} x^{3} + 690 \, d^{2} e^{13} f^{2} x^{2} + 2940 \, d^{3} e^{12} f g x^{2} + 2400 \, d^{4} e^{11} g^{2} x^{2} + 2940 \, d^{3} e^{12} f^{2} x + 9600 \, d^{4} e^{11} f g x + 6720 \, d^{5} e^{10} g^{2} x}{60 \, e^{12}} \]
16*(5*d^4*e^2*f^2 + 14*d^5*e*f*g + 9*d^6*g^2)*log(abs(e*x - d))/e^3 - 32*( d^5*e^2*f^2 + 2*d^6*e*f*g + d^7*g^2)/((e*x - d)*e^3) + 1/60*(10*e^15*g^2*x ^6 + 24*e^15*f*g*x^5 + 84*d*e^14*g^2*x^5 + 15*e^15*f^2*x^4 + 210*d*e^14*f* g*x^4 + 345*d^2*e^13*g^2*x^4 + 140*d*e^14*f^2*x^3 + 920*d^2*e^13*f*g*x^3 + 980*d^3*e^12*g^2*x^3 + 690*d^2*e^13*f^2*x^2 + 2940*d^3*e^12*f*g*x^2 + 240 0*d^4*e^11*g^2*x^2 + 2940*d^3*e^12*f^2*x + 9600*d^4*e^11*f*g*x + 6720*d^5* e^10*g^2*x)/e^12
Time = 11.99 (sec) , antiderivative size = 1029, normalized size of antiderivative = 4.72 \[ \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x^5\,\left (\frac {e^2\,g\,\left (5\,d\,g+2\,e\,f\right )}{5}+\frac {2\,d\,e^2\,g^2}{5}\right )+x^3\,\left (\frac {5\,d\,\left (2\,d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2\right )}{3}+\frac {2\,d\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{3\,e}-\frac {d^2\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{3\,e^2}\right )+x^4\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{4\,e^2}-\frac {d^2\,e\,g^2}{4}+\frac {d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{2\,e}\right )+x^2\,\left (\frac {5\,d^2\,\left (d^2\,g^2+4\,d\,e\,f\,g+2\,e^2\,f^2\right )}{2\,e}-\frac {d^2\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{2\,e^2}+\frac {d\,\left (5\,d\,\left (2\,d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2\right )+\frac {2\,d\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{e}-\frac {d^2\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e^2}\right )}{e}\right )+x\,\left (\frac {d^5\,g^2+10\,d^4\,e\,f\,g+10\,d^3\,e^2\,f^2}{e^2}-\frac {d^2\,\left (5\,d\,\left (2\,d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2\right )+\frac {2\,d\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{e}-\frac {d^2\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e^2}\right )}{e^2}+\frac {2\,d\,\left (\frac {5\,d^2\,\left (d^2\,g^2+4\,d\,e\,f\,g+2\,e^2\,f^2\right )}{e}-\frac {d^2\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{e^2}+\frac {2\,d\,\left (5\,d\,\left (2\,d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2\right )+\frac {2\,d\,\left (\frac {10\,d^2\,e^3\,g^2+10\,d\,e^4\,f\,g+e^5\,f^2}{e^2}-d^2\,e\,g^2+\frac {2\,d\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e}\right )}{e}-\frac {d^2\,\left (e^2\,g\,\left (5\,d\,g+2\,e\,f\right )+2\,d\,e^2\,g^2\right )}{e^2}\right )}{e}\right )}{e}\right )+\frac {\ln \left (e\,x-d\right )\,\left (144\,d^6\,g^2+224\,d^5\,e\,f\,g+80\,d^4\,e^2\,f^2\right )}{e^3}+\frac {32\,\left (d^7\,g^2+2\,d^6\,e\,f\,g+d^5\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {e^3\,g^2\,x^6}{6} \]
x^5*((e^2*g*(5*d*g + 2*e*f))/5 + (2*d*e^2*g^2)/5) + x^3*((5*d*(2*d^2*g^2 + e^2*f^2 + 4*d*e*f*g))/3 + (2*d*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^4*f*g) /e^2 - d^2*e*g^2 + (2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/(3*e) - (d^2*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/(3*e^2)) + x^4*((e^5*f^2 + 10 *d^2*e^3*g^2 + 10*d*e^4*f*g)/(4*e^2) - (d^2*e*g^2)/4 + (d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/(2*e)) + x^2*((5*d^2*(d^2*g^2 + 2*e^2*f^2 + 4*d*e*f *g))/(2*e) - (d^2*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^4*f*g)/e^2 - d^2*e*g ^2 + (2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/(2*e^2) + (d*(5*d*(2* d^2*g^2 + e^2*f^2 + 4*d*e*f*g) + (2*d*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^ 4*f*g)/e^2 - d^2*e*g^2 + (2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/e - (d^2*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e^2))/e) + x*((d^5*g^2 + 10 *d^3*e^2*f^2 + 10*d^4*e*f*g)/e^2 - (d^2*(5*d*(2*d^2*g^2 + e^2*f^2 + 4*d*e* f*g) + (2*d*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^4*f*g)/e^2 - d^2*e*g^2 + ( 2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/e - (d^2*(e^2*g*(5*d*g + 2* e*f) + 2*d*e^2*g^2))/e^2))/e^2 + (2*d*((5*d^2*(d^2*g^2 + 2*e^2*f^2 + 4*d*e *f*g))/e - (d^2*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^4*f*g)/e^2 - d^2*e*g^2 + (2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/e^2 + (2*d*(5*d*(2*d^2* g^2 + e^2*f^2 + 4*d*e*f*g) + (2*d*((e^5*f^2 + 10*d^2*e^3*g^2 + 10*d*e^4*f* g)/e^2 - d^2*e*g^2 + (2*d*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e))/e - ( d^2*(e^2*g*(5*d*g + 2*e*f) + 2*d*e^2*g^2))/e^2))/e))/e) + (log(e*x - d)...