Integrand size = 25, antiderivative size = 55 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {2099601}{128 (5-2 x)}+\frac {239017 x}{64}+\frac {1963 x^2}{4}+\frac {3343 x^3}{48}+\frac {123 x^4}{16}+\frac {9 x^5}{20}+\frac {807093}{64} \log (5-2 x) \] Output:
2099601/(640-256*x)+239017/64*x+1963/4*x^2+3343/48*x^3+123/16*x^4+9/20*x^5 +807093/64*ln(5-2*x)
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {202667845-177967450 x+19259640 x^2+2431760 x^3+387280 x^4+50400 x^5+3456 x^6+48425580 (-5+2 x) \log (5-2 x)}{3840 (-5+2 x)} \] Input:
Integrate[((4 + 3*x)^2*(2 + 3*x + x^2)^2)/(5 - 2*x)^2,x]
Output:
(202667845 - 177967450*x + 19259640*x^2 + 2431760*x^3 + 387280*x^4 + 50400 *x^5 + 3456*x^6 + 48425580*(-5 + 2*x)*Log[5 - 2*x])/(3840*(-5 + 2*x))
Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1195, 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.
\(\displaystyle \int \frac {(3 x+4)^2 \left (x^2+3 x+2\right )^2}{(5-2 x)^2} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {9 x^4}{4}+\frac {123 x^3}{4}+\frac {3343 x^2}{16}+\frac {1963 x}{2}+\frac {807093}{32 (2 x-5)}+\frac {2099601}{64 (2 x-5)^2}+\frac {239017}{64}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 x^5}{20}+\frac {123 x^4}{16}+\frac {3343 x^3}{48}+\frac {1963 x^2}{4}+\frac {239017 x}{64}+\frac {2099601}{128 (5-2 x)}+\frac {807093}{64} \log (5-2 x)\) |
Input:
Int[((4 + 3*x)^2*(2 + 3*x + x^2)^2)/(5 - 2*x)^2,x]
Output:
2099601/(128*(5 - 2*x)) + (239017*x)/64 + (1963*x^2)/4 + (3343*x^3)/48 + ( 123*x^4)/16 + (9*x^5)/20 + (807093*Log[5 - 2*x])/64
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 1.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {9 x^{5}}{20}+\frac {123 x^{4}}{16}+\frac {3343 x^{3}}{48}+\frac {1963 x^{2}}{4}+\frac {239017 x}{64}-\frac {2099601}{256 \left (x -\frac {5}{2}\right )}+\frac {807093 \ln \left (-5+2 x \right )}{64}\) | \(40\) |
default | \(\frac {9 x^{5}}{20}+\frac {123 x^{4}}{16}+\frac {3343 x^{3}}{48}+\frac {1963 x^{2}}{4}+\frac {239017 x}{64}-\frac {2099601}{128 \left (-5+2 x \right )}+\frac {807093 \ln \left (-5+2 x \right )}{64}\) | \(42\) |
norman | \(\frac {\frac {160497}{32} x^{2}+\frac {30397}{48} x^{3}+\frac {4841}{48} x^{4}+\frac {105}{8} x^{5}+\frac {9}{10} x^{6}-\frac {4037513}{64}}{-5+2 x}+\frac {807093 \ln \left (-5+2 x \right )}{64}\) | \(45\) |
parallelrisch | \(\frac {864 x^{6}+12600 x^{5}+96820 x^{4}+607940 x^{3}+24212790 \ln \left (x -\frac {5}{2}\right ) x +4814910 x^{2}-60562695-60531975 \ln \left (x -\frac {5}{2}\right )}{-4800+1920 x}\) | \(50\) |
meijerg | \(\frac {784 x}{25 \left (1-\frac {2 x}{5}\right )}+\frac {133 x \left (-\frac {6 x}{5}+6\right )}{3 \left (1-\frac {2 x}{5}\right )}+\frac {807093 \ln \left (1-\frac {2 x}{5}\right )}{64}+\frac {645 x \left (-\frac {8}{25} x^{2}-\frac {12}{5} x +12\right )}{8 \left (1-\frac {2 x}{5}\right )}+\frac {1385 x \left (-\frac {8}{25} x^{3}-\frac {8}{5} x^{2}-12 x +60\right )}{48 \left (1-\frac {2 x}{5}\right )}+\frac {1625 x \left (-\frac {48}{625} x^{4}-\frac {8}{25} x^{3}-\frac {8}{5} x^{2}-12 x +60\right )}{64 \left (1-\frac {2 x}{5}\right )}+\frac {1125 x \left (-\frac {448}{3125} x^{5}-\frac {336}{625} x^{4}-\frac {56}{25} x^{3}-\frac {56}{5} x^{2}-84 x +420\right )}{896 \left (1-\frac {2 x}{5}\right )}\) | \(145\) |
Input:
int((3*x+4)^2*(x^2+3*x+2)^2/(5-2*x)^2,x,method=_RETURNVERBOSE)
Output:
9/20*x^5+123/16*x^4+3343/48*x^3+1963/4*x^2+239017/64*x-2099601/256/(x-5/2) +807093/64*ln(-5+2*x)
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {1728 \, x^{6} + 25200 \, x^{5} + 193640 \, x^{4} + 1215880 \, x^{3} + 9629820 \, x^{2} + 24212790 \, {\left (2 \, x - 5\right )} \log \left (2 \, x - 5\right ) - 35852550 \, x - 31494015}{1920 \, {\left (2 \, x - 5\right )}} \] Input:
integrate((4+3*x)^2*(x^2+3*x+2)^2/(5-2*x)^2,x, algorithm="fricas")
Output:
1/1920*(1728*x^6 + 25200*x^5 + 193640*x^4 + 1215880*x^3 + 9629820*x^2 + 24 212790*(2*x - 5)*log(2*x - 5) - 35852550*x - 31494015)/(2*x - 5)
Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {9 x^{5}}{20} + \frac {123 x^{4}}{16} + \frac {3343 x^{3}}{48} + \frac {1963 x^{2}}{4} + \frac {239017 x}{64} + \frac {807093 \log {\left (2 x - 5 \right )}}{64} - \frac {2099601}{256 x - 640} \] Input:
integrate((4+3*x)**2*(x**2+3*x+2)**2/(5-2*x)**2,x)
Output:
9*x**5/20 + 123*x**4/16 + 3343*x**3/48 + 1963*x**2/4 + 239017*x/64 + 80709 3*log(2*x - 5)/64 - 2099601/(256*x - 640)
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.75 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {9}{20} \, x^{5} + \frac {123}{16} \, x^{4} + \frac {3343}{48} \, x^{3} + \frac {1963}{4} \, x^{2} + \frac {239017}{64} \, x - \frac {2099601}{128 \, {\left (2 \, x - 5\right )}} + \frac {807093}{64} \, \log \left (2 \, x - 5\right ) \] Input:
integrate((4+3*x)^2*(x^2+3*x+2)^2/(5-2*x)^2,x, algorithm="maxima")
Output:
9/20*x^5 + 123/16*x^4 + 3343/48*x^3 + 1963/4*x^2 + 239017/64*x - 2099601/1 28/(2*x - 5) + 807093/64*log(2*x - 5)
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {1}{3840} \, {\left (2 \, x - 5\right )}^{5} {\left (\frac {3195}{2 \, x - 5} + \frac {83830}{{\left (2 \, x - 5\right )}^{2}} + \frac {1316820}{{\left (2 \, x - 5\right )}^{3}} + \frac {15480210}{{\left (2 \, x - 5\right )}^{4}} + 54\right )} - \frac {2099601}{128 \, {\left (2 \, x - 5\right )}} - \frac {807093}{64} \, \log \left (\frac {{\left | 2 \, x - 5 \right |}}{2 \, {\left (2 \, x - 5\right )}^{2}}\right ) \] Input:
integrate((4+3*x)^2*(x^2+3*x+2)^2/(5-2*x)^2,x, algorithm="giac")
Output:
1/3840*(2*x - 5)^5*(3195/(2*x - 5) + 83830/(2*x - 5)^2 + 1316820/(2*x - 5) ^3 + 15480210/(2*x - 5)^4 + 54) - 2099601/128/(2*x - 5) - 807093/64*log(1/ 2*abs(2*x - 5)/(2*x - 5)^2)
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {239017\,x}{64}+\frac {807093\,\ln \left (x-\frac {5}{2}\right )}{64}-\frac {2099601}{256\,\left (x-\frac {5}{2}\right )}+\frac {1963\,x^2}{4}+\frac {3343\,x^3}{48}+\frac {123\,x^4}{16}+\frac {9\,x^5}{20} \] Input:
int(((3*x + 4)^2*(3*x + x^2 + 2)^2)/(2*x - 5)^2,x)
Output:
(239017*x)/64 + (807093*log(x - 5/2))/64 - 2099601/(256*(x - 5/2)) + (1963 *x^2)/4 + (3343*x^3)/48 + (123*x^4)/16 + (9*x^5)/20
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {(4+3 x)^2 \left (2+3 x+x^2\right )^2}{(5-2 x)^2} \, dx=\frac {24212790 \,\mathrm {log}\left (2 x -5\right ) x -60531975 \,\mathrm {log}\left (2 x -5\right )+864 x^{6}+12600 x^{5}+96820 x^{4}+607940 x^{3}+4814910 x^{2}-24225078 x}{1920 x -4800} \] Input:
int((4+3*x)^2*(x^2+3*x+2)^2/(5-2*x)^2,x)
Output:
(24212790*log(2*x - 5)*x - 60531975*log(2*x - 5) + 864*x**6 + 12600*x**5 + 96820*x**4 + 607940*x**3 + 4814910*x**2 - 24225078*x)/(960*(2*x - 5))