Integrand size = 22, antiderivative size = 64 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {8}{9317 (1-2 x)}-\frac {25}{242 (3+5 x)^2}+\frac {725}{1331 (3+5 x)}-\frac {1104 \log (1-2 x)}{717409}-\frac {81}{49} \log (2+3 x)+\frac {24225 \log (3+5 x)}{14641} \] Output:
8/(9317-18634*x)-25/242/(3+5*x)^2+725/(3993+6655*x)-1104/717409*ln(1-2*x)- 81/49*ln(2+3*x)+24225/14641*ln(3+5*x)
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {3 \left (\frac {1232}{3-6 x}-\frac {148225}{3 (3+5 x)^2}+\frac {781550}{9+15 x}-736 \log (3-6 x)-790614 \log (2+3 x)+791350 \log (-3 (3+5 x))\right )}{1434818} \] Input:
Integrate[1/((1 - 2*x)^2*(2 + 3*x)*(3 + 5*x)^3),x]
Output:
(3*(1232/(3 - 6*x) - 148225/(3*(3 + 5*x)^2) + 781550/(9 + 15*x) - 736*Log[ 3 - 6*x] - 790614*Log[2 + 3*x] + 791350*Log[-3*(3 + 5*x)]))/1434818
Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^2 (3 x+2) (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {243}{49 (3 x+2)}+\frac {121125}{14641 (5 x+3)}-\frac {3625}{1331 (5 x+3)^2}+\frac {125}{121 (5 x+3)^3}-\frac {2208}{717409 (2 x-1)}+\frac {16}{9317 (2 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{9317 (1-2 x)}+\frac {725}{1331 (5 x+3)}-\frac {25}{242 (5 x+3)^2}-\frac {1104 \log (1-2 x)}{717409}-\frac {81}{49} \log (3 x+2)+\frac {24225 \log (5 x+3)}{14641}\) |
Input:
Int[1/((1 - 2*x)^2*(2 + 3*x)*(3 + 5*x)^3),x]
Output:
8/(9317*(1 - 2*x)) - 25/(242*(3 + 5*x)^2) + 725/(1331*(3 + 5*x)) - (1104*L og[1 - 2*x])/717409 - (81*Log[2 + 3*x])/49 + (24225*Log[3 + 5*x])/14641
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\frac {50550}{9317} x^{2}+\frac {2910}{9317} x -\frac {28669}{18634}}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {1104 \ln \left (-1+2 x \right )}{717409}-\frac {81 \ln \left (2+3 x \right )}{49}+\frac {24225 \ln \left (3+5 x \right )}{14641}\) | \(52\) |
default | \(-\frac {25}{242 \left (3+5 x \right )^{2}}+\frac {725}{1331 \left (3+5 x \right )}+\frac {24225 \ln \left (3+5 x \right )}{14641}-\frac {81 \ln \left (2+3 x \right )}{49}-\frac {8}{9317 \left (-1+2 x \right )}-\frac {1104 \ln \left (-1+2 x \right )}{717409}\) | \(53\) |
norman | \(\frac {-\frac {716725}{83853} x^{3}-\frac {93515}{167706} x^{2}+\frac {66068}{27951} x}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {1104 \ln \left (-1+2 x \right )}{717409}-\frac {81 \ln \left (2+3 x \right )}{49}+\frac {24225 \ln \left (3+5 x \right )}{14641}\) | \(55\) |
parallelrisch | \(-\frac {1067328900 \ln \left (\frac {2}{3}+x \right ) x^{3}+993600 \ln \left (x -\frac {1}{2}\right ) x^{3}-1068322500 \ln \left (x +\frac {3}{5}\right ) x^{3}+747130230 \ln \left (\frac {2}{3}+x \right ) x^{2}+695520 \ln \left (x -\frac {1}{2}\right ) x^{2}-747825750 \ln \left (x +\frac {3}{5}\right ) x^{2}+110375650 x^{3}-256158936 \ln \left (\frac {2}{3}+x \right ) x -238464 \ln \left (x -\frac {1}{2}\right ) x +256397400 \ln \left (x +\frac {3}{5}\right ) x +7200655 x^{2}-192119202 \ln \left (\frac {2}{3}+x \right )-178848 \ln \left (x -\frac {1}{2}\right )+192298050 \ln \left (x +\frac {3}{5}\right )-30523416 x}{12913362 \left (-1+2 x \right ) \left (3+5 x \right )^{2}}\) | \(124\) |
Input:
int(1/(1-2*x)^2/(2+3*x)/(3+5*x)^3,x,method=_RETURNVERBOSE)
Output:
50*(1011/9317*x^2+291/46585*x-28669/931700)/(-1+2*x)/(3+5*x)^2-1104/717409 *ln(-1+2*x)-81/49*ln(2+3*x)+24225/14641*ln(3+5*x)
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {7784700 \, x^{2} + 2374050 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) - 2371842 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (3 \, x + 2\right ) - 2208 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 448140 \, x - 2207513}{1434818 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \] Input:
integrate(1/(1-2*x)^2/(2+3*x)/(3+5*x)^3,x, algorithm="fricas")
Output:
1/1434818*(7784700*x^2 + 2374050*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) - 2371842*(50*x^3 + 35*x^2 - 12*x - 9)*log(3*x + 2) - 2208*(50*x^3 + 35*x ^2 - 12*x - 9)*log(2*x - 1) + 448140*x - 2207513)/(50*x^3 + 35*x^2 - 12*x - 9)
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {101100 x^{2} + 5820 x - 28669}{931700 x^{3} + 652190 x^{2} - 223608 x - 167706} - \frac {1104 \log {\left (x - \frac {1}{2} \right )}}{717409} + \frac {24225 \log {\left (x + \frac {3}{5} \right )}}{14641} - \frac {81 \log {\left (x + \frac {2}{3} \right )}}{49} \] Input:
integrate(1/(1-2*x)**2/(2+3*x)/(3+5*x)**3,x)
Output:
(101100*x**2 + 5820*x - 28669)/(931700*x**3 + 652190*x**2 - 223608*x - 167 706) - 1104*log(x - 1/2)/717409 + 24225*log(x + 3/5)/14641 - 81*log(x + 2/ 3)/49
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {101100 \, x^{2} + 5820 \, x - 28669}{18634 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {24225}{14641} \, \log \left (5 \, x + 3\right ) - \frac {81}{49} \, \log \left (3 \, x + 2\right ) - \frac {1104}{717409} \, \log \left (2 \, x - 1\right ) \] Input:
integrate(1/(1-2*x)^2/(2+3*x)/(3+5*x)^3,x, algorithm="maxima")
Output:
1/18634*(101100*x^2 + 5820*x - 28669)/(50*x^3 + 35*x^2 - 12*x - 9) + 24225 /14641*log(5*x + 3) - 81/49*log(3*x + 2) - 1104/717409*log(2*x - 1)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=-\frac {8}{9317 \, {\left (2 \, x - 1\right )}} - \frac {250 \, {\left (\frac {297}{2 \, x - 1} + 140\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} - \frac {81}{49} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) + \frac {24225}{14641} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \] Input:
integrate(1/(1-2*x)^2/(2+3*x)/(3+5*x)^3,x, algorithm="giac")
Output:
-8/9317/(2*x - 1) - 250/14641*(297/(2*x - 1) + 140)/(11/(2*x - 1) + 5)^2 - 81/49*log(abs(-7/(2*x - 1) - 3)) + 24225/14641*log(abs(-11/(2*x - 1) - 5) )
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {24225\,\ln \left (x+\frac {3}{5}\right )}{14641}-\frac {81\,\ln \left (x+\frac {2}{3}\right )}{49}-\frac {1104\,\ln \left (x-\frac {1}{2}\right )}{717409}-\frac {\frac {1011\,x^2}{9317}+\frac {291\,x}{46585}-\frac {28669}{931700}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \] Input:
int(1/((2*x - 1)^2*(3*x + 2)*(5*x + 3)^3),x)
Output:
(24225*log(x + 3/5))/14641 - (81*log(x + 2/3))/49 - (1104*log(x - 1/2))/71 7409 - ((291*x)/46585 + (1011*x^2)/9317 - 28669/931700)/((6*x)/25 - (7*x^2 )/10 - x^3 + 9/50)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.27 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)^3} \, dx=\frac {118702500 \,\mathrm {log}\left (5 x +3\right ) x^{3}+83091750 \,\mathrm {log}\left (5 x +3\right ) x^{2}-28488600 \,\mathrm {log}\left (5 x +3\right ) x -21366450 \,\mathrm {log}\left (5 x +3\right )-118592100 \,\mathrm {log}\left (3 x +2\right ) x^{3}-83014470 \,\mathrm {log}\left (3 x +2\right ) x^{2}+28462104 \,\mathrm {log}\left (3 x +2\right ) x +21346578 \,\mathrm {log}\left (3 x +2\right )-110400 \,\mathrm {log}\left (2 x -1\right ) x^{3}-77280 \,\mathrm {log}\left (2 x -1\right ) x^{2}+26496 \,\mathrm {log}\left (2 x -1\right ) x +19872 \,\mathrm {log}\left (2 x -1\right )-11121000 x^{3}+3117180 x -205733}{71740900 x^{3}+50218630 x^{2}-17217816 x -12913362} \] Input:
int(1/(1-2*x)^2/(2+3*x)/(3+5*x)^3,x)
Output:
(118702500*log(5*x + 3)*x**3 + 83091750*log(5*x + 3)*x**2 - 28488600*log(5 *x + 3)*x - 21366450*log(5*x + 3) - 118592100*log(3*x + 2)*x**3 - 83014470 *log(3*x + 2)*x**2 + 28462104*log(3*x + 2)*x + 21346578*log(3*x + 2) - 110 400*log(2*x - 1)*x**3 - 77280*log(2*x - 1)*x**2 + 26496*log(2*x - 1)*x + 1 9872*log(2*x - 1) - 11121000*x**3 + 3117180*x - 205733)/(1434818*(50*x**3 + 35*x**2 - 12*x - 9))