Integrand size = 25, antiderivative size = 63 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {3}{(1+x)^2}+\frac {41}{1+x}-\frac {153}{10 (2+3 x)^2}+\frac {2646}{25 (2+3 x)}-233 \log (1+x)+\frac {208}{125} \log (3+2 x)+\frac {28917}{125} \log (2+3 x) \] Output:
3/(1+x)^2+41/(1+x)-153/10/(2+3*x)^2+2646/(50+75*x)-233*ln(1+x)+208/125*ln( 3+2*x)+28917/125*ln(2+3*x)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {1}{125} \left (-\frac {75 (37+47 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {47935+57210 x}{4+10 x+6 x^2}+28917 \log (-4-6 x)-29125 \log (-2 (1+x))+208 \log (3+2 x)\right ) \] Input:
Integrate[(5 - x)/((3 + 2*x)*(2 + 5*x + 3*x^2)^3),x]
Output:
((-75*(37 + 47*x))/(2*(2 + 5*x + 3*x^2)^2) + (47935 + 57210*x)/(4 + 10*x + 6*x^2) + 28917*Log[-4 - 6*x] - 29125*Log[-2*(1 + x)] + 208*Log[3 + 2*x])/ 125
Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3) \left (3 x^2+5 x+2\right )^3} \, dx\) |
\(\Big \downarrow \) 1207 |
\(\displaystyle 27 \int \left (\frac {416}{3375 (2 x+3)}+\frac {3213}{125 (3 x+2)}-\frac {294}{25 (3 x+2)^2}+\frac {17}{5 (3 x+2)^3}-\frac {233}{27 (x+1)}-\frac {41}{27 (x+1)^2}-\frac {2}{9 (x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 27 \left (\frac {41}{27 (x+1)}+\frac {98}{25 (3 x+2)}+\frac {1}{9 (x+1)^2}-\frac {17}{30 (3 x+2)^2}-\frac {233}{27} \log (x+1)+\frac {208 \log (2 x+3)}{3375}+\frac {1071}{125} \log (3 x+2)\right )\) |
Input:
Int[(5 - x)/((3 + 2*x)*(2 + 5*x + 3*x^2)^3),x]
Output:
27*(1/(9*(1 + x)^2) + 41/(27*(1 + x)) - 17/(30*(2 + 3*x)^2) + 98/(25*(2 + 3*x)) - (233*Log[1 + x])/27 + (208*Log[3 + 2*x])/3375 + (1071*Log[2 + 3*x] )/125)
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 = 1.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83
method | result | size |
norman | \(\frac {\frac {17163}{25} x^{3}+\frac {35057}{25} x +\frac {85971}{50} x^{2}+\frac {18619}{50}}{\left (3 x^{2}+5 x +2\right )^{2}}-233 \ln \left (x +1\right )+\frac {208 \ln \left (2 x +3\right )}{125}+\frac {28917 \ln \left (3 x +2\right )}{125}\) | \(52\) |
risch | \(\frac {\frac {17163}{25} x^{3}+\frac {35057}{25} x +\frac {85971}{50} x^{2}+\frac {18619}{50}}{\left (3 x^{2}+5 x +2\right )^{2}}-233 \ln \left (x +1\right )+\frac {208 \ln \left (2 x +3\right )}{125}+\frac {28917 \ln \left (3 x +2\right )}{125}\) | \(53\) |
default | \(-\frac {153}{10 \left (3 x +2\right )^{2}}+\frac {2646}{25 \left (3 x +2\right )}+\frac {28917 \ln \left (3 x +2\right )}{125}+\frac {3}{\left (x +1\right )^{2}}+\frac {41}{x +1}-233 \ln \left (x +1\right )+\frac {208 \ln \left (2 x +3\right )}{125}\) | \(56\) |
parallelrisch | \(-\frac {2097000 \ln \left (x +1\right ) x^{4}-2082024 \ln \left (x +\frac {2}{3}\right ) x^{4}-14976 \ln \left (x +\frac {3}{2}\right ) x^{4}+6990000 \ln \left (x +1\right ) x^{3}-6940080 \ln \left (x +\frac {2}{3}\right ) x^{3}-49920 \ln \left (x +\frac {3}{2}\right ) x^{3}+837855 x^{4}+8621000 \ln \left (x +1\right ) x^{2}-8559432 \ln \left (x +\frac {2}{3}\right ) x^{2}-61568 \ln \left (x +\frac {3}{2}\right ) x^{2}+2106330 x^{3}+4660000 \ln \left (x +1\right ) x -4626720 \ln \left (x +\frac {2}{3}\right ) x -33280 \ln \left (x +\frac {3}{2}\right ) x +1725095 x^{2}+932000 \ln \left (x +1\right )-925344 \ln \left (x +\frac {2}{3}\right )-6656 \ln \left (x +\frac {3}{2}\right )+459620 x}{1000 \left (3 x^{2}+5 x +2\right )^{2}}\) | \(154\) |
Input:
int((5-x)/(2*x+3)/(3*x^2+5*x+2)^3,x,method=_RETURNVERBOSE)
Output:
(17163/25*x^3+35057/25*x+85971/50*x^2+18619/50)/(3*x^2+5*x+2)^2-233*ln(x+1 )+208/125*ln(2*x+3)+28917/125*ln(3*x+2)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (55) = 110\).
Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.92 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {171630 \, x^{3} + 429855 \, x^{2} + 57834 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 416 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (2 \, x + 3\right ) - 58250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 350570 \, x + 93095}{250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+5*x+2)^3,x, algorithm="fricas")
Output:
1/250*(171630*x^3 + 429855*x^2 + 57834*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4 )*log(3*x + 2) + 416*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(2*x + 3) - 5 8250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(x + 1) + 350570*x + 93095)/( 9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=- \frac {- 34326 x^{3} - 85971 x^{2} - 70114 x - 18619}{450 x^{4} + 1500 x^{3} + 1850 x^{2} + 1000 x + 200} + \frac {28917 \log {\left (x + \frac {2}{3} \right )}}{125} - 233 \log {\left (x + 1 \right )} + \frac {208 \log {\left (x + \frac {3}{2} \right )}}{125} \] Input:
integrate((5-x)/(3+2*x)/(3*x**2+5*x+2)**3,x)
Output:
-(-34326*x**3 - 85971*x**2 - 70114*x - 18619)/(450*x**4 + 1500*x**3 + 1850 *x**2 + 1000*x + 200) + 28917*log(x + 2/3)/125 - 233*log(x + 1) + 208*log( x + 3/2)/125
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {34326 \, x^{3} + 85971 \, x^{2} + 70114 \, x + 18619}{50 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + \frac {28917}{125} \, \log \left (3 \, x + 2\right ) + \frac {208}{125} \, \log \left (2 \, x + 3\right ) - 233 \, \log \left (x + 1\right ) \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+5*x+2)^3,x, algorithm="maxima")
Output:
1/50*(34326*x^3 + 85971*x^2 + 70114*x + 18619)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 28917/125*log(3*x + 2) + 208/125*log(2*x + 3) - 233*log(x + 1)
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {34326 \, x^{3} + 85971 \, x^{2} + 70114 \, x + 18619}{50 \, {\left (3 \, x + 2\right )}^{2} {\left (x + 1\right )}^{2}} + \frac {28917}{125} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {208}{125} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - 233 \, \log \left ({\left | x + 1 \right |}\right ) \] Input:
integrate((5-x)/(3+2*x)/(3*x^2+5*x+2)^3,x, algorithm="giac")
Output:
1/50*(34326*x^3 + 85971*x^2 + 70114*x + 18619)/((3*x + 2)^2*(x + 1)^2) + 2 8917/125*log(abs(3*x + 2)) + 208/125*log(abs(2*x + 3)) - 233*log(abs(x + 1 ))
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {28917\,\ln \left (x+\frac {2}{3}\right )}{125}-233\,\ln \left (x+1\right )+\frac {208\,\ln \left (x+\frac {3}{2}\right )}{125}+\frac {\frac {1907\,x^3}{25}+\frac {28657\,x^2}{150}+\frac {35057\,x}{225}+\frac {18619}{450}}{x^4+\frac {10\,x^3}{3}+\frac {37\,x^2}{9}+\frac {20\,x}{9}+\frac {4}{9}} \] Input:
int(-(x - 5)/((2*x + 3)*(5*x + 3*x^2 + 2)^3),x)
Output:
(28917*log(x + 2/3))/125 - 233*log(x + 1) + (208*log(x + 3/2))/125 + ((350 57*x)/225 + (28657*x^2)/150 + (1907*x^3)/25 + 18619/450)/((20*x)/9 + (37*x ^2)/9 + (10*x^3)/3 + x^4 + 4/9)
Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.83 \[ \int \frac {5-x}{(3+2 x) \left (2+5 x+3 x^2\right )^3} \, dx=\frac {520506 \,\mathrm {log}\left (3 x +2\right ) x^{4}+1735020 \,\mathrm {log}\left (3 x +2\right ) x^{3}+2139858 \,\mathrm {log}\left (3 x +2\right ) x^{2}+1156680 \,\mathrm {log}\left (3 x +2\right ) x +231336 \,\mathrm {log}\left (3 x +2\right )+3744 \,\mathrm {log}\left (2 x +3\right ) x^{4}+12480 \,\mathrm {log}\left (2 x +3\right ) x^{3}+15392 \,\mathrm {log}\left (2 x +3\right ) x^{2}+8320 \,\mathrm {log}\left (2 x +3\right ) x +1664 \,\mathrm {log}\left (2 x +3\right )-524250 \,\mathrm {log}\left (x +1\right ) x^{4}-1747500 \,\mathrm {log}\left (x +1\right ) x^{3}-2155250 \,\mathrm {log}\left (x +1\right ) x^{2}-1165000 \,\mathrm {log}\left (x +1\right ) x -233000 \,\mathrm {log}\left (x +1\right )-51489 x^{4}+218178 x^{2}+236150 x +70211}{2250 x^{4}+7500 x^{3}+9250 x^{2}+5000 x +1000} \] Input:
int((5-x)/(3+2*x)/(3*x^2+5*x+2)^3,x)
Output:
(520506*log(3*x + 2)*x**4 + 1735020*log(3*x + 2)*x**3 + 2139858*log(3*x + 2)*x**2 + 1156680*log(3*x + 2)*x + 231336*log(3*x + 2) + 3744*log(2*x + 3) *x**4 + 12480*log(2*x + 3)*x**3 + 15392*log(2*x + 3)*x**2 + 8320*log(2*x + 3)*x + 1664*log(2*x + 3) - 524250*log(x + 1)*x**4 - 1747500*log(x + 1)*x* *3 - 2155250*log(x + 1)*x**2 - 1165000*log(x + 1)*x - 233000*log(x + 1) - 51489*x**4 + 218178*x**2 + 236150*x + 70211)/(250*(9*x**4 + 30*x**3 + 37*x **2 + 20*x + 4))