Integrand size = 38, antiderivative size = 233 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=\frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (423 d^2-2734 d e+293 e^2\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2} \]
4/5*x/e^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)-(40*d^5+d^4*e+28*d^3*e^2+44*d^2*e^3-2*d*e^4+e^5)*ln(e*x+d)/e^3/(5*d ^2-2*d*e+3*e^2)^2+1/25*(229*d^2-7*d*e-136*e^2)*ln(5*x^2+2*x+3)/(5*d^2-2*d* e+3*e^2)^2-1/350*(423*d^2-2734*d*e+293*e^2)*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 = 233, normalized size of antiderivative = 1.00 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=\frac {4 x}{5 e^2}+\frac {-4 d^4-5 d^3 e-3 d^2 e^2+d e^3-2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}+\frac {\left (-423 d^2+2734 d e-293 e^2\right ) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (-40 d^5-d^4 e-28 d^3 e^2-44 d^2 e^3+2 d e^4-e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2} \]
(4*x)/(5*e^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)) + ((-423*d^2 + 2734*d*e - 293*e^2)*ArcTan[(1 + 5*x)/Sqrt[14]])/(25*Sqrt[14]*(5*d^2 - 2*d*e + 3*e^2)^2) + ((-40*d^5 - d^ 4*e - 28*d^3*e^2 - 44*d^2*e^3 + 2*d*e^4 - e^5)*Log[d + e*x])/(e^3*(5*d^2 - 2*d*e + 3*e^2)^2) + ((229*d^2 - 7*d*e - 136*e^2)*Log[3 + 2*x + 5*x^2])/(2 5*(5*d^2 - 2*d*e + 3*e^2)^2)
Time = 0.48 (sec) , antiderivative size = 233, 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.
| \(\displaystyle \int \frac {4 x^4-5 x^3+3 x^2+x+2}{\left (5 x^2+2 x+3\right ) (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {2 x \left (229 d^2-7 d e-136 e^2\right )+7 d^2+544 d e-113 e^2}{5 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^2 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^2}+\frac {-40 d^5-d^4 e-28 d^3 e^2-44 d^2 e^3+2 d e^4-e^5}{e^2 \left (5 d^2-2 d e+3 e^2\right )^2 (d+e x)}+\frac {4}{5 e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {5 x+1}{\sqrt {14}}\right ) \left (423 d^2-2734 d e+293 e^2\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (5 x^2+2 x+3\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {4 x}{5 e^2}\) |
(4*x)/(5*e^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)) - ((423*d^2 - 2734*d*e + 293*e^2)*ArcTan[(1 + 5*x)/Sqrt[14]])/(25*Sqrt[14]*(5*d^2 - 2*d*e + 3*e^2)^2) - ((40*d^5 + d^4*e + 28*d^3*e^2 + 44*d^2*e^3 - 2*d*e^4 + e^5)*Log[d + e*x])/(e^3*(5*d^2 - 2* d*e + 3*e^2)^2) + ((229*d^2 - 7*d*e - 136*e^2)*Log[3 + 2*x + 5*x^2])/(25*( 5*d^2 - 2*d*e + 3*e^2)^2)
3.4.9.3.1 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 = 0.97 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(\frac {4 x}{5 e^{2}}+\frac {\frac {\left (458 d^{2}-14 d e -272 e^{2}\right ) \ln \left (5 x^{2}+2 x +3\right )}{10}+\frac {\left (-\frac {423}{5} d^{2}+\frac {2734}{5} d e -\frac {293}{5} e^{2}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{14}}{5 \left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}-\frac {4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right ) \left (e x +d \right )}+\frac {\left (-40 d^{5}-d^{4} e -28 d^{3} e^{2}-44 d^{2} e^{3}+2 d \,e^{4}-e^{5}\right ) \ln \left (e x +d \right )}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}\) | \(213\) |
| risch | \(\text {Expression too large to display}\) | \(14526\) |
4/5*x/e^2+1/5/(5*d^2-2*d*e+3*e^2)^2*(1/10*(458*d^2-14*d*e-272*e^2)*ln(5*x^ 2+2*x+3)+1/14*(-423/5*d^2+2734/5*d*e-293/5*e^2)*14^(1/2)*arctan(1/28*(10*x +2)*14^(1/2)))-1/e^3*(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)/(5*d^2-2*d*e+3* e^2)/(e*x+d)+(-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*ln(e*x+d)
Time = 0.32 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.79 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=-\frac {7000 \, d^{6} + 5950 \, d^{5} e + 5950 \, d^{4} e^{2} + 1400 \, d^{3} e^{3} + 7350 \, d^{2} e^{4} - 2450 \, d e^{5} + 2100 \, e^{6} - 280 \, {\left (25 \, d^{4} e^{2} - 20 \, d^{3} e^{3} + 34 \, d^{2} e^{4} - 12 \, d e^{5} + 9 \, e^{6}\right )} x^{2} + \sqrt {14} {\left (423 \, d^{3} e^{3} - 2734 \, d^{2} e^{4} + 293 \, d e^{5} + {\left (423 \, d^{2} e^{4} - 2734 \, d e^{5} + 293 \, e^{6}\right )} x\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) - 280 \, {\left (25 \, d^{5} e - 20 \, d^{4} e^{2} + 34 \, d^{3} e^{3} - 12 \, d^{2} e^{4} + 9 \, d e^{5}\right )} x + 350 \, {\left (40 \, d^{6} + d^{5} e + 28 \, d^{4} e^{2} + 44 \, d^{3} e^{3} - 2 \, d^{2} e^{4} + d e^{5} + {\left (40 \, d^{5} e + d^{4} e^{2} + 28 \, d^{3} e^{3} + 44 \, d^{2} e^{4} - 2 \, d e^{5} + e^{6}\right )} x\right )} \log \left (e x + d\right ) - 14 \, {\left (229 \, d^{3} e^{3} - 7 \, d^{2} e^{4} - 136 \, d e^{5} + {\left (229 \, d^{2} e^{4} - 7 \, d e^{5} - 136 \, e^{6}\right )} x\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{350 \, {\left (25 \, d^{5} e^{3} - 20 \, d^{4} e^{4} + 34 \, d^{3} e^{5} - 12 \, d^{2} e^{6} + 9 \, d e^{7} + {\left (25 \, d^{4} e^{4} - 20 \, d^{3} e^{5} + 34 \, d^{2} e^{6} - 12 \, d e^{7} + 9 \, e^{8}\right )} x\right )}} \]
-1/350*(7000*d^6 + 5950*d^5*e + 5950*d^4*e^2 + 1400*d^3*e^3 + 7350*d^2*e^4 - 2450*d*e^5 + 2100*e^6 - 280*(25*d^4*e^2 - 20*d^3*e^3 + 34*d^2*e^4 - 12* d*e^5 + 9*e^6)*x^2 + sqrt(14)*(423*d^3*e^3 - 2734*d^2*e^4 + 293*d*e^5 + (4 23*d^2*e^4 - 2734*d*e^5 + 293*e^6)*x)*arctan(1/14*sqrt(14)*(5*x + 1)) - 28 0*(25*d^5*e - 20*d^4*e^2 + 34*d^3*e^3 - 12*d^2*e^4 + 9*d*e^5)*x + 350*(40* d^6 + d^5*e + 28*d^4*e^2 + 44*d^3*e^3 - 2*d^2*e^4 + d*e^5 + (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)*log(e*x + d) - 14*(229 *d^3*e^3 - 7*d^2*e^4 - 136*d*e^5 + (229*d^2*e^4 - 7*d*e^5 - 136*e^6)*x)*lo g(5*x^2 + 2*x + 3))/(25*d^5*e^3 - 20*d^4*e^4 + 34*d^3*e^5 - 12*d^2*e^6 + 9 *d*e^7 + (25*d^4*e^4 - 20*d^3*e^5 + 34*d^2*e^6 - 12*d*e^7 + 9*e^8)*x)
Timed out. \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.26 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=-\frac {\sqrt {14} {\left (423 \, d^{2} - 2734 \, d e + 293 \, e^{2}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{350 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {{\left (40 \, d^{5} + d^{4} e + 28 \, d^{3} e^{2} + 44 \, d^{2} e^{3} - 2 \, d e^{4} + e^{5}\right )} \log \left (e x + d\right )}{25 \, d^{4} e^{3} - 20 \, d^{3} e^{4} + 34 \, d^{2} e^{5} - 12 \, d e^{6} + 9 \, e^{7}} + \frac {{\left (229 \, d^{2} - 7 \, d e - 136 \, e^{2}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{25 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}}{5 \, d^{3} e^{3} - 2 \, d^{2} e^{4} + 3 \, d e^{5} + {\left (5 \, d^{2} e^{4} - 2 \, d e^{5} + 3 \, e^{6}\right )} x} + \frac {4 \, x}{5 \, e^{2}} \]
-1/350*sqrt(14)*(423*d^2 - 2734*d*e + 293*e^2)*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) - (40*d^5 + d^4*e + 28*d^3*e^2 + 44*d^2*e^3 - 2*d*e^4 + e^5)*log(e*x + d)/(25*d^4*e^3 - 20* d^3*e^4 + 34*d^2*e^5 - 12*d*e^6 + 9*e^7) + 1/25*(229*d^2 - 7*d*e - 136*e^2 )*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)/(5*d^3*e^3 - 2*d^2*e^4 + 3*d*e^5 + (5*d^2*e^4 - 2*d*e^5 + 3*e^6)*x) + 4/5*x/e^2
Time = 0.28 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.53 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=\frac {{\left (229 \, d^{2} - 7 \, d e - 136 \, e^{2}\right )} \log \left (-\frac {10 \, d}{e x + d} + \frac {5 \, d^{2}}{{\left (e x + d\right )}^{2}} + \frac {2 \, e}{e x + d} - \frac {2 \, d e}{{\left (e x + d\right )}^{2}} + \frac {3 \, e^{2}}{{\left (e x + d\right )}^{2}} + 5\right )}{25 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {\frac {4 \, d^{4} e^{3}}{e x + d} + \frac {5 \, d^{3} e^{4}}{e x + d} + \frac {3 \, d^{2} e^{5}}{e x + d} - \frac {d e^{6}}{e x + d} + \frac {2 \, e^{7}}{e x + d}}{5 \, d^{2} e^{6} - 2 \, d e^{7} + 3 \, e^{8}} - \frac {\sqrt {14} {\left (423 \, d^{2} e^{2} - 2734 \, d e^{3} + 293 \, e^{4}\right )} \arctan \left (\frac {\sqrt {14} {\left (5 \, d - \frac {5 \, d^{2}}{e x + d} + \frac {2 \, d e}{e x + d} - e - \frac {3 \, e^{2}}{e x + d}\right )}}{14 \, e}\right )}{350 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )} e^{2}} + \frac {{\left (40 \, d + 33 \, e\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{25 \, e^{3}} + \frac {4 \, {\left (e x + d\right )}}{5 \, e^{3}} \]
1/25*(229*d^2 - 7*d*e - 136*e^2)*log(-10*d/(e*x + d) + 5*d^2/(e*x + d)^2 + 2*e/(e*x + d) - 2*d*e/(e*x + d)^2 + 3*e^2/(e*x + d)^2 + 5)/(25*d^4 - 20*d ^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4) - (4*d^4*e^3/(e*x + d) + 5*d^3*e^4/( e*x + d) + 3*d^2*e^5/(e*x + d) - d*e^6/(e*x + d) + 2*e^7/(e*x + d))/(5*d^2 *e^6 - 2*d*e^7 + 3*e^8) - 1/350*sqrt(14)*(423*d^2*e^2 - 2734*d*e^3 + 293*e ^4)*arctan(1/14*sqrt(14)*(5*d - 5*d^2/(e*x + d) + 2*d*e/(e*x + d) - e - 3* e^2/(e*x + d))/e)/((25*d^4 - 20*d^3*e + 34*d^2*e^2 - 12*d*e^3 + 9*e^4)*e^2 ) + 1/25*(40*d + 33*e)*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^3 + 4/5*(e *x + d)/e^3
Time = 14.22 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.34 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx=\frac {4\,x}{5\,e^2}-\frac {\ln \left (x+\frac {1}{5}-\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{700}-\frac {229}{25}{}\mathrm {i}\right )\,d^2+\left (-\frac {1367\,\sqrt {14}}{350}+\frac {7}{25}{}\mathrm {i}\right )\,d\,e+\left (\frac {293\,\sqrt {14}}{700}+\frac {136}{25}{}\mathrm {i}\right )\,e^2\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}+\frac {\ln \left (x+\frac {1}{5}+\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{700}+\frac {229}{25}{}\mathrm {i}\right )\,d^2+\left (-\frac {1367\,\sqrt {14}}{350}-\frac {7}{25}{}\mathrm {i}\right )\,d\,e+\left (\frac {293\,\sqrt {14}}{700}-\frac {136}{25}{}\mathrm {i}\right )\,e^2\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}-\frac {5\,\left (4\,d^4+5\,d^3\,e+3\,d^2\,e^2-d\,e^3+2\,e^4\right )}{e\,\left (5\,x\,e^3+5\,d\,e^2\right )\,\left (5\,d^2-2\,d\,e+3\,e^2\right )}-\frac {\ln \left (d+e\,x\right )\,\left (40\,d^5+d^4\,e+28\,d^3\,e^2+44\,d^2\,e^3-2\,d\,e^4+e^5\right )}{e^3\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^2} \]
(4*x)/(5*e^2) - (log(x - (14^(1/2)*1i)/5 + 1/5)*(d^2*((423*14^(1/2))/700 - 229i/25) + e^2*((293*14^(1/2))/700 + 136i/25) - d*e*((1367*14^(1/2))/350 - 7i/25)))/(d^4*25i - d^3*e*20i - d*e^3*12i + e^4*9i + d^2*e^2*34i) + (log (x + (14^(1/2)*1i)/5 + 1/5)*(d^2*((423*14^(1/2))/700 + 229i/25) + e^2*((29 3*14^(1/2))/700 - 136i/25) - d*e*((1367*14^(1/2))/350 + 7i/25)))/(d^4*25i - d^3*e*20i - d*e^3*12i + e^4*9i + d^2*e^2*34i) - (5*(5*d^3*e - d*e^3 + 4* d^4 + 2*e^4 + 3*d^2*e^2))/(e*(5*d*e^2 + 5*e^3*x)*(5*d^2 - 2*d*e + 3*e^2)) - (log(d + e*x)*(d^4*e - 2*d*e^4 + 40*d^5 + e^5 + 44*d^2*e^3 + 28*d^3*e^2) )/(e^3*(5*d^2 - 2*d*e + 3*e^2)^2)