Integrand size = 77, antiderivative size = 27 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {16 \left (x^2+\frac {-3+x^2}{-x+\log (5 x)}\right )}{225 x^6} \] Output:
16/225*((x^2-3)/(ln(5*x)-x)+x^2)/x^6
Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=-\frac {16}{225} \left (-\frac {1}{x^4}+\frac {3-x^2}{x^6 (-x+\log (5 x))}\right ) \] Input:
Integrate[(48 - 336*x - 16*x^2 + 80*x^3 - 64*x^4 + (288 - 64*x^2 + 128*x^3 )*Log[5*x] - 64*x^2*Log[5*x]^2)/(225*x^9 - 450*x^8*Log[5*x] + 225*x^7*Log[ 5*x]^2),x]
Output:
(-16*(-x^(-4) + (3 - x^2)/(x^6*(-x + Log[5*x]))))/225
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.
\(\displaystyle \int \frac {-64 x^4+80 x^3-16 x^2-64 x^2 \log ^2(5 x)+\left (128 x^3-64 x^2+288\right ) \log (5 x)-336 x+48}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-64 x^4+80 x^3-16 x^2-64 x^2 \log ^2(5 x)+\left (128 x^3-64 x^2+288\right ) \log (5 x)-336 x+48}{225 x^7 (x-\log (5 x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{225} \int \frac {16 \left (-4 x^4+5 x^3-4 \log ^2(5 x) x^2-x^2-21 x+2 \left (4 x^3-2 x^2+9\right ) \log (5 x)+3\right )}{x^7 (x-\log (5 x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16}{225} \int \frac {-4 x^4+5 x^3-4 \log ^2(5 x) x^2-x^2-21 x+2 \left (4 x^3-2 x^2+9\right ) \log (5 x)+3}{x^7 (x-\log (5 x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {16}{225} \int \left (\frac {2 \left (2 x^2-9\right )}{x^7 (x-\log (5 x))}+\frac {x^3-x^2-3 x+3}{x^7 (x-\log (5 x))^2}-\frac {4}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16}{225} \left (3 \int \frac {1}{x^7 (x-\log (5 x))^2}dx-18 \int \frac {1}{x^7 (x-\log (5 x))}dx-3 \int \frac {1}{x^6 (x-\log (5 x))^2}dx-\int \frac {1}{x^5 (x-\log (5 x))^2}dx+4 \int \frac {1}{x^5 (x-\log (5 x))}dx+\int \frac {1}{x^4 (x-\log (5 x))^2}dx+\frac {1}{x^4}\right )\) |
Input:
Int[(48 - 336*x - 16*x^2 + 80*x^3 - 64*x^4 + (288 - 64*x^2 + 128*x^3)*Log[ 5*x] - 64*x^2*Log[5*x]^2)/(225*x^9 - 450*x^8*Log[5*x] + 225*x^7*Log[5*x]^2 ),x]
Output:
$Aborted
Time = 0.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {16}{225 x^{4}}-\frac {16 \left (x^{2}-3\right )}{225 x^{6} \left (x -\ln \left (5 x \right )\right )}\) | \(27\) |
parallelrisch | \(\frac {48+16 x^{3}-16 x^{2} \ln \left (5 x \right )-16 x^{2}}{225 x^{6} \left (x -\ln \left (5 x \right )\right )}\) | \(37\) |
derivativedivides | \(-\frac {16 \left (375+125 x^{3}-125 x^{2}-125 x^{2} \ln \left (5 x \right )\right )}{5625 x^{6} \left (5 \ln \left (5 x \right )-5 x \right )}\) | \(39\) |
default | \(-\frac {16 \left (375+125 x^{3}-125 x^{2}-125 x^{2} \ln \left (5 x \right )\right )}{5625 x^{6} \left (5 \ln \left (5 x \right )-5 x \right )}\) | \(39\) |
Input:
int((-64*x^2*ln(5*x)^2+(128*x^3-64*x^2+288)*ln(5*x)-64*x^4+80*x^3-16*x^2-3 36*x+48)/(225*x^7*ln(5*x)^2-450*x^8*ln(5*x)+225*x^9),x,method=_RETURNVERBO SE)
Output:
16/225/x^4-16/225*(x^2-3)/x^6/(x-ln(5*x))
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {16 \, {\left (x^{3} - x^{2} \log \left (5 \, x\right ) - x^{2} + 3\right )}}{225 \, {\left (x^{7} - x^{6} \log \left (5 \, x\right )\right )}} \] Input:
integrate((-64*x^2*log(5*x)^2+(128*x^3-64*x^2+288)*log(5*x)-64*x^4+80*x^3- 16*x^2-336*x+48)/(225*x^7*log(5*x)^2-450*x^8*log(5*x)+225*x^9),x, algorith m="fricas")
Output:
16/225*(x^3 - x^2*log(5*x) - x^2 + 3)/(x^7 - x^6*log(5*x))
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {16 x^{2} - 48}{- 225 x^{7} + 225 x^{6} \log {\left (5 x \right )}} + \frac {16}{225 x^{4}} \] Input:
integrate((-64*x**2*ln(5*x)**2+(128*x**3-64*x**2+288)*ln(5*x)-64*x**4+80*x **3-16*x**2-336*x+48)/(225*x**7*ln(5*x)**2-450*x**8*ln(5*x)+225*x**9),x)
Output:
(16*x**2 - 48)/(-225*x**7 + 225*x**6*log(5*x)) + 16/(225*x**4)
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {16 \, {\left (x^{3} - x^{2} {\left (\log \left (5\right ) + 1\right )} - x^{2} \log \left (x\right ) + 3\right )}}{225 \, {\left (x^{7} - x^{6} \log \left (5\right ) - x^{6} \log \left (x\right )\right )}} \] Input:
integrate((-64*x^2*log(5*x)^2+(128*x^3-64*x^2+288)*log(5*x)-64*x^4+80*x^3- 16*x^2-336*x+48)/(225*x^7*log(5*x)^2-450*x^8*log(5*x)+225*x^9),x, algorith m="maxima")
Output:
16/225*(x^3 - x^2*(log(5) + 1) - x^2*log(x) + 3)/(x^7 - x^6*log(5) - x^6*l og(x))
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=-\frac {16 \, {\left (x^{2} - 3\right )}}{225 \, {\left (x^{7} - x^{6} \log \left (5 \, x\right )\right )}} + \frac {16}{225 \, x^{4}} \] Input:
integrate((-64*x^2*log(5*x)^2+(128*x^3-64*x^2+288)*log(5*x)-64*x^4+80*x^3- 16*x^2-336*x+48)/(225*x^7*log(5*x)^2-450*x^8*log(5*x)+225*x^9),x, algorith m="giac")
Output:
-16/225*(x^2 - 3)/(x^7 - x^6*log(5*x)) + 16/225/x^4
Time = 1.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {16}{225\,x^4}-\frac {\frac {16\,x^2}{225}-\frac {16}{75}}{x^6\,\left (x-\ln \left (5\,x\right )\right )} \] Input:
int(-(336*x - log(5*x)*(128*x^3 - 64*x^2 + 288) + 16*x^2 - 80*x^3 + 64*x^4 + 64*x^2*log(5*x)^2 - 48)/(225*x^9 - 450*x^8*log(5*x) + 225*x^7*log(5*x)^ 2),x)
Output:
16/(225*x^4) - ((16*x^2)/225 - 16/75)/(x^6*(x - log(5*x)))
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {48-336 x-16 x^2+80 x^3-64 x^4+\left (288-64 x^2+128 x^3\right ) \log (5 x)-64 x^2 \log ^2(5 x)}{225 x^9-450 x^8 \log (5 x)+225 x^7 \log ^2(5 x)} \, dx=\frac {\frac {16 \,\mathrm {log}\left (5 x \right ) x^{2}}{225}-\frac {16 x^{3}}{225}+\frac {16 x^{2}}{225}-\frac {16}{75}}{x^{6} \left (\mathrm {log}\left (5 x \right )-x \right )} \] Input:
int((-64*x^2*log(5*x)^2+(128*x^3-64*x^2+288)*log(5*x)-64*x^4+80*x^3-16*x^2 -336*x+48)/(225*x^7*log(5*x)^2-450*x^8*log(5*x)+225*x^9),x)
Output:
(16*(log(5*x)*x**2 - x**3 + x**2 - 3))/(225*x**6*(log(5*x) - x))