Integrand size = 15, antiderivative size = 54 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {4}{1331 (1-2 x)}-\frac {5}{242 (3+5 x)^2}-\frac {20}{1331 (3+5 x)}-\frac {60 \log (1-2 x)}{14641}+\frac {60 \log (3+5 x)}{14641} \] Output:
4/(1331-2662*x)-5/242/(3+5*x)^2-20/(3993+6655*x)-60/14641*ln(1-2*x)+60/146 41*ln(3+5*x)
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {-\frac {11 \left (-103+390 x+600 x^2\right )}{(-1+2 x) (3+5 x)^2}-120 \log (1-2 x)+120 \log (6+10 x)}{29282} \] Input:
Integrate[1/((1 - 2*x)^2*(3 + 5*x)^3),x]
Output:
((-11*(-103 + 390*x + 600*x^2))/((-1 + 2*x)*(3 + 5*x)^2) - 120*Log[1 - 2*x ] + 120*Log[6 + 10*x])/29282
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {54, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^2 (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (\frac {300}{14641 (5 x+3)}+\frac {100}{1331 (5 x+3)^2}+\frac {25}{121 (5 x+3)^3}-\frac {120}{14641 (2 x-1)}+\frac {8}{1331 (2 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{1331 (1-2 x)}-\frac {20}{1331 (5 x+3)}-\frac {5}{242 (5 x+3)^2}-\frac {60 \log (1-2 x)}{14641}+\frac {60 \log (5 x+3)}{14641}\) |
Input:
Int[1/((1 - 2*x)^2*(3 + 5*x)^3),x]
Output:
4/(1331*(1 - 2*x)) - 5/(242*(3 + 5*x)^2) - 20/(1331*(3 + 5*x)) - (60*Log[1 - 2*x])/14641 + (60*Log[3 + 5*x])/14641
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {-\frac {300}{1331} x^{2}-\frac {195}{1331} x +\frac {103}{2662}}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {60 \ln \left (-1+2 x \right )}{14641}+\frac {60 \ln \left (3+5 x \right )}{14641}\) | \(44\) |
default | \(-\frac {5}{242 \left (3+5 x \right )^{2}}-\frac {20}{1331 \left (3+5 x \right )}+\frac {60 \ln \left (3+5 x \right )}{14641}-\frac {4}{1331 \left (-1+2 x \right )}-\frac {60 \ln \left (-1+2 x \right )}{14641}\) | \(45\) |
norman | \(\frac {\frac {2575}{11979} x^{3}-\frac {1795}{23958} x^{2}-\frac {791}{3993} x}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {60 \ln \left (-1+2 x \right )}{14641}+\frac {60 \ln \left (3+5 x \right )}{14641}\) | \(47\) |
parallelrisch | \(-\frac {54000 \ln \left (x -\frac {1}{2}\right ) x^{3}-54000 \ln \left (x +\frac {3}{5}\right ) x^{3}+37800 \ln \left (x -\frac {1}{2}\right ) x^{2}-37800 \ln \left (x +\frac {3}{5}\right ) x^{2}-56650 x^{3}-12960 \ln \left (x -\frac {1}{2}\right ) x +12960 \ln \left (x +\frac {3}{5}\right ) x +19745 x^{2}-9720 \ln \left (x -\frac {1}{2}\right )+9720 \ln \left (x +\frac {3}{5}\right )+52206 x}{263538 \left (-1+2 x \right ) \left (3+5 x \right )^{2}}\) | \(93\) |
Input:
int(1/(1-2*x)^2/(3+5*x)^3,x,method=_RETURNVERBOSE)
Output:
50*(-6/1331*x^2-39/13310*x+103/133100)/(-1+2*x)/(3+5*x)^2-60/14641*ln(-1+2 *x)+60/14641*ln(3+5*x)
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {6600 \, x^{2} - 120 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) + 120 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 4290 \, x - 1133}{29282 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \] Input:
integrate(1/(1-2*x)^2/(3+5*x)^3,x, algorithm="fricas")
Output:
-1/29282*(6600*x^2 - 120*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) + 120*( 50*x^3 + 35*x^2 - 12*x - 9)*log(2*x - 1) + 4290*x - 1133)/(50*x^3 + 35*x^2 - 12*x - 9)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {- 600 x^{2} - 390 x + 103}{133100 x^{3} + 93170 x^{2} - 31944 x - 23958} - \frac {60 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {60 \log {\left (x + \frac {3}{5} \right )}}{14641} \] Input:
integrate(1/(1-2*x)**2/(3+5*x)**3,x)
Output:
(-600*x**2 - 390*x + 103)/(133100*x**3 + 93170*x**2 - 31944*x - 23958) - 6 0*log(x - 1/2)/14641 + 60*log(x + 3/5)/14641
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {600 \, x^{2} + 390 \, x - 103}{2662 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {60}{14641} \, \log \left (5 \, x + 3\right ) - \frac {60}{14641} \, \log \left (2 \, x - 1\right ) \] Input:
integrate(1/(1-2*x)^2/(3+5*x)^3,x, algorithm="maxima")
Output:
-1/2662*(600*x^2 + 390*x - 103)/(50*x^3 + 35*x^2 - 12*x - 9) + 60/14641*lo g(5*x + 3) - 60/14641*log(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {4}{1331 \, {\left (2 \, x - 1\right )}} + \frac {50 \, {\left (\frac {66}{2 \, x - 1} + 25\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} + \frac {60}{14641} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \] Input:
integrate(1/(1-2*x)^2/(3+5*x)^3,x, algorithm="giac")
Output:
-4/1331/(2*x - 1) + 50/14641*(66/(2*x - 1) + 25)/(11/(2*x - 1) + 5)^2 + 60 /14641*log(abs(-11/(2*x - 1) - 5))
Time = 1.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {120\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}+\frac {\frac {6\,x^2}{1331}+\frac {39\,x}{13310}-\frac {103}{133100}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \] Input:
int(1/((2*x - 1)^2*(5*x + 3)^3),x)
Output:
(120*atanh((20*x)/11 + 1/11))/14641 + ((39*x)/13310 + (6*x^2)/1331 - 103/1 33100)/((6*x)/25 - (7*x^2)/10 - x^3 + 9/50)
Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {42000 \,\mathrm {log}\left (5 x +3\right ) x^{3}+29400 \,\mathrm {log}\left (5 x +3\right ) x^{2}-10080 \,\mathrm {log}\left (5 x +3\right ) x -7560 \,\mathrm {log}\left (5 x +3\right )-42000 \,\mathrm {log}\left (2 x -1\right ) x^{3}-29400 \,\mathrm {log}\left (2 x -1\right ) x^{2}+10080 \,\mathrm {log}\left (2 x -1\right ) x +7560 \,\mathrm {log}\left (2 x -1\right )+66000 x^{3}-45870 x -3949}{10248700 x^{3}+7174090 x^{2}-2459688 x -1844766} \] Input:
int(1/(1-2*x)^2/(3+5*x)^3,x)
Output:
(42000*log(5*x + 3)*x**3 + 29400*log(5*x + 3)*x**2 - 10080*log(5*x + 3)*x - 7560*log(5*x + 3) - 42000*log(2*x - 1)*x**3 - 29400*log(2*x - 1)*x**2 + 10080*log(2*x - 1)*x + 7560*log(2*x - 1) + 66000*x**3 - 45870*x - 3949)/(2 04974*(50*x**3 + 35*x**2 - 12*x - 9))