Integrand size = 22, antiderivative size = 64 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {8}{3773 (1-2 x)}+\frac {9}{98 (2+3 x)^2}+\frac {351}{343 (2+3 x)}-\frac {1072 \log (1-2 x)}{290521}-\frac {12393 \log (2+3 x)}{2401}+\frac {625}{121} \log (3+5 x) \] Output:
8/(3773-7546*x)+9/98/(2+3*x)^2+351/(686+1029*x)-1072/290521*ln(1-2*x)-1239 3/2401*ln(2+3*x)+625/121*ln(3+5*x)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {-2144 \log (5-10 x)-2999106 \log (5 (2+3 x))+7 \left (\frac {176}{1-2 x}+\frac {7623}{(2+3 x)^2}+\frac {84942}{2+3 x}+428750 \log (3+5 x)\right )}{581042} \] Input:
Integrate[1/((1 - 2*x)^2*(2 + 3*x)^3*(3 + 5*x)),x]
Output:
(-2144*Log[5 - 10*x] - 2999106*Log[5*(2 + 3*x)] + 7*(176/(1 - 2*x) + 7623/ (2 + 3*x)^2 + 84942/(2 + 3*x) + 428750*Log[3 + 5*x]))/581042
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.
\(\displaystyle \int \frac {1}{(1-2 x)^2 (3 x+2)^3 (5 x+3)} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {37179}{2401 (3 x+2)}+\frac {3125}{121 (5 x+3)}-\frac {1053}{343 (3 x+2)^2}-\frac {27}{49 (3 x+2)^3}-\frac {2144}{290521 (2 x-1)}+\frac {16}{3773 (2 x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{3773 (1-2 x)}+\frac {351}{343 (3 x+2)}+\frac {9}{98 (3 x+2)^2}-\frac {1072 \log (1-2 x)}{290521}-\frac {12393 \log (3 x+2)}{2401}+\frac {625}{121} \log (5 x+3)\) |
Input:
Int[1/((1 - 2*x)^2*(2 + 3*x)^3*(3 + 5*x)),x]
Output:
8/(3773*(1 - 2*x)) + 9/(98*(2 + 3*x)^2) + 351/(343*(2 + 3*x)) - (1072*Log[ 1 - 2*x])/290521 - (12393*Log[2 + 3*x])/2401 + (625*Log[3 + 5*x])/121
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 = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\frac {23094}{3773} x^{2}+\frac {4458}{3773} x -\frac {16201}{7546}}{\left (-1+2 x \right ) \left (2+3 x \right )^{2}}-\frac {1072 \ln \left (-1+2 x \right )}{290521}-\frac {12393 \ln \left (2+3 x \right )}{2401}+\frac {625 \ln \left (3+5 x \right )}{121}\) | \(52\) |
default | \(\frac {625 \ln \left (3+5 x \right )}{121}+\frac {9}{98 \left (2+3 x \right )^{2}}+\frac {351}{343 \left (2+3 x \right )}-\frac {12393 \ln \left (2+3 x \right )}{2401}-\frac {8}{3773 \left (-1+2 x \right )}-\frac {1072 \ln \left (-1+2 x \right )}{290521}\) | \(53\) |
norman | \(\frac {\frac {79623}{7546} x^{2}+\frac {20061}{3773} x^{3}-\frac {25117}{7546}}{\left (-1+2 x \right ) \left (2+3 x \right )^{2}}-\frac {1072 \ln \left (-1+2 x \right )}{290521}-\frac {12393 \ln \left (2+3 x \right )}{2401}+\frac {625 \ln \left (3+5 x \right )}{121}\) | \(53\) |
parallelrisch | \(-\frac {215935632 \ln \left (\frac {2}{3}+x \right ) x^{3}+154368 \ln \left (x -\frac {1}{2}\right ) x^{3}-216090000 \ln \left (x +\frac {3}{5}\right ) x^{3}+179946360 \ln \left (\frac {2}{3}+x \right ) x^{2}+128640 \ln \left (x -\frac {1}{2}\right ) x^{2}-180075000 \ln \left (x +\frac {3}{5}\right ) x^{2}+22454586 x^{3}-47985696 \ln \left (\frac {2}{3}+x \right ) x -34304 \ln \left (x -\frac {1}{2}\right ) x +48020000 \ln \left (x +\frac {3}{5}\right ) x +4486251 x^{2}-47985696 \ln \left (\frac {2}{3}+x \right )-34304 \ln \left (x -\frac {1}{2}\right )+48020000 \ln \left (x +\frac {3}{5}\right )-7736036 x}{2324168 \left (-1+2 x \right ) \left (2+3 x \right )^{2}}\) | \(124\) |
Input:
int(1/(1-2*x)^2/(2+3*x)^3/(3+5*x),x,method=_RETURNVERBOSE)
Output:
18*(1283/3773*x^2+743/11319*x-16201/135828)/(-1+2*x)/(2+3*x)^2-1072/290521 *ln(-1+2*x)-12393/2401*ln(2+3*x)+625/121*ln(3+5*x)
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {3556476 \, x^{2} + 3001250 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (5 \, x + 3\right ) - 2999106 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (3 \, x + 2\right ) - 2144 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (2 \, x - 1\right ) + 686532 \, x - 1247477}{581042 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \] Input:
integrate(1/(1-2*x)^2/(2+3*x)^3/(3+5*x),x, algorithm="fricas")
Output:
1/581042*(3556476*x^2 + 3001250*(18*x^3 + 15*x^2 - 4*x - 4)*log(5*x + 3) - 2999106*(18*x^3 + 15*x^2 - 4*x - 4)*log(3*x + 2) - 2144*(18*x^3 + 15*x^2 - 4*x - 4)*log(2*x - 1) + 686532*x - 1247477)/(18*x^3 + 15*x^2 - 4*x - 4)
Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {46188 x^{2} + 8916 x - 16201}{135828 x^{3} + 113190 x^{2} - 30184 x - 30184} - \frac {1072 \log {\left (x - \frac {1}{2} \right )}}{290521} + \frac {625 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {12393 \log {\left (x + \frac {2}{3} \right )}}{2401} \] Input:
integrate(1/(1-2*x)**2/(2+3*x)**3/(3+5*x),x)
Output:
(46188*x**2 + 8916*x - 16201)/(135828*x**3 + 113190*x**2 - 30184*x - 30184 ) - 1072*log(x - 1/2)/290521 + 625*log(x + 3/5)/121 - 12393*log(x + 2/3)/2 401
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {46188 \, x^{2} + 8916 \, x - 16201}{7546 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} + \frac {625}{121} \, \log \left (5 \, x + 3\right ) - \frac {12393}{2401} \, \log \left (3 \, x + 2\right ) - \frac {1072}{290521} \, \log \left (2 \, x - 1\right ) \] Input:
integrate(1/(1-2*x)^2/(2+3*x)^3/(3+5*x),x, algorithm="maxima")
Output:
1/7546*(46188*x^2 + 8916*x - 16201)/(18*x^3 + 15*x^2 - 4*x - 4) + 625/121* log(5*x + 3) - 12393/2401*log(3*x + 2) - 1072/290521*log(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=-\frac {8}{3773 \, {\left (2 \, x - 1\right )}} - \frac {54 \, {\left (\frac {287}{2 \, x - 1} + 120\right )}}{2401 \, {\left (\frac {7}{2 \, x - 1} + 3\right )}^{2}} - \frac {12393}{2401} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) + \frac {625}{121} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \] Input:
integrate(1/(1-2*x)^2/(2+3*x)^3/(3+5*x),x, algorithm="giac")
Output:
-8/3773/(2*x - 1) - 54/2401*(287/(2*x - 1) + 120)/(7/(2*x - 1) + 3)^2 - 12 393/2401*log(abs(-7/(2*x - 1) - 3)) + 625/121*log(abs(-11/(2*x - 1) - 5))
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {625\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {12393\,\ln \left (x+\frac {2}{3}\right )}{2401}-\frac {1072\,\ln \left (x-\frac {1}{2}\right )}{290521}-\frac {\frac {1283\,x^2}{3773}+\frac {743\,x}{11319}-\frac {16201}{135828}}{-x^3-\frac {5\,x^2}{6}+\frac {2\,x}{9}+\frac {2}{9}} \] Input:
int(1/((2*x - 1)^2*(3*x + 2)^3*(5*x + 3)),x)
Output:
(625*log(x + 3/5))/121 - (12393*log(x + 2/3))/2401 - (1072*log(x - 1/2))/2 90521 - ((743*x)/11319 + (1283*x^2)/3773 - 16201/135828)/((2*x)/9 - (5*x^2 )/6 - x^3 + 2/9)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.27 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^3 (3+5 x)} \, dx=\frac {270112500 \,\mathrm {log}\left (5 x +3\right ) x^{3}+225093750 \,\mathrm {log}\left (5 x +3\right ) x^{2}-60025000 \,\mathrm {log}\left (5 x +3\right ) x -60025000 \,\mathrm {log}\left (5 x +3\right )-269919540 \,\mathrm {log}\left (3 x +2\right ) x^{3}-224932950 \,\mathrm {log}\left (3 x +2\right ) x^{2}+59982120 \,\mathrm {log}\left (3 x +2\right ) x +59982120 \,\mathrm {log}\left (3 x +2\right )-192960 \,\mathrm {log}\left (2 x -1\right ) x^{3}-160800 \,\mathrm {log}\left (2 x -1\right ) x^{2}+42880 \,\mathrm {log}\left (2 x -1\right ) x +42880 \,\mathrm {log}\left (2 x -1\right )-21338856 x^{3}+8174628 x -1495417}{52293780 x^{3}+43578150 x^{2}-11620840 x -11620840} \] Input:
int(1/(1-2*x)^2/(2+3*x)^3/(3+5*x),x)
Output:
(270112500*log(5*x + 3)*x**3 + 225093750*log(5*x + 3)*x**2 - 60025000*log( 5*x + 3)*x - 60025000*log(5*x + 3) - 269919540*log(3*x + 2)*x**3 - 2249329 50*log(3*x + 2)*x**2 + 59982120*log(3*x + 2)*x + 59982120*log(3*x + 2) - 1 92960*log(2*x - 1)*x**3 - 160800*log(2*x - 1)*x**2 + 42880*log(2*x - 1)*x + 42880*log(2*x - 1) - 21338856*x**3 + 8174628*x - 1495417)/(2905210*(18*x **3 + 15*x**2 - 4*x - 4))