Integrand size = 38, antiderivative size = 168 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=-\frac {(20 d+33 e) x}{25 e^2}+\frac {2 x^2}{5 e}-\frac {(423 d-1367 e) \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )}{125 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )}+\frac {\left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )}+\frac {(458 d-7 e) \log \left (3+2 x+5 x^2\right )}{250 \left (5 d^2-2 d e+3 e^2\right )} \]
-1/25*(20*d+33*e)*x/e^2+2/5*x^2/e+(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln (e*x+d)/e^3/(5*d^2-2*d*e+3*e^2)+1/250*(458*d-7*e)*ln(5*x^2+2*x+3)/(5*d^2-2 *d*e+3*e^2)-1/1750*(423*d-1367*e)*arctan(1/14*(1+5*x)*14^(1/2))/(5*d^2-2*d *e+3*e^2)*14^(1/2)
Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=\frac {70 e \left (5 d^2-2 d e+3 e^2\right ) x (-20 d+e (-33+10 x))-\sqrt {14} (423 d-1367 e) e^3 \arctan \left (\frac {1+5 x}{\sqrt {14}}\right )+1750 \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)+7 (458 d-7 e) e^3 \log \left (3+2 x+5 x^2\right )}{1750 e^3 \left (5 d^2-2 d e+3 e^2\right )} \]
(70*e*(5*d^2 - 2*d*e + 3*e^2)*x*(-20*d + e*(-33 + 10*x)) - Sqrt[14]*(423*d - 1367*e)*e^3*ArcTan[(1 + 5*x)/Sqrt[14]] + 1750*(4*d^4 + 5*d^3*e + 3*d^2* e^2 - d*e^3 + 2*e^4)*Log[d + e*x] + 7*(458*d - 7*e)*e^3*Log[3 + 2*x + 5*x^ 2])/(1750*e^3*(5*d^2 - 2*d*e + 3*e^2))
Time = 0.40 (sec) , antiderivative size = 168, 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.
| \(\displaystyle \int \frac {4 x^4-5 x^3+3 x^2+x+2}{\left (5 x^2+2 x+3\right ) (d+e x)} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {x (458 d-7 e)+7 d+272 e}{25 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}+\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^2 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}+\frac {-20 d-33 e}{25 e^2}+\frac {4 x}{5 e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {5 x+1}{\sqrt {14}}\right ) (423 d-1367 e)}{125 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )}+\frac {(458 d-7 e) \log \left (5 x^2+2 x+3\right )}{250 \left (5 d^2-2 d e+3 e^2\right )}+\frac {\left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )}-\frac {x (20 d+33 e)}{25 e^2}+\frac {2 x^2}{5 e}\) |
-1/25*((20*d + 33*e)*x)/e^2 + (2*x^2)/(5*e) - ((423*d - 1367*e)*ArcTan[(1 + 5*x)/Sqrt[14]])/(125*Sqrt[14]*(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)*Log[d + e*x])/(e^3*(5*d^2 - 2*d*e + 3*e^2) ) + ((458*d - 7*e)*Log[3 + 2*x + 5*x^2])/(250*(5*d^2 - 2*d*e + 3*e^2))
3.4.8.3.1 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 = 0.78 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.85
| method | result | size |
| default | \(-\frac {-10 e \,x^{2}+20 d x +33 e x}{25 e^{2}}+\frac {\frac {\left (458 d -7 e \right ) \ln \left (5 x^{2}+2 x +3\right )}{10}+\frac {\left (-\frac {423 d}{5}+\frac {1367 e}{5}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{14}}{125 d^{2}-50 d e +75 e^{2}}+\frac {\left (4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}\right ) \ln \left (e x +d \right )}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right )}\) | \(142\) |
| risch | \(\text {Expression too large to display}\) | \(7288\) |
-1/25/e^2*(-10*e*x^2+20*d*x+33*e*x)+1/(125*d^2-50*d*e+75*e^2)*(1/10*(458*d -7*e)*ln(5*x^2+2*x+3)+1/14*(-423/5*d+1367/5*e)*14^(1/2)*arctan(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^3/(5*d^2-2 *d*e+3*e^2)
Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=\frac {700 \, {\left (5 \, d^{2} e^{2} - 2 \, d e^{3} + 3 \, e^{4}\right )} x^{2} - \sqrt {14} {\left (423 \, d e^{3} - 1367 \, e^{4}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) - 70 \, {\left (100 \, d^{3} e + 125 \, d^{2} e^{2} - 6 \, d e^{3} + 99 \, e^{4}\right )} x + 1750 \, {\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} \log \left (e x + d\right ) + 7 \, {\left (458 \, d e^{3} - 7 \, e^{4}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{1750 \, {\left (5 \, d^{2} e^{3} - 2 \, d e^{4} + 3 \, e^{5}\right )}} \]
1/1750*(700*(5*d^2*e^2 - 2*d*e^3 + 3*e^4)*x^2 - sqrt(14)*(423*d*e^3 - 1367 *e^4)*arctan(1/14*sqrt(14)*(5*x + 1)) - 70*(100*d^3*e + 125*d^2*e^2 - 6*d* e^3 + 99*e^4)*x + 1750*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*log(e *x + d) + 7*(458*d*e^3 - 7*e^4)*log(5*x^2 + 2*x + 3))/(5*d^2*e^3 - 2*d*e^4 + 3*e^5)
Timed out. \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=-\frac {\sqrt {14} {\left (423 \, d - 1367 \, e\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{1750 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}} + \frac {{\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} \log \left (e x + d\right )}{5 \, d^{2} e^{3} - 2 \, d e^{4} + 3 \, e^{5}} + \frac {{\left (458 \, d - 7 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{250 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}} + \frac {10 \, e x^{2} - {\left (20 \, d + 33 \, e\right )} x}{25 \, e^{2}} \]
-1/1750*sqrt(14)*(423*d - 1367*e)*arctan(1/14*sqrt(14)*(5*x + 1))/(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)*log(e*x + d)/(5*d^2*e^3 - 2*d*e^4 + 3*e^5) + 1/250*(458*d - 7*e)*log(5*x^2 + 2*x + 3 )/(5*d^2 - 2*d*e + 3*e^2) + 1/25*(10*e*x^2 - (20*d + 33*e)*x)/e^2
Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.95 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=-\frac {\sqrt {14} {\left (423 \, d - 1367 \, e\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{1750 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}} + \frac {{\left (458 \, d - 7 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{250 \, {\left (5 \, d^{2} - 2 \, d e + 3 \, e^{2}\right )}} + \frac {{\left (4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{5 \, d^{2} e^{3} - 2 \, d e^{4} + 3 \, e^{5}} + \frac {10 \, e x^{2} - 20 \, d x - 33 \, e x}{25 \, e^{2}} \]
-1/1750*sqrt(14)*(423*d - 1367*e)*arctan(1/14*sqrt(14)*(5*x + 1))/(5*d^2 - 2*d*e + 3*e^2) + 1/250*(458*d - 7*e)*log(5*x^2 + 2*x + 3)/(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)*log(abs(e*x + d)) /(5*d^2*e^3 - 2*d*e^4 + 3*e^5) + 1/25*(10*e*x^2 - 20*d*x - 33*e*x)/e^2
Time = 15.97 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.24 \[ \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x) \left (3+2 x+5 x^2\right )} \, dx=\frac {2\,x^2}{5\,e}-\ln \left (d+e\,x\right )\,\left (\frac {\frac {458\,d}{125}-\frac {7\,e}{125}}{5\,d^2-2\,d\,e+3\,e^2}-\frac {100\,d^2+165\,d\,e+81\,e^2}{125\,e^3}\right )-x\,\left (\frac {4\,\left (5\,d+2\,e\right )}{25\,e^2}+\frac {1}{e}\right )-\frac {\ln \left (\frac {-28\,d^3+1053\,d^2\,e+1791\,d\,e^2+916\,e^3}{25\,e^2}-\frac {x\,\left (1832\,d^3+2318\,d^2\,e+321\,d\,e^2-2249\,e^3\right )}{25\,e^2}+\frac {\left (d\,\left (\frac {423\,\sqrt {14}}{3500}-\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}-\frac {7}{250}{}\mathrm {i}\right )\right )\,\left (\frac {-1000\,d^4+4350\,d^3\,e+8490\,d^2\,e^2+4751\,d\,e^3+874\,e^4}{25\,e^2}+\frac {x\,\left (-5000\,d^4-6250\,d^3\,e+1850\,d^2\,e^2+8200\,d\,e^3+2917\,e^4\right )}{25\,e^2}-\frac {\left (\frac {1250\,d^2\,e^3-14500\,d\,e^4+750\,e^5}{25\,e^2}-\frac {x\,\left (-6250\,d^2\,e^3+2500\,d\,e^4+10250\,e^5\right )}{25\,e^2}\right )\,\left (d\,\left (\frac {423\,\sqrt {14}}{3500}-\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}-\frac {7}{250}{}\mathrm {i}\right )\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}}\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}}\right )\,\left (d\,\left (\frac {423\,\sqrt {14}}{3500}-\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}-\frac {7}{250}{}\mathrm {i}\right )\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}}+\frac {\ln \left (\frac {-28\,d^3+1053\,d^2\,e+1791\,d\,e^2+916\,e^3}{25\,e^2}-\frac {x\,\left (1832\,d^3+2318\,d^2\,e+321\,d\,e^2-2249\,e^3\right )}{25\,e^2}-\frac {\left (d\,\left (\frac {423\,\sqrt {14}}{3500}+\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}+\frac {7}{250}{}\mathrm {i}\right )\right )\,\left (\frac {-1000\,d^4+4350\,d^3\,e+8490\,d^2\,e^2+4751\,d\,e^3+874\,e^4}{25\,e^2}+\frac {x\,\left (-5000\,d^4-6250\,d^3\,e+1850\,d^2\,e^2+8200\,d\,e^3+2917\,e^4\right )}{25\,e^2}+\frac {\left (\frac {1250\,d^2\,e^3-14500\,d\,e^4+750\,e^5}{25\,e^2}-\frac {x\,\left (-6250\,d^2\,e^3+2500\,d\,e^4+10250\,e^5\right )}{25\,e^2}\right )\,\left (d\,\left (\frac {423\,\sqrt {14}}{3500}+\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}+\frac {7}{250}{}\mathrm {i}\right )\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}}\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}}\right )\,\left (d\,\left (\frac {423\,\sqrt {14}}{3500}+\frac {229}{125}{}\mathrm {i}\right )-e\,\left (\frac {1367\,\sqrt {14}}{3500}+\frac {7}{250}{}\mathrm {i}\right )\right )}{d^2\,5{}\mathrm {i}-d\,e\,2{}\mathrm {i}+e^2\,3{}\mathrm {i}} \]
(2*x^2)/(5*e) - log(d + e*x)*(((458*d)/125 - (7*e)/125)/(5*d^2 - 2*d*e + 3 *e^2) - (165*d*e + 100*d^2 + 81*e^2)/(125*e^3)) - x*((4*(5*d + 2*e))/(25*e ^2) + 1/e) - (log((1791*d*e^2 + 1053*d^2*e - 28*d^3 + 916*e^3)/(25*e^2) - (x*(321*d*e^2 + 2318*d^2*e + 1832*d^3 - 2249*e^3))/(25*e^2) + ((d*((423*14 ^(1/2))/3500 - 229i/125) - e*((1367*14^(1/2))/3500 - 7i/250))*((4751*d*e^3 + 4350*d^3*e - 1000*d^4 + 874*e^4 + 8490*d^2*e^2)/(25*e^2) + (x*(8200*d*e ^3 - 6250*d^3*e - 5000*d^4 + 2917*e^4 + 1850*d^2*e^2))/(25*e^2) - (((750*e ^5 - 14500*d*e^4 + 1250*d^2*e^3)/(25*e^2) - (x*(2500*d*e^4 + 10250*e^5 - 6 250*d^2*e^3))/(25*e^2))*(d*((423*14^(1/2))/3500 - 229i/125) - e*((1367*14^ (1/2))/3500 - 7i/250)))/(d^2*5i - d*e*2i + e^2*3i)))/(d^2*5i - d*e*2i + e^ 2*3i))*(d*((423*14^(1/2))/3500 - 229i/125) - e*((1367*14^(1/2))/3500 - 7i/ 250)))/(d^2*5i - d*e*2i + e^2*3i) + (log((1791*d*e^2 + 1053*d^2*e - 28*d^3 + 916*e^3)/(25*e^2) - (x*(321*d*e^2 + 2318*d^2*e + 1832*d^3 - 2249*e^3))/ (25*e^2) - ((d*((423*14^(1/2))/3500 + 229i/125) - e*((1367*14^(1/2))/3500 + 7i/250))*((4751*d*e^3 + 4350*d^3*e - 1000*d^4 + 874*e^4 + 8490*d^2*e^2)/ (25*e^2) + (x*(8200*d*e^3 - 6250*d^3*e - 5000*d^4 + 2917*e^4 + 1850*d^2*e^ 2))/(25*e^2) + (((750*e^5 - 14500*d*e^4 + 1250*d^2*e^3)/(25*e^2) - (x*(250 0*d*e^4 + 10250*e^5 - 6250*d^2*e^3))/(25*e^2))*(d*((423*14^(1/2))/3500 + 2 29i/125) - e*((1367*14^(1/2))/3500 + 7i/250)))/(d^2*5i - d*e*2i + e^2*3i)) )/(d^2*5i - d*e*2i + e^2*3i))*(d*((423*14^(1/2))/3500 + 229i/125) - e*(...