Integrand size = 38, antiderivative size = 224 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=-\frac {1367 d-293 e+(423 d-1367 e) x}{700 \left (5 d^2-2 d e+3 e^2\right ) \left (3+2 x+5 x^2\right )}+\frac {\left (6565 d^3-26423 d^2 e+11089 d e^2-6623 e^3\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{700 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {\left (205 d^3-61 d^2 e+23 d e^2+14 e^3\right ) \log \left (3+2 x+5 x^2\right )}{50 \left (5 d^2-2 d e+3 e^2\right )^2} \]
1/700*(-1367*d+293*e-(423*d-1367*e)*x)/(5*d^2-2*d*e+3*e^2)/(5*x^2+2*x+3)+( 4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln(e*x+d)/e/(5*d^2-2*d*e+3*e^2)^2-1/5 0*(205*d^3-61*d^2*e+23*d*e^2+14*e^3)*ln(5*x^2+2*x+3)/(5*d^2-2*d*e+3*e^2)^2 +1/9800*(6565*d^3-26423*d^2*e+11089*d*e^2-6623*e^3)*arctan(1/14*(1+5*x)*14 ^(1/2))/(5*d^2-2*d*e+3*e^2)^2*14^(1/2)
Time = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.83 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=\frac {\frac {14 \left (5 d^2-2 d e+3 e^2\right ) (-d (1367+423 x)+e (293+1367 x))}{3+2 x+5 x^2}+\sqrt {14} \left (6565 d^3-26423 d^2 e+11089 d e^2-6623 e^3\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )+\frac {9800 \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e}-196 \left (205 d^3-61 d^2 e+23 d e^2+14 e^3\right ) \log \left (3+2 x+5 x^2\right )}{9800 \left (5 d^2-2 d e+3 e^2\right )^2} \]
((14*(5*d^2 - 2*d*e + 3*e^2)*(-(d*(1367 + 423*x)) + e*(293 + 1367*x)))/(3 + 2*x + 5*x^2) + Sqrt[14]*(6565*d^3 - 26423*d^2*e + 11089*d*e^2 - 6623*e^3 )*ArcTan[(1 + 5*x)/Sqrt[14]] + (9800*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[d + e*x])/e - 196*(205*d^3 - 61*d^2*e + 23*d*e^2 + 14*e^3)*Lo g[3 + 2*x + 5*x^2])/(9800*(5*d^2 - 2*d*e + 3*e^2)^2)
Time = 0.60 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2177, 27, 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.
| \(\displaystyle \int \frac {4 x^4-5 x^3+3 x^2+x+2}{\left (5 x^2+2 x+3\right )^2 (d+e x)} \, dx\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {1}{56} \int \frac {2 \left (112 x^2-\frac {\left (924 d^2-285 e d+281 e^2\right ) x}{5 d^2-2 e d+3 e^2}+\frac {369 d^2-421 e d+280 e^2}{5 d^2-2 e d+3 e^2}\right )}{5 (d+e x) \left (5 x^2+2 x+3\right )}dx-\frac {x (423 d-1367 e)+1367 d-293 e}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{140} \int \frac {112 x^2-\frac {\left (924 d^2-285 e d+281 e^2\right ) x}{5 d^2-2 e d+3 e^2}+\frac {369 d^2-421 e d+280 e^2}{5 d^2-2 e d+3 e^2}}{(d+e x) \left (5 x^2+2 x+3\right )}dx-\frac {x (423 d-1367 e)+1367 d-293 e}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \frac {1}{140} \int \left (\frac {140 \left (4 d^4+5 e d^3+3 e^2 d^2-e^3 d+2 e^4\right )}{\left (5 d^2-2 e d+3 e^2\right )^2 (d+e x)}+\frac {165 d^3-4943 e d^2+2089 e^2 d-1403 e^3-28 \left (205 d^3-61 e d^2+23 e^2 d+14 e^3\right ) x}{\left (5 d^2-2 e d+3 e^2\right )^2 \left (5 x^2+2 x+3\right )}\right )dx-\frac {x (423 d-1367 e)+1367 d-293 e}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{140} \left (\frac {\arctan \left (\frac {5 x+1}{\sqrt {14}}\right ) \left (6565 d^3-26423 d^2 e+11089 d e^2-6623 e^3\right )}{5 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {14 \left (205 d^3-61 d^2 e+23 d e^2+14 e^3\right ) \log \left (5 x^2+2 x+3\right )}{5 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {140 \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e \left (5 d^2-2 d e+3 e^2\right )^2}\right )-\frac {x (423 d-1367 e)+1367 d-293 e}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
-1/700*(1367*d - 293*e + (423*d - 1367*e)*x)/((5*d^2 - 2*d*e + 3*e^2)*(3 + 2*x + 5*x^2)) + (((6565*d^3 - 26423*d^2*e + 11089*d*e^2 - 6623*e^3)*ArcTa n[(1 + 5*x)/Sqrt[14]])/(5*Sqrt[14]*(5*d^2 - 2*d*e + 3*e^2)^2) + (140*(4*d^ 4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[d + e*x])/(e*(5*d^2 - 2*d*e + 3*e^2)^2) - (14*(205*d^3 - 61*d^2*e + 23*d*e^2 + 14*e^3)*Log[3 + 2*x + 5* x^2])/(5*(5*d^2 - 2*d*e + 3*e^2)^2))/140
3.4.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x + c* x^2, x], R = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + (2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^ m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Qx)/(d + e*x )^m - ((2*p + 3)*(2*c*R - b*S))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a* e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Time = 0.95 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(-\frac {\frac {\left (\frac {423}{700} d^{3}-\frac {7681}{3500} d^{2} e +\frac {4003}{3500} d \,e^{2}-\frac {4101}{3500} e^{3}\right ) x +\frac {1367 d^{3}}{700}-\frac {4199 d^{2} e}{3500}+\frac {4687 d \,e^{2}}{3500}-\frac {879 e^{3}}{3500}}{x^{2}+\frac {2}{5} x +\frac {3}{5}}+\frac {\left (5740 d^{3}-1708 d^{2} e +644 d \,e^{2}+392 e^{3}\right ) \ln \left (5 x^{2}+2 x +3\right )}{1400}+\frac {\left (-1313 d^{3}+\frac {26423}{5} d^{2} e -\frac {11089}{5} d \,e^{2}+\frac {6623}{5} e^{3}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{1960}}{\left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}+\frac {\left (4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}\right ) \ln \left (e x +d \right )}{e \left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}\) | \(214\) |
| risch | \(\text {Expression too large to display}\) | \(14754\) |
-1/(5*d^2-2*d*e+3*e^2)^2*(((423/700*d^3-7681/3500*d^2*e+4003/3500*d*e^2-41 01/3500*e^3)*x+1367/700*d^3-4199/3500*d^2*e+4687/3500*d*e^2-879/3500*e^3)/ (x^2+2/5*x+3/5)+1/1400*(5740*d^3-1708*d^2*e+644*d*e^2+392*e^3)*ln(5*x^2+2* x+3)+1/1960*(-1313*d^3+26423/5*d^2*e-11089/5*d*e^2+6623/5*e^3)*14^(1/2)*ar ctan(1/28*(10*x+2)*14^(1/2)))+(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln(e*x +d)/e/(5*d^2-2*d*e+3*e^2)^2
Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (215) = 430\).
Time = 0.32 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.14 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=-\frac {95690 \, d^{3} e - 58786 \, d^{2} e^{2} + 65618 \, d e^{3} - 12306 \, e^{4} - \sqrt {14} {\left (19695 \, d^{3} e - 79269 \, d^{2} e^{2} + 33267 \, d e^{3} - 19869 \, e^{4} + 5 \, {\left (6565 \, d^{3} e - 26423 \, d^{2} e^{2} + 11089 \, d e^{3} - 6623 \, e^{4}\right )} x^{2} + 2 \, {\left (6565 \, d^{3} e - 26423 \, d^{2} e^{2} + 11089 \, d e^{3} - 6623 \, e^{4}\right )} x\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + 14 \, {\left (2115 \, d^{3} e - 7681 \, d^{2} e^{2} + 4003 \, d e^{3} - 4101 \, e^{4}\right )} x - 9800 \, {\left (12 \, d^{4} + 15 \, d^{3} e + 9 \, d^{2} e^{2} - 3 \, d e^{3} + 6 \, e^{4} + 5 \, {\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} x^{2} + 2 \, {\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} x\right )} \log \left (e x + d\right ) + 196 \, {\left (615 \, d^{3} e - 183 \, d^{2} e^{2} + 69 \, d e^{3} + 42 \, e^{4} + 5 \, {\left (205 \, d^{3} e - 61 \, d^{2} e^{2} + 23 \, d e^{3} + 14 \, e^{4}\right )} x^{2} + 2 \, {\left (205 \, d^{3} e - 61 \, d^{2} e^{2} + 23 \, d e^{3} + 14 \, e^{4}\right )} x\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{9800 \, {\left (75 \, d^{4} e - 60 \, d^{3} e^{2} + 102 \, d^{2} e^{3} - 36 \, d e^{4} + 27 \, e^{5} + 5 \, {\left (25 \, d^{4} e - 20 \, d^{3} e^{2} + 34 \, d^{2} e^{3} - 12 \, d e^{4} + 9 \, e^{5}\right )} x^{2} + 2 \, {\left (25 \, d^{4} e - 20 \, d^{3} e^{2} + 34 \, d^{2} e^{3} - 12 \, d e^{4} + 9 \, e^{5}\right )} x\right )}} \]
-1/9800*(95690*d^3*e - 58786*d^2*e^2 + 65618*d*e^3 - 12306*e^4 - sqrt(14)* (19695*d^3*e - 79269*d^2*e^2 + 33267*d*e^3 - 19869*e^4 + 5*(6565*d^3*e - 2 6423*d^2*e^2 + 11089*d*e^3 - 6623*e^4)*x^2 + 2*(6565*d^3*e - 26423*d^2*e^2 + 11089*d*e^3 - 6623*e^4)*x)*arctan(1/14*sqrt(14)*(5*x + 1)) + 14*(2115*d ^3*e - 7681*d^2*e^2 + 4003*d*e^3 - 4101*e^4)*x - 9800*(12*d^4 + 15*d^3*e + 9*d^2*e^2 - 3*d*e^3 + 6*e^4 + 5*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2* e^4)*x^2 + 2*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*x)*log(e*x + d) + 196*(615*d^3*e - 183*d^2*e^2 + 69*d*e^3 + 42*e^4 + 5*(205*d^3*e - 61*d^ 2*e^2 + 23*d*e^3 + 14*e^4)*x^2 + 2*(205*d^3*e - 61*d^2*e^2 + 23*d*e^3 + 14 *e^4)*x)*log(5*x^2 + 2*x + 3))/(75*d^4*e - 60*d^3*e^2 + 102*d^2*e^3 - 36*d *e^4 + 27*e^5 + 5*(25*d^4*e - 20*d^3*e^2 + 34*d^2*e^3 - 12*d*e^4 + 9*e^5)* x^2 + 2*(25*d^4*e - 20*d^3*e^2 + 34*d^2*e^3 - 12*d*e^4 + 9*e^5)*x)
Timed out. \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.29 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=\frac {\sqrt {14} {\left (6565 \, d^{3} - 26423 \, d^{2} e + 11089 \, d e^{2} - 6623 \, e^{3}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{9800 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} + \frac {{\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} \log \left (e x + d\right )}{25 \, d^{4} e - 20 \, d^{3} e^{2} + 34 \, d^{2} e^{3} - 12 \, d e^{4} + 9 \, e^{5}} - \frac {{\left (205 \, d^{3} - 61 \, d^{2} e + 23 \, d e^{2} + 14 \, e^{3}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{50 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {{\left (423 \, d - 1367 \, e\right )} x + 1367 \, d - 293 \, e}{700 \, {\left (5 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )} x^{2} + 15 \, d^{2} - 6 \, d e + 9 \, e^{2} + 2 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )} x\right )}} \]
1/9800*sqrt(14)*(6565*d^3 - 26423*d^2*e + 11089*d*e^2 - 6623*e^3)*arctan(1 /14*sqrt(14)*(5*x + 1))/(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4 ) + (4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*log(e*x + d)/(25*d^4*e - 20*d^3*e^2 + 34*d^2*e^3 - 12*d*e^4 + 9*e^5) - 1/50*(205*d^3 - 61*d^2*e + 23*d*e^2 + 14*e^3)*log(5*x^2 + 2*x + 3)/(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4) - 1/700*((423*d - 1367*e)*x + 1367*d - 293*e)/(5*(5*d^2 - 2*d*e + 3*e^2)*x^2 + 15*d^2 - 6*d*e + 9*e^2 + 2*(5*d^2 - 2*d*e + 3*e^2)* x)
Time = 0.28 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.32 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=\frac {\sqrt {14} {\left (6565 \, d^{3} - 26423 \, d^{2} e + 11089 \, d e^{2} - 6623 \, e^{3}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{9800 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {{\left (205 \, d^{3} - 61 \, d^{2} e + 23 \, d e^{2} + 14 \, e^{3}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{50 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} + \frac {{\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{25 \, d^{4} e - 20 \, d^{3} e^{2} + 34 \, d^{2} e^{3} - 12 \, d e^{4} + 9 \, e^{5}} - \frac {6835 \, d^{3} - 4199 \, d^{2} e + 4687 \, d e^{2} - 879 \, e^{3} + {\left (2115 \, d^{3} - 7681 \, d^{2} e + 4003 \, d e^{2} - 4101 \, e^{3}\right )} x}{700 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}^{2} {\left (5 \, x^{2} + 2 \, x + 3\right )}} \]
1/9800*sqrt(14)*(6565*d^3 - 26423*d^2*e + 11089*d*e^2 - 6623*e^3)*arctan(1 /14*sqrt(14)*(5*x + 1))/(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4 ) - 1/50*(205*d^3 - 61*d^2*e + 23*d*e^2 + 14*e^3)*log(5*x^2 + 2*x + 3)/(25 *d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4) + (4*d^4 + 5*d^3*e + 3*d^ 2*e^2 - d*e^3 + 2*e^4)*log(abs(e*x + d))/(25*d^4*e - 20*d^3*e^2 + 34*d^2*e ^3 - 12*d*e^4 + 9*e^5) - 1/700*(6835*d^3 - 4199*d^2*e + 4687*d*e^2 - 879*e ^3 + (2115*d^3 - 7681*d^2*e + 4003*d*e^2 - 4101*e^3)*x)/((5*d^2 - 2*d*e + 3*e^2)^2*(5*x^2 + 2*x + 3))
Time = 14.22 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.47 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^2} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (4\,d^4+5\,d^3\,e+3\,d^2\,e^2-d\,e^3+2\,e^4\right )}{e\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^2}+\frac {\ln \left (x+\frac {1}{5}-\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {1313\,\sqrt {14}}{3920}-\frac {41}{10}{}\mathrm {i}\right )\,d^3+\left (-\frac {26423\,\sqrt {14}}{19600}+\frac {61}{50}{}\mathrm {i}\right )\,d^2\,e+\left (\frac {11089\,\sqrt {14}}{19600}-\frac {23}{50}{}\mathrm {i}\right )\,d\,e^2+\left (-\frac {6623\,\sqrt {14}}{19600}-\frac {7}{25}{}\mathrm {i}\right )\,e^3\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}-\frac {\ln \left (x+\frac {1}{5}+\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {1313\,\sqrt {14}}{3920}+\frac {41}{10}{}\mathrm {i}\right )\,d^3+\left (-\frac {26423\,\sqrt {14}}{19600}-\frac {61}{50}{}\mathrm {i}\right )\,d^2\,e+\left (\frac {11089\,\sqrt {14}}{19600}+\frac {23}{50}{}\mathrm {i}\right )\,d\,e^2+\left (-\frac {6623\,\sqrt {14}}{19600}+\frac {7}{25}{}\mathrm {i}\right )\,e^3\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}-\frac {\frac {1367\,d-293\,e}{700\,\left (5\,d^2-2\,d\,e+3\,e^2\right )}+\frac {x\,\left (423\,d-1367\,e\right )}{700\,\left (5\,d^2-2\,d\,e+3\,e^2\right )}}{5\,x^2+2\,x+3} \]
(log(x - (14^(1/2)*1i)/5 + 1/5)*(d^3*((1313*14^(1/2))/3920 - 41i/10) - e^3 *((6623*14^(1/2))/19600 + 7i/25) + d*e^2*((11089*14^(1/2))/19600 - 23i/50) - d^2*e*((26423*14^(1/2))/19600 - 61i/50)))/(d^4*25i - d^3*e*20i - d*e^3* 12i + e^4*9i + d^2*e^2*34i) - ((1367*d - 293*e)/(700*(5*d^2 - 2*d*e + 3*e^ 2)) + (x*(423*d - 1367*e))/(700*(5*d^2 - 2*d*e + 3*e^2)))/(2*x + 5*x^2 + 3 ) - (log(x + (14^(1/2)*1i)/5 + 1/5)*(d^3*((1313*14^(1/2))/3920 + 41i/10) - e^3*((6623*14^(1/2))/19600 - 7i/25) + d*e^2*((11089*14^(1/2))/19600 + 23i /50) - d^2*e*((26423*14^(1/2))/19600 + 61i/50)))/(d^4*25i - d^3*e*20i - d* e^3*12i + e^4*9i + d^2*e^2*34i) + (log(d + e*x)*(5*d^3*e - d*e^3 + 4*d^4 + 2*e^4 + 3*d^2*e^2))/(e*(5*d^2 - 2*d*e + 3*e^2)^2)