Integrand size = 38, antiderivative size = 329 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=-\frac {1367 d-293 e+(423 d-1367 e) x}{1400 \left (5 d^2-2 d e+3 e^2\right ) \left (3+2 x+5 x^2\right )^2}+\frac {171735 d^3-92989 d^2 e+36207 d e^2+1831 e^3+25 \left (2203 d^3-9033 d^2 e+3635 d e^2-1829 e^3\right ) x}{39200 \left (5 d^2-2 d e+3 e^2\right )^2 \left (3+2 x+5 x^2\right )}+\frac {\left (42375 d^5-16643 d^4 e+58530 d^3 e^2-56058 d^2 e^3+31811 d e^4-8623 e^5\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{1568 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^3}+\frac {e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{\left (5 d^2-2 d e+3 e^2\right )^3}-\frac {e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log \left (3+2 x+5 x^2\right )}{2 \left (5 d^2-2 d e+3 e^2\right )^3} \]
1/1400*(-1367*d+293*e-(423*d-1367*e)*x)/(5*d^2-2*d*e+3*e^2)/(5*x^2+2*x+3)^ 2+1/39200*(171735*d^3-92989*d^2*e+36207*d*e^2+1831*e^3+25*(2203*d^3-9033*d ^2*e+3635*d*e^2-1829*e^3)*x)/(5*d^2-2*d*e+3*e^2)^2/(5*x^2+2*x+3)+e*(4*d^4+ 5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln(e*x+d)/(5*d^2-2*d*e+3*e^2)^3-1/2*e*(4*d^ 4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln(5*x^2+2*x+3)/(5*d^2-2*d*e+3*e^2)^3+1/2 1952*(42375*d^5-16643*d^4*e+58530*d^3*e^2-56058*d^2*e^3+31811*d*e^4-8623*e ^5)*arctan(1/14*(1+5*x)*14^(1/2))/(5*d^2-2*d*e+3*e^2)^3*14^(1/2)
Time = 0.20 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.86 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\frac {\frac {392 \left (5 d^2-2 d e+3 e^2\right )^2 (-d (1367+423 x)+e (293+1367 x))}{\left (3+2 x+5 x^2\right )^2}+\frac {14 \left (5 d^2-2 d e+3 e^2\right ) \left (e^3 (1831-45725 x)+5 d^3 (34347+11015 x)+d e^2 (36207+90875 x)-d^2 e (92989+225825 x)\right )}{3+2 x+5 x^2}+25 \sqrt {14} \left (42375 d^5-16643 d^4 e+58530 d^3 e^2-56058 d^2 e^3+31811 d e^4-8623 e^5\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )+548800 e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)-274400 e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log \left (3+2 x+5 x^2\right )}{548800 \left (5 d^2-2 d e+3 e^2\right )^3} \]
((392*(5*d^2 - 2*d*e + 3*e^2)^2*(-(d*(1367 + 423*x)) + e*(293 + 1367*x)))/ (3 + 2*x + 5*x^2)^2 + (14*(5*d^2 - 2*d*e + 3*e^2)*(e^3*(1831 - 45725*x) + 5*d^3*(34347 + 11015*x) + d*e^2*(36207 + 90875*x) - d^2*e*(92989 + 225825* x)))/(3 + 2*x + 5*x^2) + 25*Sqrt[14]*(42375*d^5 - 16643*d^4*e + 58530*d^3* e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*ArcTan[(1 + 5*x)/Sqrt[14]] + 548800*e*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[d + e*x] - 274 400*e*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[3 + 2*x + 5*x^2])/ (548800*(5*d^2 - 2*d*e + 3*e^2)^3)
Time = 0.81 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2177, 27, 2177, 27, 1200, 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.
\(\displaystyle \int \frac {4 x^4-5 x^3+3 x^2+x+2}{\left (5 x^2+2 x+3\right )^3 (d+e x)} \, dx\) |
\(\Big \downarrow \) 2177 |
\(\displaystyle \frac {1}{112} \int \frac {2 \left (1120 x^2-\frac {3 \left (3080 d^2-809 e d+481 e^2\right ) x}{5 d^2-2 e d+3 e^2}+\frac {3267 d^2-2843 e d+2800 e^2}{5 d^2-2 e d+3 e^2}\right )}{25 (d+e x) \left (5 x^2+2 x+3\right )^2}dx-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1120 x^2-\frac {3 \left (3080 d^2-809 e d+481 e^2\right ) x}{5 d^2-2 e d+3 e^2}+\frac {3267 d^2-2843 e d+2800 e^2}{5 d^2-2 e d+3 e^2}}{(d+e x) \left (5 x^2+2 x+3\right )^2}dx}{1400}-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
\(\Big \downarrow \) 2177 |
\(\displaystyle \frac {\frac {1}{56} \int \frac {50 \left (8475 d^4-1193 e d^3+8339 e^2 d^2-3397 e^3 d+3136 e^4+e \left (2203 d^3-9033 e d^2+3635 e^2 d-1829 e^3\right ) x\right )}{\left (5 d^2-2 e d+3 e^2\right )^2 (d+e x) \left (5 x^2+2 x+3\right )}dx+\frac {171735 d^3-92989 d^2 e+25 x \left (2203 d^3-9033 d^2 e+3635 d e^2-1829 e^3\right )+36207 d e^2+1831 e^3}{28 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^2}}{1400}-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {25 \int \frac {8475 d^4-1193 e d^3+8339 e^2 d^2-3397 e^3 d+3136 e^4+e \left (2203 d^3-9033 e d^2+3635 e^2 d-1829 e^3\right ) x}{(d+e x) \left (5 x^2+2 x+3\right )}dx}{28 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {171735 d^3-92989 d^2 e+25 x \left (2203 d^3-9033 d^2 e+3635 d e^2-1829 e^3\right )+36207 d e^2+1831 e^3}{28 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^2}}{1400}-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {\frac {25 \int \left (\frac {1568 \left (4 d^4+5 e d^3+3 e^2 d^2-e^3 d+2 e^4\right ) e^2}{\left (5 d^2-2 e d+3 e^2\right ) (d+e x)}+\frac {42375 d^5-22915 e d^4+50690 e^2 d^3-60762 e^3 d^2+33379 e^4 d-11759 e^5-7840 e \left (4 d^4+5 e d^3+3 e^2 d^2-e^3 d+2 e^4\right ) x}{\left (5 d^2-2 e d+3 e^2\right ) \left (5 x^2+2 x+3\right )}\right )dx}{28 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {171735 d^3-92989 d^2 e+25 x \left (2203 d^3-9033 d^2 e+3635 d e^2-1829 e^3\right )+36207 d e^2+1831 e^3}{28 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^2}}{1400}-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {25 \left (\frac {\arctan \left (\frac {5 x+1}{\sqrt {14}}\right ) \left (42375 d^5-16643 d^4 e+58530 d^3 e^2-56058 d^2 e^3+31811 d e^4-8623 e^5\right )}{\sqrt {14} \left (5 d^2-2 d e+3 e^2\right )}-\frac {784 e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log \left (5 x^2+2 x+3\right )}{5 d^2-2 d e+3 e^2}+\frac {1568 e \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{5 d^2-2 d e+3 e^2}\right )}{28 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {171735 d^3-92989 d^2 e+25 x \left (2203 d^3-9033 d^2 e+3635 d e^2-1829 e^3\right )+36207 d e^2+1831 e^3}{28 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^2}}{1400}-\frac {x (423 d-1367 e)+1367 d-293 e}{1400 \left (5 x^2+2 x+3\right )^2 \left (5 d^2-2 d e+3 e^2\right )}\) |
-1/1400*(1367*d - 293*e + (423*d - 1367*e)*x)/((5*d^2 - 2*d*e + 3*e^2)*(3 + 2*x + 5*x^2)^2) + ((171735*d^3 - 92989*d^2*e + 36207*d*e^2 + 1831*e^3 + 25*(2203*d^3 - 9033*d^2*e + 3635*d*e^2 - 1829*e^3)*x)/(28*(5*d^2 - 2*d*e + 3*e^2)^2*(3 + 2*x + 5*x^2)) + (25*(((42375*d^5 - 16643*d^4*e + 58530*d^3* e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*ArcTan[(1 + 5*x)/Sqrt[14]])/ (Sqrt[14]*(5*d^2 - 2*d*e + 3*e^2)) + (1568*e*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[d + e*x])/(5*d^2 - 2*d*e + 3*e^2) - (784*e*(4*d^4 + 5 *d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[3 + 2*x + 5*x^2])/(5*d^2 - 2*d*e + 3*e^2)))/(28*(5*d^2 - 2*d*e + 3*e^2)^2))/1400
3.4.22.3.1 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[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
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 = 1.05 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {\frac {25 \left (\frac {2203}{1568} d^{5}-\frac {49571}{7840} d^{4} e +\frac {4285}{784} d^{3} e^{2}-\frac {21757}{3920} d^{2} e^{3}+\frac {14563}{7840} d \,e^{4}-\frac {5487}{7840} e^{5}\right ) x^{3}+25 \left (\frac {38753}{7840} d^{5}-\frac {10433}{1568} d^{4} e +\frac {655359}{98000} d^{3} e^{2}-\frac {388683}{98000} d^{2} e^{3}+\frac {250589}{196000} d \,e^{4}-\frac {49377}{196000} e^{5}\right ) x^{2}+25 \left (\frac {17979}{7840} d^{5}-\frac {33127}{7840} d^{4} e +\frac {380997}{98000} d^{3} e^{2}-\frac {250449}{98000} d^{2} e^{3}+\frac {147247}{196000} d \,e^{4}-\frac {11211}{196000} e^{5}\right ) x +\frac {64765 d^{5}}{1568}-\frac {58185 d^{4} e}{1568}+\frac {118119 d^{3} e^{2}}{3920}-\frac {28843 d^{2} e^{3}}{3920}-\frac {25611 d \,e^{4}}{7840}+\frac {18063 e^{5}}{7840}}{\left (5 x^{2}+2 x +3\right )^{2}}+\frac {\left (-31360 d^{4} e -39200 d^{3} e^{2}-23520 d^{2} e^{3}+7840 d \,e^{4}-15680 e^{5}\right ) \ln \left (5 x^{2}+2 x +3\right )}{15680}+\frac {\left (42375 d^{5}-16643 d^{4} e +58530 d^{3} e^{2}-56058 d^{2} e^{3}+31811 d \,e^{4}-8623 e^{5}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{21952}}{\left (5 d^{2}-2 d e +3 e^{2}\right )^{3}}+\frac {e \left (4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}\right ) \ln \left (e x +d \right )}{\left (5 d^{2}-2 d e +3 e^{2}\right )^{3}}\) | \(359\) |
risch | \(\text {Expression too large to display}\) | \(28636\) |
1/(5*d^2-2*d*e+3*e^2)^3*(25*((2203/1568*d^5-49571/7840*d^4*e+4285/784*d^3* e^2-21757/3920*d^2*e^3+14563/7840*d*e^4-5487/7840*e^5)*x^3+(38753/7840*d^5 -10433/1568*d^4*e+655359/98000*d^3*e^2-388683/98000*d^2*e^3+250589/196000* d*e^4-49377/196000*e^5)*x^2+(17979/7840*d^5-33127/7840*d^4*e+380997/98000* d^3*e^2-250449/98000*d^2*e^3+147247/196000*d*e^4-11211/196000*e^5)*x+12953 /7840*d^5-11637/7840*d^4*e+118119/98000*d^3*e^2-28843/98000*d^2*e^3-25611/ 196000*d*e^4+18063/196000*e^5)/(5*x^2+2*x+3)^2+1/15680*(-31360*d^4*e-39200 *d^3*e^2-23520*d^2*e^3+7840*d*e^4-15680*e^5)*ln(5*x^2+2*x+3)+1/21952*(4237 5*d^5-16643*d^4*e+58530*d^3*e^2-56058*d^2*e^3+31811*d*e^4-8623*e^5)*14^(1/ 2)*arctan(1/28*(10*x+2)*14^(1/2)))+e*(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4) *ln(e*x+d)/(5*d^2-2*d*e+3*e^2)^3
Leaf count of result is larger than twice the leaf count of optimal. 1052 vs. \(2 (318) = 636\).
Time = 0.39 (sec) , antiderivative size = 1052, normalized size of antiderivative = 3.20 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\text {Too large to display} \]
1/109760*(4533550*d^5 - 4072950*d^4*e + 3307332*d^3*e^2 - 807604*d^2*e^3 - 358554*d*e^4 + 252882*e^5 + 350*(11015*d^5 - 49571*d^4*e + 42850*d^3*e^2 - 43514*d^2*e^3 + 14563*d*e^4 - 5487*e^5)*x^3 + 14*(968825*d^5 - 1304125*d ^4*e + 1310718*d^3*e^2 - 777366*d^2*e^3 + 250589*d*e^4 - 49377*e^5)*x^2 + 5*sqrt(14)*(381375*d^5 - 149787*d^4*e + 526770*d^3*e^2 - 504522*d^2*e^3 + 286299*d*e^4 - 77607*e^5 + 25*(42375*d^5 - 16643*d^4*e + 58530*d^3*e^2 - 5 6058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*x^4 + 20*(42375*d^5 - 16643*d^4*e + 58530*d^3*e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*x^3 + 34*(42375*d ^5 - 16643*d^4*e + 58530*d^3*e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5) *x^2 + 12*(42375*d^5 - 16643*d^4*e + 58530*d^3*e^2 - 56058*d^2*e^3 + 31811 *d*e^4 - 8623*e^5)*x)*arctan(1/14*sqrt(14)*(5*x + 1)) + 14*(449475*d^5 - 8 28175*d^4*e + 761994*d^3*e^2 - 500898*d^2*e^3 + 147247*d*e^4 - 11211*e^5)* x + 109760*(36*d^4*e + 45*d^3*e^2 + 27*d^2*e^3 - 9*d*e^4 + 18*e^5 + 25*(4* d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^4 + 20*(4*d^4*e + 5*d^3*e ^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^3 + 34*(4*d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^2 + 12*(4*d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5 )*x)*log(e*x + d) - 54880*(36*d^4*e + 45*d^3*e^2 + 27*d^2*e^3 - 9*d*e^4 + 18*e^5 + 25*(4*d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^4 + 20*(4* d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^3 + 34*(4*d^4*e + 5*d^3*e ^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*x^2 + 12*(4*d^4*e + 5*d^3*e^2 + 3*d^2*e...
Timed out. \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 571, normalized size of antiderivative = 1.74 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\frac {\sqrt {14} {\left (42375 \, d^{5} - 16643 \, d^{4} e + 58530 \, d^{3} e^{2} - 56058 \, d^{2} e^{3} + 31811 \, d e^{4} - 8623 \, e^{5}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{21952 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {{\left (4 \, d^{4} e + 5 \, d^{3} e^{2} + 3 \, d^{2} e^{3} - d e^{4} + 2 \, e^{5}\right )} \log \left (e x + d\right )}{125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}} - \frac {{\left (4 \, d^{4} e + 5 \, d^{3} e^{2} + 3 \, d^{2} e^{3} - d e^{4} + 2 \, e^{5}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{2 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {25 \, {\left (2203 \, d^{3} - 9033 \, d^{2} e + 3635 \, d e^{2} - 1829 \, e^{3}\right )} x^{3} + 64765 \, d^{3} - 32279 \, d^{2} e - 4523 \, d e^{2} + 6021 \, e^{3} + {\left (193765 \, d^{3} - 183319 \, d^{2} e + 72557 \, d e^{2} - 16459 \, e^{3}\right )} x^{2} + {\left (89895 \, d^{3} - 129677 \, d^{2} e + 46591 \, d e^{2} - 3737 \, e^{3}\right )} x}{7840 \, {\left (25 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )} x^{4} + 225 \, d^{4} - 180 \, d^{3} e + 306 \, d^{2} e^{2} - 108 \, d e^{3} + 81 \, e^{4} + 20 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )} x^{3} + 34 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )} x^{2} + 12 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )} x\right )}} \]
1/21952*sqrt(14)*(42375*d^5 - 16643*d^4*e + 58530*d^3*e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*arctan(1/14*sqrt(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) + (4*d ^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*log(e*x + d)/(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 *(4*d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*log(5*x^2 + 2*x + 3)/(1 25*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) + 1/7840*(25*(2203*d^3 - 9033*d^2*e + 3635*d*e^2 - 1829*e^3)*x^3 + 64765*d^3 - 32279*d^2*e - 4523*d*e^2 + 6021*e^3 + (193765*d^3 - 183319*d^ 2*e + 72557*d*e^2 - 16459*e^3)*x^2 + (89895*d^3 - 129677*d^2*e + 46591*d*e ^2 - 3737*e^3)*x)/(25*(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4)* x^4 + 225*d^4 - 180*d^3*e + 306*d^2*e^2 - 108*d*e^3 + 81*e^4 + 20*(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4)*x^3 + 34*(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4)*x^2 + 12*(25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4)*x)
Time = 0.28 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.50 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\frac {\sqrt {14} {\left (42375 \, d^{5} - 16643 \, d^{4} e + 58530 \, d^{3} e^{2} - 56058 \, d^{2} e^{3} + 31811 \, d e^{4} - 8623 \, e^{5}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{21952 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} - \frac {{\left (4 \, d^{4} e + 5 \, d^{3} e^{2} + 3 \, d^{2} e^{3} - d e^{4} + 2 \, e^{5}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{2 \, {\left (125 \, d^{6} - 150 \, d^{5} e + 285 \, d^{4} e^{2} - 188 \, d^{3} e^{3} + 171 \, d^{2} e^{4} - 54 \, d e^{5} + 27 \, e^{6}\right )}} + \frac {{\left (4 \, d^{4} e^{2} + 5 \, d^{3} e^{3} + 3 \, d^{2} e^{4} - d e^{5} + 2 \, e^{6}\right )} \log \left ({\left | e x + d \right |}\right )}{125 \, d^{6} e - 150 \, d^{5} e^{2} + 285 \, d^{4} e^{3} - 188 \, d^{3} e^{4} + 171 \, d^{2} e^{5} - 54 \, d e^{6} + 27 \, e^{7}} + \frac {323825 \, d^{5} - 290925 \, d^{4} e + 236238 \, d^{3} e^{2} - 57686 \, d^{2} e^{3} - 25611 \, d e^{4} + 18063 \, e^{5} + 25 \, {\left (11015 \, d^{5} - 49571 \, d^{4} e + 42850 \, d^{3} e^{2} - 43514 \, d^{2} e^{3} + 14563 \, d e^{4} - 5487 \, e^{5}\right )} x^{3} + {\left (968825 \, d^{5} - 1304125 \, d^{4} e + 1310718 \, d^{3} e^{2} - 777366 \, d^{2} e^{3} + 250589 \, d e^{4} - 49377 \, e^{5}\right )} x^{2} + {\left (449475 \, d^{5} - 828175 \, d^{4} e + 761994 \, d^{3} e^{2} - 500898 \, d^{2} e^{3} + 147247 \, d e^{4} - 11211 \, e^{5}\right )} x}{7840 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}^{3} {\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \]
1/21952*sqrt(14)*(42375*d^5 - 16643*d^4*e + 58530*d^3*e^2 - 56058*d^2*e^3 + 31811*d*e^4 - 8623*e^5)*arctan(1/14*sqrt(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) - 1/2* (4*d^4*e + 5*d^3*e^2 + 3*d^2*e^3 - d*e^4 + 2*e^5)*log(5*x^2 + 2*x + 3)/(12 5*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 + 2 7*e^6) + (4*d^4*e^2 + 5*d^3*e^3 + 3*d^2*e^4 - d*e^5 + 2*e^6)*log(abs(e*x + d))/(125*d^6*e - 150*d^5*e^2 + 285*d^4*e^3 - 188*d^3*e^4 + 171*d^2*e^5 - 54*d*e^6 + 27*e^7) + 1/7840*(323825*d^5 - 290925*d^4*e + 236238*d^3*e^2 - 57686*d^2*e^3 - 25611*d*e^4 + 18063*e^5 + 25*(11015*d^5 - 49571*d^4*e + 42 850*d^3*e^2 - 43514*d^2*e^3 + 14563*d*e^4 - 5487*e^5)*x^3 + (968825*d^5 - 1304125*d^4*e + 1310718*d^3*e^2 - 777366*d^2*e^3 + 250589*d*e^4 - 49377*e^ 5)*x^2 + (449475*d^5 - 828175*d^4*e + 761994*d^3*e^2 - 500898*d^2*e^3 + 14 7247*d*e^4 - 11211*e^5)*x)/((5*d^2 - 2*d*e + 3*e^2)^3*(5*x^2 + 2*x + 3)^2)
Time = 13.84 (sec) , antiderivative size = 641, normalized size of antiderivative = 1.95 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )^3} \, dx=\frac {\frac {x\,\left (89895\,d^3-129677\,d^2\,e+46591\,d\,e^2-3737\,e^3\right )}{7840\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}-\frac {-64765\,d^3+32279\,d^2\,e+4523\,d\,e^2-6021\,e^3}{7840\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}+\frac {5\,x^3\,\left (2203\,d^3-9033\,d^2\,e+3635\,d\,e^2-1829\,e^3\right )}{1568\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}+\frac {x^2\,\left (193765\,d^3-183319\,d^2\,e+72557\,d\,e^2-16459\,e^3\right )}{7840\,\left (25\,d^4-20\,d^3\,e+34\,d^2\,e^2-12\,d\,e^3+9\,e^4\right )}}{25\,x^4+20\,x^3+34\,x^2+12\,x+9}+\ln \left (d+e\,x\right )\,\left (\frac {4\,e}{25\,\left (5\,d^2-2\,d\,e+3\,e^2\right )}+\frac {e^2\,\left (205\,d+21\,e\right )}{125\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^2}-\frac {e^4\,\left (458\,d-7\,e\right )}{125\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^3}\right )-\frac {\ln \left (x+\frac {1}{5}-\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (-\frac {42375\,\sqrt {14}\,d^5}{43904}+\left (\frac {16643\,\sqrt {14}}{43904}+2{}\mathrm {i}\right )\,d^4\,e+\left (-\frac {29265\,\sqrt {14}}{21952}+\frac {5}{2}{}\mathrm {i}\right )\,d^3\,e^2+\left (\frac {28029\,\sqrt {14}}{21952}+\frac {3}{2}{}\mathrm {i}\right )\,d^2\,e^3+\left (-\frac {31811\,\sqrt {14}}{43904}-\frac {1}{2}{}\mathrm {i}\right )\,d\,e^4+\left (\frac {8623\,\sqrt {14}}{43904}+1{}\mathrm {i}\right )\,e^5\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 (-\frac {42375\,\sqrt {14}\,d^5}{43904}+\left (\frac {16643\,\sqrt {14}}{43904}-2{}\mathrm {i}\right )\,d^4\,e+\left (-\frac {29265\,\sqrt {14}}{21952}-\frac {5}{2}{}\mathrm {i}\right )\,d^3\,e^2+\left (\frac {28029\,\sqrt {14}}{21952}-\frac {3}{2}{}\mathrm {i}\right )\,d^2\,e^3+\left (-\frac {31811\,\sqrt {14}}{43904}+\frac {1}{2}{}\mathrm {i}\right )\,d\,e^4+\left (\frac {8623\,\sqrt {14}}{43904}-\mathrm {i}\right )\,e^5\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}} \]
((x*(46591*d*e^2 - 129677*d^2*e + 89895*d^3 - 3737*e^3))/(7840*(25*d^4 - 2 0*d^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e^2)) - (4523*d*e^2 + 32279*d^2*e - 64 765*d^3 - 6021*e^3)/(7840*(25*d^4 - 20*d^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e ^2)) + (5*x^3*(3635*d*e^2 - 9033*d^2*e + 2203*d^3 - 1829*e^3))/(1568*(25*d ^4 - 20*d^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e^2)) + (x^2*(72557*d*e^2 - 1833 19*d^2*e + 193765*d^3 - 16459*e^3))/(7840*(25*d^4 - 20*d^3*e - 12*d*e^3 + 9*e^4 + 34*d^2*e^2)))/(12*x + 34*x^2 + 20*x^3 + 25*x^4 + 9) + log(d + e*x) *((4*e)/(25*(5*d^2 - 2*d*e + 3*e^2)) + (e^2*(205*d + 21*e))/(125*(5*d^2 - 2*d*e + 3*e^2)^2) - (e^4*(458*d - 7*e))/(125*(5*d^2 - 2*d*e + 3*e^2)^3)) - (log(x - (14^(1/2)*1i)/5 + 1/5)*(e^5*((8623*14^(1/2))/43904 + 1i) - (4237 5*14^(1/2)*d^5)/43904 + d^2*e^3*((28029*14^(1/2))/21952 + 3i/2) - d^3*e^2* ((29265*14^(1/2))/21952 - 5i/2) + d^4*e*((16643*14^(1/2))/43904 + 2i) - d* e^4*((31811*14^(1/2))/43904 + 1i/2)))/(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)*(e^5*((8623*14^(1/2))/43904 - 1i) - (42375*14^(1/2)*d^5)/43 904 + d^2*e^3*((28029*14^(1/2))/21952 - 3i/2) - d^3*e^2*((29265*14^(1/2))/ 21952 + 5i/2) + d^4*e*((16643*14^(1/2))/43904 - 2i) - d*e^4*((31811*14^(1/ 2))/43904 - 1i/2)))/(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)