Integrand size = 54, antiderivative size = 30 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\log \left (\frac {1}{3} x \left ((-3-x)^2-x \left (-5-\frac {4 x}{5}+\frac {2}{\log (x)}\right )\right )\right ) \] Output:
ln(x*(-1/3*x*(-4/5*x+2/ln(x)-5)+1/3*(-3-x)^2))
Time = 7.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\log (x)-\log (\log (x))+\log \left (-10 x+45 \log (x)+55 x \log (x)+9 x^2 \log (x)\right ) \] Input:
Integrate[(10*x - 20*x*Log[x] + (45 + 110*x + 27*x^2)*Log[x]^2)/(-10*x^2*L og[x] + (45*x + 55*x^2 + 9*x^3)*Log[x]^2),x]
Output:
Log[x] - Log[Log[x]] + Log[-10*x + 45*Log[x] + 55*x*Log[x] + 9*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.
\(\displaystyle \int \frac {\left (27 x^2+110 x+45\right ) \log ^2(x)+10 x-20 x \log (x)}{\left (9 x^3+55 x^2+45 x\right ) \log ^2(x)-10 x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (27 x^2+110 x+45\right ) \log ^2(x)\right )-10 x+20 x \log (x)}{x \log (x) \left (-9 x^2 \log (x)+10 x-55 x \log (x)-45 \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {27 x^2+110 x+45}{x \left (9 x^2+55 x+45\right )}+\frac {90 \left (x^2-5\right )}{\left (9 x^2+55 x+45\right ) \left (9 x^2 \log (x)-10 x+55 x \log (x)+45 \log (x)\right )}+\frac {9 x^2+55 x+45}{x \left (9 x^2 \log (x)-10 x+55 x \log (x)+45 \log (x)\right )}-\frac {1}{x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 65 \int \frac {1}{9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)}dx+3240 \sqrt {\frac {5}{281}} \int \frac {1}{\left (-18 x+\sqrt {1405}-55\right ) \left (9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)\right )}dx+45 \int \frac {1}{x \left (9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)\right )}dx+9 \int \frac {x}{9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)}dx-\frac {550}{281} \left (281-11 \sqrt {1405}\right ) \int \frac {1}{\left (18 x-\sqrt {1405}+55\right ) \left (9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)\right )}dx-\frac {550}{281} \left (281+11 \sqrt {1405}\right ) \int \frac {1}{\left (18 x+\sqrt {1405}+55\right ) \left (9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)\right )}dx+3240 \sqrt {\frac {5}{281}} \int \frac {1}{\left (18 x+\sqrt {1405}+55\right ) \left (9 \log (x) x^2+55 \log (x) x-10 x+45 \log (x)\right )}dx+\log \left (9 x^2+55 x+45\right )+\log (x)-\log (\log (x))\) |
Input:
Int[(10*x - 20*x*Log[x] + (45 + 110*x + 27*x^2)*Log[x]^2)/(-10*x^2*Log[x] + (45*x + 55*x^2 + 9*x^3)*Log[x]^2),x]
Output:
$Aborted
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (x \right )\right )+\ln \left (x^{2} \ln \left (x \right )+\frac {55 x \ln \left (x \right )}{9}-\frac {10 x}{9}+5 \ln \left (x \right )\right )+\ln \left (x \right )\) | \(29\) |
default | \(\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (9 x^{2} \ln \left (x \right )+55 x \ln \left (x \right )+45 \ln \left (x \right )-10 x \right )\) | \(30\) |
norman | \(\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (9 x^{2} \ln \left (x \right )+55 x \ln \left (x \right )+45 \ln \left (x \right )-10 x \right )\) | \(30\) |
risch | \(\ln \left (9 x^{3}+55 x^{2}+45 x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )-\frac {10 x}{9 x^{2}+55 x +45}\right )\) | \(41\) |
Input:
int(((27*x^2+110*x+45)*ln(x)^2-20*x*ln(x)+10*x)/((9*x^3+55*x^2+45*x)*ln(x) ^2-10*x^2*ln(x)),x,method=_RETURNVERBOSE)
Output:
-ln(ln(x))+ln(x^2*ln(x)+55/9*x*ln(x)-10/9*x+5*ln(x))+ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (25) = 50\).
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\log \left (9 \, x^{3} + 55 \, x^{2} + 45 \, x\right ) + \log \left (\frac {{\left (9 \, x^{2} + 55 \, x + 45\right )} \log \left (x\right ) - 10 \, x}{9 \, x^{2} + 55 \, x + 45}\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate(((27*x^2+110*x+45)*log(x)^2-20*x*log(x)+10*x)/((9*x^3+55*x^2+45* x)*log(x)^2-10*x^2*log(x)),x, algorithm="fricas")
Output:
log(9*x^3 + 55*x^2 + 45*x) + log(((9*x^2 + 55*x + 45)*log(x) - 10*x)/(9*x^ 2 + 55*x + 45)) - log(log(x))
Exception generated. \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \] Input:
integrate(((27*x**2+110*x+45)*ln(x)**2-20*x*ln(x)+10*x)/((9*x**3+55*x**2+4 5*x)*ln(x)**2-10*x**2*ln(x)),x)
Output:
Exception raised: PolynomialError >> 1/(81*x**5 + 990*x**4 + 3835*x**3 + 4 950*x**2 + 2025*x) contains an element of the set of generators.
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\log \left (9 \, x^{2} + 55 \, x + 45\right ) + \log \left (x\right ) + \log \left (\frac {{\left (9 \, x^{2} + 55 \, x + 45\right )} \log \left (x\right ) - 10 \, x}{9 \, x^{2} + 55 \, x + 45}\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate(((27*x^2+110*x+45)*log(x)^2-20*x*log(x)+10*x)/((9*x^3+55*x^2+45* x)*log(x)^2-10*x^2*log(x)),x, algorithm="maxima")
Output:
log(9*x^2 + 55*x + 45) + log(x) + log(((9*x^2 + 55*x + 45)*log(x) - 10*x)/ (9*x^2 + 55*x + 45)) - log(log(x))
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\log \left (9 \, x^{2} \log \left (x\right ) + 55 \, x \log \left (x\right ) - 10 \, x + 45 \, \log \left (x\right )\right ) + \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate(((27*x^2+110*x+45)*log(x)^2-20*x*log(x)+10*x)/((9*x^3+55*x^2+45* x)*log(x)^2-10*x^2*log(x)),x, algorithm="giac")
Output:
log(9*x^2*log(x) + 55*x*log(x) - 10*x + 45*log(x)) + log(x) - log(log(x))
Time = 3.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=\ln \left (45\,\ln \left (x\right )-10\,x+9\,x^2\,\ln \left (x\right )+55\,x\,\ln \left (x\right )\right )-\ln \left (\ln \left (x\right )\right )+\ln \left (x\right ) \] Input:
int(-(10*x + log(x)^2*(110*x + 27*x^2 + 45) - 20*x*log(x))/(10*x^2*log(x) - log(x)^2*(45*x + 55*x^2 + 9*x^3)),x)
Output:
log(45*log(x) - 10*x + 9*x^2*log(x) + 55*x*log(x)) - log(log(x)) + log(x)
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {10 x-20 x \log (x)+\left (45+110 x+27 x^2\right ) \log ^2(x)}{-10 x^2 \log (x)+\left (45 x+55 x^2+9 x^3\right ) \log ^2(x)} \, dx=-\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (9 \,\mathrm {log}\left (x \right ) x^{2}+55 \,\mathrm {log}\left (x \right ) x +45 \,\mathrm {log}\left (x \right )-10 x \right )+\mathrm {log}\left (x \right ) \] Input:
int(((27*x^2+110*x+45)*log(x)^2-20*x*log(x)+10*x)/((9*x^3+55*x^2+45*x)*log (x)^2-10*x^2*log(x)),x)
Output:
- log(log(x)) + log(9*log(x)*x**2 + 55*log(x)*x + 45*log(x) - 10*x) + log (x)