Integrand size = 25, antiderivative size = 88 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {1258}{75 (3+2 x)^3}-\frac {2212}{125 (3+2 x)^2}-\frac {16522}{625 (3+2 x)}-\frac {3 (37+47 x)}{5 (3+2 x)^3 \left (2+5 x+3 x^2\right )}-13 \log (1+x)+\frac {65816 \log (3+2 x)}{3125}-\frac {25191 \log (2+3 x)}{3125} \]
-1258/75/(3+2*x)^3-2212/125/(3+2*x)^2-16522/625/(3+2*x)-3/5*(37+47*x)/(3+2 *x)^3/(3*x^2+5*x+2)-13*ln(1+x)+65816/3125*ln(3+2*x)-25191/3125*ln(2+3*x)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=\frac {-\frac {6500}{(3+2 x)^3}-\frac {30450}{(3+2 x)^2}-\frac {121560}{3+2 x}-\frac {45 (2959+4209 x)}{2+5 x+3 x^2}-75573 \log (-4-6 x)-121875 \log (-2 (1+x))+197448 \log (3+2 x)}{9375} \]
(-6500/(3 + 2*x)^3 - 30450/(3 + 2*x)^2 - 121560/(3 + 2*x) - (45*(2959 + 42 09*x))/(2 + 5*x + 3*x^2) - 75573*Log[-4 - 6*x] - 121875*Log[-2*(1 + x)] + 197448*Log[3 + 2*x])/9375
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3)^4 \left (3 x^2+5 x+2\right )^2} \, dx\) |
\(\Big \downarrow \) 1207 |
\(\displaystyle 9 \int \left (\frac {131632}{28125 (2 x+3)}-\frac {8397}{3125 (3 x+2)}+\frac {16208}{5625 (2 x+3)^2}+\frac {459}{625 (3 x+2)^2}+\frac {1624}{1125 (2 x+3)^3}+\frac {104}{225 (2 x+3)^4}-\frac {13}{9 (x+1)}+\frac {2}{3 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 \left (-\frac {2}{3 (x+1)}-\frac {8104}{5625 (2 x+3)}-\frac {153}{625 (3 x+2)}-\frac {406}{1125 (2 x+3)^2}-\frac {52}{675 (2 x+3)^3}-\frac {13}{9} \log (x+1)+\frac {65816 \log (2 x+3)}{28125}-\frac {2799 \log (3 x+2)}{3125}\right )\) |
9*(-2/(3*(1 + x)) - 52/(675*(3 + 2*x)^3) - 406/(1125*(3 + 2*x)^2) - 8104/( 5625*(3 + 2*x)) - 153/(625*(2 + 3*x)) - (13*Log[1 + x])/9 + (65816*Log[3 + 2*x])/28125 - (2799*Log[2 + 3*x])/3125)
3.24.95.3.1 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 = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.73
method | result | size |
norman | \(\frac {-\frac {4260599}{1875} x -\frac {991592}{625} x^{3}-\frac {362236}{125} x^{2}-\frac {198264}{625} x^{4}-\frac {1195793}{1875}}{\left (3+2 x \right )^{3} \left (3 x^{2}+5 x +2\right )}+\frac {65816 \ln \left (3+2 x \right )}{3125}-13 \ln \left (1+x \right )-\frac {25191 \ln \left (2+3 x \right )}{3125}\) | \(64\) |
risch | \(\frac {-\frac {4260599}{1875} x -\frac {991592}{625} x^{3}-\frac {362236}{125} x^{2}-\frac {198264}{625} x^{4}-\frac {1195793}{1875}}{\left (3+2 x \right )^{3} \left (3 x^{2}+5 x +2\right )}+\frac {65816 \ln \left (3+2 x \right )}{3125}-13 \ln \left (1+x \right )-\frac {25191 \ln \left (2+3 x \right )}{3125}\) | \(65\) |
default | \(-\frac {1377}{625 \left (2+3 x \right )}-\frac {25191 \ln \left (2+3 x \right )}{3125}-\frac {52}{75 \left (3+2 x \right )^{3}}-\frac {406}{125 \left (3+2 x \right )^{2}}-\frac {8104}{625 \left (3+2 x \right )}+\frac {65816 \ln \left (3+2 x \right )}{3125}-\frac {6}{1+x}-13 \ln \left (1+x \right )\) | \(67\) |
parallelrisch | \(-\frac {47831720+170423960 x +52650000 \ln \left (1+x \right )+217341600 x^{2}+23791680 x^{4}+118991040 x^{3}-85297536 \ln \left (x +\frac {3}{2}\right )+32647536 \ln \left (x +\frac {2}{3}\right )+412425000 \ln \left (1+x \right ) x^{2}-668164032 \ln \left (x +\frac {3}{2}\right ) x^{2}+255739032 \ln \left (x +\frac {2}{3}\right ) x^{2}+23400000 \ln \left (1+x \right ) x^{5}-37910016 \ln \left (x +\frac {3}{2}\right ) x^{5}+14510016 \ln \left (x +\frac {2}{3}\right ) x^{5}+349050000 \ln \left (1+x \right ) x^{3}-565491072 \ln \left (x +\frac {3}{2}\right ) x^{3}+216441072 \ln \left (x +\frac {2}{3}\right ) x^{3}+144300000 \ln \left (1+x \right ) x^{4}-233778432 \ln \left (x +\frac {3}{2}\right ) x^{4}+89478432 \ln \left (x +\frac {2}{3}\right ) x^{4}+236925000 \ln \left (1+x \right ) x -383838912 \ln \left (x +\frac {3}{2}\right ) x +146913912 \ln \left (x +\frac {2}{3}\right ) x}{75000 \left (3+2 x \right )^{3} \left (3 x^{2}+5 x +2\right )}\) | \(189\) |
(-4260599/1875*x-991592/625*x^3-362236/125*x^2-198264/625*x^4-1195793/1875 )/(3+2*x)^3/(3*x^2+5*x+2)+65816/3125*ln(3+2*x)-13*ln(1+x)-25191/3125*ln(2+ 3*x)
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.66 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {2973960 \, x^{4} + 14873880 \, x^{3} + 27167700 \, x^{2} + 75573 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (3 \, x + 2\right ) - 197448 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (2 \, x + 3\right ) + 121875 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (x + 1\right ) + 21302995 \, x + 5978965}{9375 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )}} \]
-1/9375*(2973960*x^4 + 14873880*x^3 + 27167700*x^2 + 75573*(24*x^5 + 148*x ^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(3*x + 2) - 197448*(24*x^5 + 148*x ^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(2*x + 3) + 121875*(24*x^5 + 148*x ^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(x + 1) + 21302995*x + 5978965)/(2 4*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=- \frac {594792 x^{4} + 2974776 x^{3} + 5433540 x^{2} + 4260599 x + 1195793}{45000 x^{5} + 277500 x^{4} + 671250 x^{3} + 793125 x^{2} + 455625 x + 101250} - \frac {25191 \log {\left (x + \frac {2}{3} \right )}}{3125} - 13 \log {\left (x + 1 \right )} + \frac {65816 \log {\left (x + \frac {3}{2} \right )}}{3125} \]
-(594792*x**4 + 2974776*x**3 + 5433540*x**2 + 4260599*x + 1195793)/(45000* x**5 + 277500*x**4 + 671250*x**3 + 793125*x**2 + 455625*x + 101250) - 2519 1*log(x + 2/3)/3125 - 13*log(x + 1) + 65816*log(x + 3/2)/3125
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {594792 \, x^{4} + 2974776 \, x^{3} + 5433540 \, x^{2} + 4260599 \, x + 1195793}{1875 \, {\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )}} - \frac {25191}{3125} \, \log \left (3 \, x + 2\right ) + \frac {65816}{3125} \, \log \left (2 \, x + 3\right ) - 13 \, \log \left (x + 1\right ) \]
-1/1875*(594792*x^4 + 2974776*x^3 + 5433540*x^2 + 4260599*x + 1195793)/(24 *x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54) - 25191/3125*log(3*x + 2) + 65816/3125*log(2*x + 3) - 13*log(x + 1)
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {594792 \, x^{4} + 2974776 \, x^{3} + 5433540 \, x^{2} + 4260599 \, x + 1195793}{1875 \, {\left (3 \, x + 2\right )} {\left (2 \, x + 3\right )}^{3} {\left (x + 1\right )}} - \frac {25191}{3125} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {65816}{3125} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - 13 \, \log \left ({\left | x + 1 \right |}\right ) \]
-1/1875*(594792*x^4 + 2974776*x^3 + 5433540*x^2 + 4260599*x + 1195793)/((3 *x + 2)*(2*x + 3)^3*(x + 1)) - 25191/3125*log(abs(3*x + 2)) + 65816/3125*l og(abs(2*x + 3)) - 13*log(abs(x + 1))
Time = 11.75 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {5-x}{(3+2 x)^4 \left (2+5 x+3 x^2\right )^2} \, dx=\frac {65816\,\ln \left (x+\frac {3}{2}\right )}{3125}-\frac {25191\,\ln \left (x+\frac {2}{3}\right )}{3125}-13\,\ln \left (x+1\right )-\frac {\frac {8261\,x^4}{625}+\frac {123949\,x^3}{1875}+\frac {90559\,x^2}{750}+\frac {4260599\,x}{45000}+\frac {1195793}{45000}}{x^5+\frac {37\,x^4}{6}+\frac {179\,x^3}{12}+\frac {141\,x^2}{8}+\frac {81\,x}{8}+\frac {9}{4}} \]