Integrand size = 22, antiderivative size = 55 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=\frac {16807}{704 (1-2 x)^2}-\frac {156065}{1936 (1-2 x)}-\frac {10287 x}{400}-\frac {243 x^2}{80}-\frac {543655 \log (1-2 x)}{10648}+\frac {\log (3+5 x)}{166375} \] Output:
16807/704/(1-2*x)^2-156065/(1936-3872*x)-10287/400*x-243/80*x^2-543655/106 48*ln(1-2*x)+1/166375*ln(3+5*x)
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=\frac {\frac {254205875}{(1-2 x)^2}+\frac {858357500}{-1+2 x}-47005596 (3+5 x)-1293732 (3+5 x)^2-543655000 \log (5-10 x)+64 \log (3+5 x)}{10648000} \] Input:
Integrate[(2 + 3*x)^5/((1 - 2*x)^3*(3 + 5*x)),x]
Output:
(254205875/(1 - 2*x)^2 + 858357500/(-1 + 2*x) - 47005596*(3 + 5*x) - 12937 32*(3 + 5*x)^2 - 543655000*Log[5 - 10*x] + 64*Log[3 + 5*x])/10648000
Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
\(\displaystyle \int \frac {(3 x+2)^5}{(1-2 x)^3 (5 x+3)} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {243 x}{40}-\frac {543655}{5324 (2 x-1)}+\frac {1}{33275 (5 x+3)}-\frac {156065}{968 (2 x-1)^2}-\frac {16807}{176 (2 x-1)^3}-\frac {10287}{400}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {243 x^2}{80}-\frac {10287 x}{400}-\frac {156065}{1936 (1-2 x)}+\frac {16807}{704 (1-2 x)^2}-\frac {543655 \log (1-2 x)}{10648}+\frac {\log (5 x+3)}{166375}\) |
Input:
Int[(2 + 3*x)^5/((1 - 2*x)^3*(3 + 5*x)),x]
Output:
16807/(704*(1 - 2*x)^2) - 156065/(1936*(1 - 2*x)) - (10287*x)/400 - (243*x ^2)/80 - (543655*Log[1 - 2*x])/10648 + Log[3 + 5*x]/166375
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 = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {243 x^{2}}{80}-\frac {10287 x}{400}+\frac {\frac {156065 x}{968}-\frac {439383}{7744}}{\left (-1+2 x \right )^{2}}-\frac {543655 \ln \left (-1+2 x \right )}{10648}+\frac {\ln \left (3+5 x \right )}{166375}\) | \(40\) |
default | \(-\frac {243 x^{2}}{80}-\frac {10287 x}{400}+\frac {\ln \left (3+5 x \right )}{166375}+\frac {16807}{704 \left (-1+2 x \right )^{2}}+\frac {156065}{1936 \left (-1+2 x \right )}-\frac {543655 \ln \left (-1+2 x \right )}{10648}\) | \(44\) |
norman | \(\frac {-\frac {1106513}{12100} x +\frac {3954117}{12100} x^{2}-\frac {2268}{25} x^{3}-\frac {243}{20} x^{4}}{\left (-1+2 x \right )^{2}}-\frac {543655 \ln \left (-1+2 x \right )}{10648}+\frac {\ln \left (3+5 x \right )}{166375}\) | \(45\) |
parallelrisch | \(-\frac {16171650 x^{4}+271827500 \ln \left (x -\frac {1}{2}\right ) x^{2}-32 \ln \left (x +\frac {3}{5}\right ) x^{2}+120748320 x^{3}-271827500 \ln \left (x -\frac {1}{2}\right ) x +32 \ln \left (x +\frac {3}{5}\right ) x -434952870 x^{2}+67956875 \ln \left (x -\frac {1}{2}\right )-8 \ln \left (x +\frac {3}{5}\right )+121716430 x}{1331000 \left (-1+2 x \right )^{2}}\) | \(73\) |
Input:
int((2+3*x)^5/(1-2*x)^3/(3+5*x),x,method=_RETURNVERBOSE)
Output:
-243/80*x^2-10287/400*x+4*(156065/3872*x-439383/30976)/(-1+2*x)^2-543655/1 0648*ln(-1+2*x)+1/166375*ln(3+5*x)
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=-\frac {129373200 \, x^{4} + 965986560 \, x^{3} - 1063016460 \, x^{2} - 64 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (5 \, x + 3\right ) + 543655000 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 1442875060 \, x + 604151625}{10648000 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] Input:
integrate((2+3*x)^5/(1-2*x)^3/(3+5*x),x, algorithm="fricas")
Output:
-1/10648000*(129373200*x^4 + 965986560*x^3 - 1063016460*x^2 - 64*(4*x^2 - 4*x + 1)*log(5*x + 3) + 543655000*(4*x^2 - 4*x + 1)*log(2*x - 1) - 1442875 060*x + 604151625)/(4*x^2 - 4*x + 1)
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=- \frac {243 x^{2}}{80} - \frac {10287 x}{400} - \frac {439383 - 1248520 x}{30976 x^{2} - 30976 x + 7744} - \frac {543655 \log {\left (x - \frac {1}{2} \right )}}{10648} + \frac {\log {\left (x + \frac {3}{5} \right )}}{166375} \] Input:
integrate((2+3*x)**5/(1-2*x)**3/(3+5*x),x)
Output:
-243*x**2/80 - 10287*x/400 - (439383 - 1248520*x)/(30976*x**2 - 30976*x + 7744) - 543655*log(x - 1/2)/10648 + log(x + 3/5)/166375
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=-\frac {243}{80} \, x^{2} - \frac {10287}{400} \, x + \frac {2401 \, {\left (520 \, x - 183\right )}}{7744 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {1}{166375} \, \log \left (5 \, x + 3\right ) - \frac {543655}{10648} \, \log \left (2 \, x - 1\right ) \] Input:
integrate((2+3*x)^5/(1-2*x)^3/(3+5*x),x, algorithm="maxima")
Output:
-243/80*x^2 - 10287/400*x + 2401/7744*(520*x - 183)/(4*x^2 - 4*x + 1) + 1/ 166375*log(5*x + 3) - 543655/10648*log(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=-\frac {243}{80} \, x^{2} - \frac {10287}{400} \, x + \frac {2401 \, {\left (520 \, x - 183\right )}}{7744 \, {\left (2 \, x - 1\right )}^{2}} + \frac {1}{166375} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {543655}{10648} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \] Input:
integrate((2+3*x)^5/(1-2*x)^3/(3+5*x),x, algorithm="giac")
Output:
-243/80*x^2 - 10287/400*x + 2401/7744*(520*x - 183)/(2*x - 1)^2 + 1/166375 *log(abs(5*x + 3)) - 543655/10648*log(abs(2*x - 1))
Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=\frac {\ln \left (x+\frac {3}{5}\right )}{166375}-\frac {543655\,\ln \left (x-\frac {1}{2}\right )}{10648}-\frac {10287\,x}{400}+\frac {\frac {156065\,x}{3872}-\frac {439383}{30976}}{x^2-x+\frac {1}{4}}-\frac {243\,x^2}{80} \] Input:
int(-(3*x + 2)^5/((2*x - 1)^3*(5*x + 3)),x)
Output:
log(x + 3/5)/166375 - (543655*log(x - 1/2))/10648 - (10287*x)/400 + ((1560 65*x)/3872 - 439383/30976)/(x^2 - x + 1/4) - (243*x^2)/80
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56 \[ \int \frac {(2+3 x)^5}{(1-2 x)^3 (3+5 x)} \, dx=\frac {64 \,\mathrm {log}\left (5 x +3\right ) x^{2}-64 \,\mathrm {log}\left (5 x +3\right ) x +16 \,\mathrm {log}\left (5 x +3\right )-543655000 \,\mathrm {log}\left (2 x -1\right ) x^{2}+543655000 \,\mathrm {log}\left (2 x -1\right ) x -135913750 \,\mathrm {log}\left (2 x -1\right )-32343300 x^{4}-241496640 x^{3}+626472880 x^{2}-60858215}{10648000 x^{2}-10648000 x +2662000} \] Input:
int((2+3*x)^5/(1-2*x)^3/(3+5*x),x)
Output:
(64*log(5*x + 3)*x**2 - 64*log(5*x + 3)*x + 16*log(5*x + 3) - 543655000*lo g(2*x - 1)*x**2 + 543655000*log(2*x - 1)*x - 135913750*log(2*x - 1) - 3234 3300*x**4 - 241496640*x**3 + 626472880*x**2 - 60858215)/(2662000*(4*x**2 - 4*x + 1))