Integrand size = 84, antiderivative size = 26 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {2}{5} x \left (x+\frac {3}{8 x \left (-3+x+x^2\right ) \log (x)}\right ) \] Output:
2/5*x*(3/8/x/ln(x)/(x^2+x-3)+x)
Time = 5.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {1}{20} \left (8 x^2+\frac {3}{\left (-3+x+x^2\right ) \log (x)}\right ) \] Input:
Integrate[(9 - 3*x - 3*x^2 + (-3*x - 6*x^2)*Log[x] + (144*x^2 - 96*x^3 - 8 0*x^4 + 32*x^5 + 16*x^6)*Log[x]^2)/((180*x - 120*x^2 - 100*x^3 + 40*x^4 + 20*x^5)*Log[x]^2),x]
Output:
(8*x^2 + 3/((-3 + x + x^2)*Log[x]))/20
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.
\(\displaystyle \int \frac {-3 x^2+\left (-6 x^2-3 x\right ) \log (x)+\left (16 x^6+32 x^5-80 x^4-96 x^3+144 x^2\right ) \log ^2(x)-3 x+9}{\left (20 x^5+40 x^4-100 x^3-120 x^2+180 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-3 x^2+\left (-6 x^2-3 x\right ) \log (x)+\left (16 x^6+32 x^5-80 x^4-96 x^3+144 x^2\right ) \log ^2(x)-3 x+9}{x \left (20 x^4+40 x^3-100 x^2-120 x+180\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {-3 x^2+\left (-6 x^2-3 x\right ) \log (x)+\left (16 x^6+32 x^5-80 x^4-96 x^3+144 x^2\right ) \log ^2(x)-3 x+9}{20 x \left (x^2+x-3\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \int \frac {-3 x^2-3 x+16 \left (x^6+2 x^5-5 x^4-6 x^3+9 x^2\right ) \log ^2(x)-3 \left (2 x^2+x\right ) \log (x)+9}{x \left (-x^2-x+3\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {1}{20} \int \left (16 x-\frac {3 (2 x+1)}{\left (x^2+x-3\right )^2 \log (x)}-\frac {3}{\left (x^2+x-3\right ) \log ^2(x) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{20} \left (-3 \int \frac {1}{x \left (x^2+x-3\right ) \log ^2(x)}dx-3 \int \frac {2 x+1}{\left (x^2+x-3\right )^2 \log (x)}dx+8 x^2\right )\) |
Input:
Int[(9 - 3*x - 3*x^2 + (-3*x - 6*x^2)*Log[x] + (144*x^2 - 96*x^3 - 80*x^4 + 32*x^5 + 16*x^6)*Log[x]^2)/((180*x - 120*x^2 - 100*x^3 + 40*x^4 + 20*x^5 )*Log[x]^2),x]
Output:
$Aborted
Time = 1.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {2 x^{2}}{5}+\frac {3}{20 \left (x^{2}+x -3\right ) \ln \left (x \right )}\) | \(21\) |
norman | \(\frac {\frac {3}{20}-\frac {18 \ln \left (x \right )}{5}+\frac {6 x \ln \left (x \right )}{5}+\frac {2 x^{3} \ln \left (x \right )}{5}+\frac {2 x^{4} \ln \left (x \right )}{5}}{\left (x^{2}+x -3\right ) \ln \left (x \right )}\) | \(39\) |
default | \(\frac {3-72 \ln \left (x \right )+24 x \ln \left (x \right )+8 x^{3} \ln \left (x \right )+8 x^{4} \ln \left (x \right )}{20 \left (x^{2}+x -3\right ) \ln \left (x \right )}\) | \(40\) |
parallelrisch | \(\frac {3-72 \ln \left (x \right )+24 x \ln \left (x \right )+8 x^{3} \ln \left (x \right )+8 x^{4} \ln \left (x \right )}{20 \left (x^{2}+x -3\right ) \ln \left (x \right )}\) | \(40\) |
Input:
int(((16*x^6+32*x^5-80*x^4-96*x^3+144*x^2)*ln(x)^2+(-6*x^2-3*x)*ln(x)-3*x^ 2-3*x+9)/(20*x^5+40*x^4-100*x^3-120*x^2+180*x)/ln(x)^2,x,method=_RETURNVER BOSE)
Output:
2/5*x^2+3/20/(x^2+x-3)/ln(x)
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {8 \, {\left (x^{4} + x^{3} - 3 \, x^{2}\right )} \log \left (x\right ) + 3}{20 \, {\left (x^{2} + x - 3\right )} \log \left (x\right )} \] Input:
integrate(((16*x^6+32*x^5-80*x^4-96*x^3+144*x^2)*log(x)^2+(-6*x^2-3*x)*log (x)-3*x^2-3*x+9)/(20*x^5+40*x^4-100*x^3-120*x^2+180*x)/log(x)^2,x, algorit hm="fricas")
Output:
1/20*(8*(x^4 + x^3 - 3*x^2)*log(x) + 3)/((x^2 + x - 3)*log(x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {2 x^{2}}{5} + \frac {3}{\left (20 x^{2} + 20 x - 60\right ) \log {\left (x \right )}} \] Input:
integrate(((16*x**6+32*x**5-80*x**4-96*x**3+144*x**2)*ln(x)**2+(-6*x**2-3* x)*ln(x)-3*x**2-3*x+9)/(20*x**5+40*x**4-100*x**3-120*x**2+180*x)/ln(x)**2, x)
Output:
2*x**2/5 + 3/((20*x**2 + 20*x - 60)*log(x))
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {8 \, {\left (x^{4} + x^{3} - 3 \, x^{2}\right )} \log \left (x\right ) + 3}{20 \, {\left (x^{2} + x - 3\right )} \log \left (x\right )} \] Input:
integrate(((16*x^6+32*x^5-80*x^4-96*x^3+144*x^2)*log(x)^2+(-6*x^2-3*x)*log (x)-3*x^2-3*x+9)/(20*x^5+40*x^4-100*x^3-120*x^2+180*x)/log(x)^2,x, algorit hm="maxima")
Output:
1/20*(8*(x^4 + x^3 - 3*x^2)*log(x) + 3)/((x^2 + x - 3)*log(x))
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {2}{5} \, x^{2} + \frac {3}{20 \, {\left (x^{2} \log \left (x\right ) + x \log \left (x\right ) - 3 \, \log \left (x\right )\right )}} \] Input:
integrate(((16*x^6+32*x^5-80*x^4-96*x^3+144*x^2)*log(x)^2+(-6*x^2-3*x)*log (x)-3*x^2-3*x+9)/(20*x^5+40*x^4-100*x^3-120*x^2+180*x)/log(x)^2,x, algorit hm="giac")
Output:
2/5*x^2 + 3/20/(x^2*log(x) + x*log(x) - 3*log(x))
Time = 2.95 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {\frac {2\,x^4}{5}+\frac {2\,x^3}{5}-\frac {6\,x^2}{5}}{x^2+x-3}+\frac {3}{20\,\ln \left (x\right )\,\left (x^2+x-3\right )} \] Input:
int(-(3*x - log(x)^2*(144*x^2 - 96*x^3 - 80*x^4 + 32*x^5 + 16*x^6) + log(x )*(3*x + 6*x^2) + 3*x^2 - 9)/(log(x)^2*(180*x - 120*x^2 - 100*x^3 + 40*x^4 + 20*x^5)),x)
Output:
((2*x^3)/5 - (6*x^2)/5 + (2*x^4)/5)/(x + x^2 - 3) + 3/(20*log(x)*(x + x^2 - 3))
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {9-3 x-3 x^2+\left (-3 x-6 x^2\right ) \log (x)+\left (144 x^2-96 x^3-80 x^4+32 x^5+16 x^6\right ) \log ^2(x)}{\left (180 x-120 x^2-100 x^3+40 x^4+20 x^5\right ) \log ^2(x)} \, dx=\frac {8 \,\mathrm {log}\left (x \right ) x^{4}+8 \,\mathrm {log}\left (x \right ) x^{3}-24 \,\mathrm {log}\left (x \right ) x^{2}+3}{20 \,\mathrm {log}\left (x \right ) \left (x^{2}+x -3\right )} \] Input:
int(((16*x^6+32*x^5-80*x^4-96*x^3+144*x^2)*log(x)^2+(-6*x^2-3*x)*log(x)-3* x^2-3*x+9)/(20*x^5+40*x^4-100*x^3-120*x^2+180*x)/log(x)^2,x)
Output:
(8*log(x)*x**4 + 8*log(x)*x**3 - 24*log(x)*x**2 + 3)/(20*log(x)*(x**2 + x - 3))