Integrand size = 20, antiderivative size = 102 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {(1+2 x) (7+6 x)}{4 \left (3+5 x+2 x^2\right )^4}+\frac {73+62 x}{3 \left (3+5 x+2 x^2\right )^3}-\frac {155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}+\frac {620 (5+4 x)}{3+5 x+2 x^2}+2480 \log (1+x)-2480 \log (3+2 x) \]
1/4*(1+2*x)*(7+6*x)/(2*x^2+5*x+3)^4+1/3*(73+62*x)/(2*x^2+5*x+3)^3-155/3*(5 +4*x)/(2*x^2+5*x+3)^2+620*(5+4*x)/(2*x^2+5*x+3)+2480*ln(1+x)-2480*ln(3+2*x )
Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=-\frac {11+10 x}{4 \left (3+5 x+2 x^2\right )^4}+\frac {31 (5+4 x)}{6 \left (3+5 x+2 x^2\right )^3}-\frac {155 (5+4 x)}{3 \left (3+5 x+2 x^2\right )^2}+\frac {620 (5+4 x)}{3+5 x+2 x^2}+2480 \log (2 (1+x))-2480 \log (3+2 x) \]
-1/4*(11 + 10*x)/(3 + 5*x + 2*x^2)^4 + (31*(5 + 4*x))/(6*(3 + 5*x + 2*x^2) ^3) - (155*(5 + 4*x))/(3*(3 + 5*x + 2*x^2)^2) + (620*(5 + 4*x))/(3 + 5*x + 2*x^2) + 2480*Log[2*(1 + x)] - 2480*Log[3 + 2*x]
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1141, 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.
\(\displaystyle \int \frac {(2 x+1)^2}{\left (2 x^2+5 x+3\right )^5} \, dx\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle 32 \int \left (-\frac {155}{2 x+3}-\frac {85}{(2 x+3)^2}-\frac {41}{(2 x+3)^3}-\frac {16}{(2 x+3)^4}-\frac {4}{(2 x+3)^5}+\frac {155}{2 (x+1)}-\frac {35}{2 (x+1)^2}+\frac {13}{4 (x+1)^3}-\frac {7}{16 (x+1)^4}+\frac {1}{32 (x+1)^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 32 \left (\frac {35}{2 (x+1)}+\frac {85}{2 (2 x+3)}-\frac {13}{8 (x+1)^2}+\frac {41}{4 (2 x+3)^2}+\frac {7}{48 (x+1)^3}+\frac {8}{3 (2 x+3)^3}-\frac {1}{128 (x+1)^4}+\frac {1}{2 (2 x+3)^4}+\frac {155}{2} \log (x+1)-\frac {155}{2} \log (2 x+3)\right )\) |
32*(-1/128*1/(1 + x)^4 + 7/(48*(1 + x)^3) - 13/(8*(1 + x)^2) + 35/(2*(1 + x)) + 1/(2*(3 + 2*x)^4) + 8/(3*(3 + 2*x)^3) + 41/(4*(3 + 2*x)^2) + 85/(2*( 3 + 2*x)) + (155*Log[1 + x])/2 - (155*Log[3 + 2*x])/2)
3.3.5.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63
method | result | size |
norman | \(\frac {19840 x^{7}+173600 x^{6}+1624648 x^{3}+\frac {1428116}{3} x +\frac {1939360}{3} x^{5}+\frac {3552290}{3} x^{2}+\frac {3983500}{3} x^{4}+\frac {325799}{4}}{\left (2 x^{2}+5 x +3\right )^{4}}+2480 \ln \left (1+x \right )-2480 \ln \left (3+2 x \right )\) | \(64\) |
risch | \(\frac {19840 x^{7}+173600 x^{6}+1624648 x^{3}+\frac {1428116}{3} x +\frac {1939360}{3} x^{5}+\frac {3552290}{3} x^{2}+\frac {3983500}{3} x^{4}+\frac {325799}{4}}{\left (2 x^{2}+5 x +3\right )^{4}}+2480 \ln \left (1+x \right )-2480 \ln \left (3+2 x \right )\) | \(65\) |
default | \(\frac {16}{\left (3+2 x \right )^{4}}+\frac {256}{3 \left (3+2 x \right )^{3}}+\frac {328}{\left (3+2 x \right )^{2}}+\frac {1360}{3+2 x}-2480 \ln \left (3+2 x \right )-\frac {1}{4 \left (1+x \right )^{4}}+\frac {14}{3 \left (1+x \right )^{3}}-\frac {52}{\left (1+x \right )^{2}}+\frac {560}{1+x}+2480 \ln \left (1+x \right )\) | \(80\) |
parallelrisch | \(\frac {3909588+8332800 x^{6}+22849856 x +31029760 x^{5}+9642240 \ln \left (1+x \right )+63736000 x^{4}+77983104 x^{3}+56836640 x^{2}+952320 x^{7}+82851840 \ln \left (1+x \right ) x^{6}+204748800 \ln \left (1+x \right ) x^{5}+1904640 \ln \left (1+x \right ) x^{8}+19046400 \ln \left (1+x \right ) x^{7}-1904640 \ln \left (\frac {3}{2}+x \right ) x^{8}-19046400 \ln \left (\frac {3}{2}+x \right ) x^{7}-82851840 \ln \left (\frac {3}{2}+x \right ) x^{6}-204748800 \ln \left (\frac {3}{2}+x \right ) x^{5}-314384640 \ln \left (\frac {3}{2}+x \right ) x^{4}-307123200 \ln \left (\frac {3}{2}+x \right ) x^{3}-186416640 \ln \left (\frac {3}{2}+x \right ) x^{2}-64281600 \ln \left (\frac {3}{2}+x \right ) x -9642240 \ln \left (\frac {3}{2}+x \right )+314384640 \ln \left (1+x \right ) x^{4}+307123200 \ln \left (1+x \right ) x^{3}+186416640 \ln \left (1+x \right ) x^{2}+64281600 \ln \left (1+x \right ) x}{48 \left (2 x^{2}+5 x +3\right )^{4}}\) | \(202\) |
(19840*x^7+173600*x^6+1624648*x^3+1428116/3*x+1939360/3*x^5+3552290/3*x^2+ 3983500/3*x^4+325799/4)/(2*x^2+5*x+3)^4+2480*ln(1+x)-2480*ln(3+2*x)
Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.70 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} - 29760 \, {\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )} \log \left (2 \, x + 3\right ) + 29760 \, {\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )} \log \left (x + 1\right ) + 5712464 \, x + 977397}{12 \, {\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )}} \]
1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 - 29760*(16*x^8 + 160*x^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)*log(2*x + 3) + 29760*(16*x^8 + 160*x^7 + 696*x^6 + 1720*x^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)*log(x + 1) + 5712464*x + 977397)/(16*x^8 + 160*x^7 + 696*x^6 + 1720*x^5 + 2641* x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81)
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {238080 x^{7} + 2083200 x^{6} + 7757440 x^{5} + 15934000 x^{4} + 19495776 x^{3} + 14209160 x^{2} + 5712464 x + 977397}{192 x^{8} + 1920 x^{7} + 8352 x^{6} + 20640 x^{5} + 31692 x^{4} + 30960 x^{3} + 18792 x^{2} + 6480 x + 972} + 2480 \log {\left (x + 1 \right )} - 2480 \log {\left (x + \frac {3}{2} \right )} \]
(238080*x**7 + 2083200*x**6 + 7757440*x**5 + 15934000*x**4 + 19495776*x**3 + 14209160*x**2 + 5712464*x + 977397)/(192*x**8 + 1920*x**7 + 8352*x**6 + 20640*x**5 + 31692*x**4 + 30960*x**3 + 18792*x**2 + 6480*x + 972) + 2480* log(x + 1) - 2480*log(x + 3/2)
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} + 5712464 \, x + 977397}{12 \, {\left (16 \, x^{8} + 160 \, x^{7} + 696 \, x^{6} + 1720 \, x^{5} + 2641 \, x^{4} + 2580 \, x^{3} + 1566 \, x^{2} + 540 \, x + 81\right )}} - 2480 \, \log \left (2 \, x + 3\right ) + 2480 \, \log \left (x + 1\right ) \]
1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 + 5712464*x + 977397)/(16*x^8 + 160*x^7 + 696*x^6 + 1720*x ^5 + 2641*x^4 + 2580*x^3 + 1566*x^2 + 540*x + 81) - 2480*log(2*x + 3) + 24 80*log(x + 1)
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {238080 \, x^{7} + 2083200 \, x^{6} + 7757440 \, x^{5} + 15934000 \, x^{4} + 19495776 \, x^{3} + 14209160 \, x^{2} + 5712464 \, x + 977397}{12 \, {\left (2 \, x^{2} + 5 \, x + 3\right )}^{4}} - 2480 \, \log \left ({\left | 2 \, x + 3 \right |}\right ) + 2480 \, \log \left ({\left | x + 1 \right |}\right ) \]
1/12*(238080*x^7 + 2083200*x^6 + 7757440*x^5 + 15934000*x^4 + 19495776*x^3 + 14209160*x^2 + 5712464*x + 977397)/(2*x^2 + 5*x + 3)^4 - 2480*log(abs(2 *x + 3)) + 2480*log(abs(x + 1))
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int \frac {(1+2 x)^2}{\left (3+5 x+2 x^2\right )^5} \, dx=\frac {1240\,x^7+10850\,x^6+\frac {121210\,x^5}{3}+\frac {995875\,x^4}{12}+\frac {203081\,x^3}{2}+\frac {1776145\,x^2}{24}+\frac {357029\,x}{12}+\frac {325799}{64}}{x^8+10\,x^7+\frac {87\,x^6}{2}+\frac {215\,x^5}{2}+\frac {2641\,x^4}{16}+\frac {645\,x^3}{4}+\frac {783\,x^2}{8}+\frac {135\,x}{4}+\frac {81}{16}}-4960\,\mathrm {atanh}\left (4\,x+5\right ) \]