Integrand size = 79, antiderivative size = 24 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 x^2 \log ^2(3+x)}{144 (5-x)^2 \log (x)} \] Output:
25/144*x^2/(5-x)^2*ln(3+x)^2/ln(x)
Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 x^2 \log ^2(3+x)}{144 (-5+x)^2 \log (x)} \] Input:
Integrate[((-250*x^2 + 50*x^3)*Log[x]*Log[3 + x] + (375*x + 50*x^2 - 25*x^ 3 + (-750*x - 250*x^2)*Log[x])*Log[3 + x]^2)/((-54000 + 14400*x + 4320*x^2 - 1728*x^3 + 144*x^4)*Log[x]^2),x]
Output:
(25*x^2*Log[3 + x]^2)/(144*(-5 + x)^2*Log[x])
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.
\(\displaystyle \int \frac {\left (-25 x^3+50 x^2+\left (-250 x^2-750 x\right ) \log (x)+375 x\right ) \log ^2(x+3)+\left (50 x^3-250 x^2\right ) \log (x) \log (x+3)}{\left (144 x^4-1728 x^3+4320 x^2+14400 x-54000\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-25 x^3+50 x^2+\left (-250 x^2-750 x\right ) \log (x)+375 x\right ) \log ^2(x+3)+\left (50 x^3-250 x^2\right ) \log (x) \log (x+3)}{73728 (x-5) \log ^2(x)}-\frac {\left (-25 x^3+50 x^2+\left (-250 x^2-750 x\right ) \log (x)+375 x\right ) \log ^2(x+3)+\left (50 x^3-250 x^2\right ) \log (x) \log (x+3)}{73728 (x+3) \log ^2(x)}-\frac {\left (-25 x^3+50 x^2+\left (-250 x^2-750 x\right ) \log (x)+375 x\right ) \log ^2(x+3)+\left (50 x^3-250 x^2\right ) \log (x) \log (x+3)}{9216 (x-5)^2 \log ^2(x)}+\frac {\left (-25 x^3+50 x^2+\left (-250 x^2-750 x\right ) \log (x)+375 x\right ) \log ^2(x+3)+\left (50 x^3-250 x^2\right ) \log (x) \log (x+3)}{1152 (x-5)^3 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {125}{144} \int \frac {\log ^2(x+3)}{(x-5)^2 \log ^2(x)}dx-\frac {25}{144} \int \frac {\log ^2(x+3)}{(x-5) \log ^2(x)}dx-\frac {625}{72} \int \frac {\log ^2(x+3)}{(x-5)^3 \log (x)}dx-\frac {125}{72} \int \frac {\log ^2(x+3)}{(x-5)^2 \log (x)}dx+\frac {625}{576} \int \frac {\log (x+3)}{(x-5)^2 \log (x)}dx+\frac {1375 \int \frac {\log (x+3)}{(x-5) \log (x)}dx}{4608}+\frac {25}{512} \int \frac {\log (x+3)}{(x+3) \log (x)}dx\) |
Input:
Int[((-250*x^2 + 50*x^3)*Log[x]*Log[3 + x] + (375*x + 50*x^2 - 25*x^3 + (- 750*x - 250*x^2)*Log[x])*Log[3 + x]^2)/((-54000 + 14400*x + 4320*x^2 - 172 8*x^3 + 144*x^4)*Log[x]^2),x]
Output:
$Aborted
Time = 81.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {25 x^{2} \ln \left (3+x \right )^{2}}{144 \left (x^{2}-10 x +25\right ) \ln \left (x \right )}\) | \(26\) |
parallelrisch | \(\frac {25 x^{2} \ln \left (3+x \right )^{2}}{144 \left (x^{2}-10 x +25\right ) \ln \left (x \right )}\) | \(26\) |
Input:
int((((-250*x^2-750*x)*ln(x)-25*x^3+50*x^2+375*x)*ln(3+x)^2+(50*x^3-250*x^ 2)*ln(x)*ln(3+x))/(144*x^4-1728*x^3+4320*x^2+14400*x-54000)/ln(x)^2,x,meth od=_RETURNVERBOSE)
Output:
25/144*x^2/(x^2-10*x+25)/ln(x)*ln(3+x)^2
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 \, x^{2} \log \left (x + 3\right )^{2}}{144 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (x\right )} \] Input:
integrate((((-250*x^2-750*x)*log(x)-25*x^3+50*x^2+375*x)*log(3+x)^2+(50*x^ 3-250*x^2)*log(x)*log(3+x))/(144*x^4-1728*x^3+4320*x^2+14400*x-54000)/log( x)^2,x, algorithm="fricas")
Output:
25/144*x^2*log(x + 3)^2/((x^2 - 10*x + 25)*log(x))
Exception generated. \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((-250*x**2-750*x)*ln(x)-25*x**3+50*x**2+375*x)*ln(3+x)**2+(50* x**3-250*x**2)*ln(x)*ln(3+x))/(144*x**4-1728*x**3+4320*x**2+14400*x-54000) /ln(x)**2,x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 \, x^{2} \log \left (x + 3\right )^{2}}{144 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (x\right )} \] Input:
integrate((((-250*x^2-750*x)*log(x)-25*x^3+50*x^2+375*x)*log(3+x)^2+(50*x^ 3-250*x^2)*log(x)*log(3+x))/(144*x^4-1728*x^3+4320*x^2+14400*x-54000)/log( x)^2,x, algorithm="maxima")
Output:
25/144*x^2*log(x + 3)^2/((x^2 - 10*x + 25)*log(x))
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 \, x^{2} \log \left (x + 3\right )^{2}}{144 \, {\left (x^{2} \log \left (x\right ) - 10 \, x \log \left (x\right ) + 25 \, \log \left (x\right )\right )}} \] Input:
integrate((((-250*x^2-750*x)*log(x)-25*x^3+50*x^2+375*x)*log(3+x)^2+(50*x^ 3-250*x^2)*log(x)*log(3+x))/(144*x^4-1728*x^3+4320*x^2+14400*x-54000)/log( x)^2,x, algorithm="giac")
Output:
25/144*x^2*log(x + 3)^2/(x^2*log(x) - 10*x*log(x) + 25*log(x))
Time = 2.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25\,x^2\,{\ln \left (x+3\right )}^2}{144\,\ln \left (x\right )\,{\left (x-5\right )}^2} \] Input:
int((log(x + 3)^2*(375*x - log(x)*(750*x + 250*x^2) + 50*x^2 - 25*x^3) - l og(x + 3)*log(x)*(250*x^2 - 50*x^3))/(log(x)^2*(14400*x + 4320*x^2 - 1728* x^3 + 144*x^4 - 54000)),x)
Output:
(25*x^2*log(x + 3)^2)/(144*log(x)*(x - 5)^2)
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-250 x^2+50 x^3\right ) \log (x) \log (3+x)+\left (375 x+50 x^2-25 x^3+\left (-750 x-250 x^2\right ) \log (x)\right ) \log ^2(3+x)}{\left (-54000+14400 x+4320 x^2-1728 x^3+144 x^4\right ) \log ^2(x)} \, dx=\frac {25 \mathrm {log}\left (x +3\right )^{2} x^{2}}{144 \,\mathrm {log}\left (x \right ) \left (x^{2}-10 x +25\right )} \] Input:
int((((-250*x^2-750*x)*log(x)-25*x^3+50*x^2+375*x)*log(3+x)^2+(50*x^3-250* x^2)*log(x)*log(3+x))/(144*x^4-1728*x^3+4320*x^2+14400*x-54000)/log(x)^2,x )
Output:
(25*log(x + 3)**2*x**2)/(144*log(x)*(x**2 - 10*x + 25))