Integrand size = 25, antiderivative size = 94 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=\frac {12946}{125 (3+2 x)}-\frac {3 (37+47 x)}{10 (3+2 x) \left (2+5 x+3 x^2\right )^2}+\frac {9293+10848 x}{50 (3+2 x) \left (2+5 x+3 x^2\right )}-175 \log (1+x)+\frac {4912}{625} \log (3+2 x)+\frac {104463}{625} \log (2+3 x) \]
12946/125/(3+2*x)-3/10*(37+47*x)/(3+2*x)/(3*x^2+5*x+2)^2+1/50*(9293+10848* x)/(3+2*x)/(3*x^2+5*x+2)-175*ln(1+x)+4912/625*ln(3+2*x)+104463/625*ln(2+3* x)
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=\frac {1}{625} \left (-\frac {1040}{3+2 x}-\frac {75 (151+201 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {5 (33697+39462 x)}{4+10 x+6 x^2}+104463 \log (-4-6 x)-109375 \log (-2 (1+x))+4912 \log (3+2 x)\right ) \]
(-1040/(3 + 2*x) - (75*(151 + 201*x))/(2*(2 + 5*x + 3*x^2)^2) + (5*(33697 + 39462*x))/(4 + 10*x + 6*x^2) + 104463*Log[-4 - 6*x] - 109375*Log[-2*(1 + x)] + 4912*Log[3 + 2*x])/625
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1207, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3)^2 \left (3 x^2+5 x+2\right )^3} \, dx\) |
\(\Big \downarrow \) 1207 |
\(\displaystyle 27 \int \left (\frac {9824}{16875 (2 x+3)}+\frac {11607}{625 (3 x+2)}+\frac {416}{3375 (2 x+3)^2}-\frac {984}{125 (3 x+2)^2}+\frac {51}{25 (3 x+2)^3}-\frac {175}{27 (x+1)}-\frac {29}{27 (x+1)^2}-\frac {2}{9 (x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 27 \left (\frac {29}{27 (x+1)}-\frac {208}{3375 (2 x+3)}+\frac {328}{125 (3 x+2)}+\frac {1}{9 (x+1)^2}-\frac {17}{50 (3 x+2)^2}-\frac {175}{27} \log (x+1)+\frac {4912 \log (2 x+3)}{16875}+\frac {3869}{625} \log (3 x+2)\right )\) |
27*(1/(9*(1 + x)^2) + 29/(27*(1 + x)) - 208/(3375*(3 + 2*x)) - 17/(50*(2 + 3*x)^2) + 328/(125*(2 + 3*x)) - (175*Log[1 + x])/27 + (4912*Log[3 + 2*x]) /16875 + (3869*Log[2 + 3*x])/625)
3.25.3.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((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*(f + g*x)^n*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[p, -1] && IntegersQ[m, n] && NiceSqrtQ[b^2 - 4* a*c]
Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68
method | result | size |
norman | \(\frac {\frac {85512}{25} x +\frac {93948}{25} x^{3}+\frac {116514}{125} x^{4}+\frac {1368599}{250} x^{2}+\frac {193723}{250}}{\left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )^{2}}+\frac {4912 \ln \left (3+2 x \right )}{625}-175 \ln \left (1+x \right )+\frac {104463 \ln \left (2+3 x \right )}{625}\) | \(64\) |
default | \(-\frac {459}{50 \left (2+3 x \right )^{2}}+\frac {8856}{125 \left (2+3 x \right )}+\frac {104463 \ln \left (2+3 x \right )}{625}-\frac {208}{125 \left (3+2 x \right )}+\frac {4912 \ln \left (3+2 x \right )}{625}+\frac {3}{\left (1+x \right )^{2}}+\frac {29}{1+x}-175 \ln \left (1+x \right )\) | \(65\) |
risch | \(\frac {\frac {85512}{25} x +\frac {93948}{25} x^{3}+\frac {116514}{125} x^{4}+\frac {1368599}{250} x^{2}+\frac {193723}{250}}{\left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )^{2}}+\frac {4912 \ln \left (3+2 x \right )}{625}-175 \ln \left (1+x \right )+\frac {104463 \ln \left (2+3 x \right )}{625}\) | \(65\) |
parallelrisch | \(-\frac {14558620 x +31500000 \ln \left (1+x \right )+64144925 x^{2}+17435070 x^{5}+70287825 x^{4}+102484060 x^{3}-1414656 \ln \left (x +\frac {3}{2}\right )-30085344 \ln \left (x +\frac {2}{3}\right )+396375000 \ln \left (1+x \right ) x^{2}-17801088 \ln \left (x +\frac {3}{2}\right ) x^{2}-378573912 \ln \left (x +\frac {2}{3}\right ) x^{2}+47250000 \ln \left (1+x \right ) x^{5}-2121984 \ln \left (x +\frac {3}{2}\right ) x^{5}-45128016 \ln \left (x +\frac {2}{3}\right ) x^{5}+430500000 \ln \left (1+x \right ) x^{3}-19333632 \ln \left (x +\frac {3}{2}\right ) x^{3}-411166368 \ln \left (x +\frac {2}{3}\right ) x^{3}+228375000 \ln \left (1+x \right ) x^{4}-10256256 \ln \left (x +\frac {3}{2}\right ) x^{4}-218118744 \ln \left (x +\frac {2}{3}\right ) x^{4}+178500000 \ln \left (1+x \right ) x -8016384 \ln \left (x +\frac {3}{2}\right ) x -170483616 \ln \left (x +\frac {2}{3}\right ) x}{15000 \left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )^{2}}\) | \(193\) |
(85512/25*x+93948/25*x^3+116514/125*x^4+1368599/250*x^2+193723/250)/(3+2*x )/(3*x^2+5*x+2)^2+4912/625*ln(3+2*x)-175*ln(1+x)+104463/625*ln(2+3*x)
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.55 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=\frac {1165140 \, x^{4} + 4697400 \, x^{3} + 6842995 \, x^{2} + 208926 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (3 \, x + 2\right ) + 9824 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (2 \, x + 3\right ) - 218750 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )} \log \left (x + 1\right ) + 4275600 \, x + 968615}{1250 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )}} \]
1/1250*(1165140*x^4 + 4697400*x^3 + 6842995*x^2 + 208926*(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12)*log(3*x + 2) + 9824*(18*x^5 + 87*x^4 + 16 4*x^3 + 151*x^2 + 68*x + 12)*log(2*x + 3) - 218750*(18*x^5 + 87*x^4 + 164* x^3 + 151*x^2 + 68*x + 12)*log(x + 1) + 4275600*x + 968615)/(18*x^5 + 87*x ^4 + 164*x^3 + 151*x^2 + 68*x + 12)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=- \frac {- 233028 x^{4} - 939480 x^{3} - 1368599 x^{2} - 855120 x - 193723}{4500 x^{5} + 21750 x^{4} + 41000 x^{3} + 37750 x^{2} + 17000 x + 3000} + \frac {104463 \log {\left (x + \frac {2}{3} \right )}}{625} - 175 \log {\left (x + 1 \right )} + \frac {4912 \log {\left (x + \frac {3}{2} \right )}}{625} \]
-(-233028*x**4 - 939480*x**3 - 1368599*x**2 - 855120*x - 193723)/(4500*x** 5 + 21750*x**4 + 41000*x**3 + 37750*x**2 + 17000*x + 3000) + 104463*log(x + 2/3)/625 - 175*log(x + 1) + 4912*log(x + 3/2)/625
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=\frac {233028 \, x^{4} + 939480 \, x^{3} + 1368599 \, x^{2} + 855120 \, x + 193723}{250 \, {\left (18 \, x^{5} + 87 \, x^{4} + 164 \, x^{3} + 151 \, x^{2} + 68 \, x + 12\right )}} + \frac {104463}{625} \, \log \left (3 \, x + 2\right ) + \frac {4912}{625} \, \log \left (2 \, x + 3\right ) - 175 \, \log \left (x + 1\right ) \]
1/250*(233028*x^4 + 939480*x^3 + 1368599*x^2 + 855120*x + 193723)/(18*x^5 + 87*x^4 + 164*x^3 + 151*x^2 + 68*x + 12) + 104463/625*log(3*x + 2) + 4912 /625*log(2*x + 3) - 175*log(x + 1)
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=-\frac {208}{125 \, {\left (2 \, x + 3\right )}} - \frac {2 \, {\left (\frac {168231}{2 \, x + 3} - \frac {211036}{{\left (2 \, x + 3\right )}^{2}} + \frac {82447}{{\left (2 \, x + 3\right )}^{3}} - 42642\right )}}{125 \, {\left (\frac {5}{2 \, x + 3} - 3\right )}^{2} {\left (\frac {1}{2 \, x + 3} - 1\right )}^{2}} - 175 \, \log \left ({\left | -\frac {1}{2 \, x + 3} + 1 \right |}\right ) + \frac {104463}{625} \, \log \left ({\left | -\frac {5}{2 \, x + 3} + 3 \right |}\right ) \]
-208/125/(2*x + 3) - 2/125*(168231/(2*x + 3) - 211036/(2*x + 3)^2 + 82447/ (2*x + 3)^3 - 42642)/((5/(2*x + 3) - 3)^2*(1/(2*x + 3) - 1)^2) - 175*log(a bs(-1/(2*x + 3) + 1)) + 104463/625*log(abs(-5/(2*x + 3) + 3))
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69 \[ \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^3} \, dx=\frac {104463\,\ln \left (x+\frac {2}{3}\right )}{625}-175\,\ln \left (x+1\right )+\frac {4912\,\ln \left (x+\frac {3}{2}\right )}{625}+\frac {\frac {6473\,x^4}{125}+\frac {15658\,x^3}{75}+\frac {1368599\,x^2}{4500}+\frac {14252\,x}{75}+\frac {193723}{4500}}{x^5+\frac {29\,x^4}{6}+\frac {82\,x^3}{9}+\frac {151\,x^2}{18}+\frac {34\,x}{9}+\frac {2}{3}} \]