Integrand size = 162, antiderivative size = 24 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {(4+x)^2}{x^2 \left (3+\left (-\log \left (\frac {1}{x}\right )+\log (x)\right )^2\right )} \] Output:
(4+x)^2/x^2/(3+(ln(x)-ln(1/x))^2)
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {(4+x)^2}{x^2 \left (3+\log ^2\left (\frac {1}{x}\right )-2 \log \left (\frac {1}{x}\right ) \log (x)+\log ^2(x)\right )} \] Input:
Integrate[(-96 - 24*x + (64 + 32*x + 4*x^2)*Log[x^(-1)] + (-32 - 8*x)*Log[ x^(-1)]^2 + (-64 - 32*x - 4*x^2 + (64 + 16*x)*Log[x^(-1)])*Log[x] + (-32 - 8*x)*Log[x]^2)/(9*x^3 + 6*x^3*Log[x^(-1)]^2 + x^3*Log[x^(-1)]^4 + (-12*x^ 3*Log[x^(-1)] - 4*x^3*Log[x^(-1)]^3)*Log[x] + (6*x^3 + 6*x^3*Log[x^(-1)]^2 )*Log[x]^2 - 4*x^3*Log[x^(-1)]*Log[x]^3 + x^3*Log[x]^4),x]
Output:
(4 + x)^2/(x^2*(3 + Log[x^(-1)]^2 - 2*Log[x^(-1)]*Log[x] + Log[x]^2))
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.
| \(\displaystyle \int \frac {\left (4 x^2+32 x+64\right ) \log \left (\frac {1}{x}\right )+\left (-4 x^2-32 x+(16 x+64) \log \left (\frac {1}{x}\right )-64\right ) \log (x)-24 x+(-8 x-32) \log ^2\left (\frac {1}{x}\right )+(-8 x-32) \log ^2(x)-96}{9 x^3+x^3 \log ^4\left (\frac {1}{x}\right )+x^3 \log ^4(x)-4 x^3 \log ^3(x) \log \left (\frac {1}{x}\right )+\left (-4 x^3 \log ^3\left (\frac {1}{x}\right )-12 x^3 \log \left (\frac {1}{x}\right )\right ) \log (x)+6 x^3 \log ^2\left (\frac {1}{x}\right )+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {4 (x+4) \left (-2 \log ^2\left (\frac {1}{x}\right )-2 \log ^2(x)+(x+4 \log (x)+4) \log \left (\frac {1}{x}\right )-(x+4) \log (x)-6\right )}{x^3 \left (\log ^2\left (\frac {1}{x}\right )+\log ^2(x)-2 \log (x) \log \left (\frac {1}{x}\right )+3\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int -\frac {(x+4) \left (2 \log ^2\left (\frac {1}{x}\right )-(x+4 \log (x)+4) \log \left (\frac {1}{x}\right )+2 \log ^2(x)+(x+4) \log (x)+6\right )}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \int \frac {(x+4) \left (2 \log ^2\left (\frac {1}{x}\right )-(x+4 \log (x)+4) \log \left (\frac {1}{x}\right )+2 \log ^2(x)+(x+4) \log (x)+6\right )}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -4 \int \left (\frac {2 (x+4)}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )}-\frac {(x+4)^2 \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \left (-16 \int \frac {\log \left (\frac {1}{x}\right )}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx+16 \int \frac {\log (x)}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx+8 \int \frac {1}{x^3 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )}dx-8 \int \frac {\log \left (\frac {1}{x}\right )}{x^2 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx+8 \int \frac {\log (x)}{x^2 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx+2 \int \frac {1}{x^2 \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )}dx-\int \frac {\log \left (\frac {1}{x}\right )}{x \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx+\int \frac {\log (x)}{x \left (\log ^2\left (\frac {1}{x}\right )-2 \log (x) \log \left (\frac {1}{x}\right )+\log ^2(x)+3\right )^2}dx\right )\) |
Input:
Int[(-96 - 24*x + (64 + 32*x + 4*x^2)*Log[x^(-1)] + (-32 - 8*x)*Log[x^(-1) ]^2 + (-64 - 32*x - 4*x^2 + (64 + 16*x)*Log[x^(-1)])*Log[x] + (-32 - 8*x)* Log[x]^2)/(9*x^3 + 6*x^3*Log[x^(-1)]^2 + x^3*Log[x^(-1)]^4 + (-12*x^3*Log[ x^(-1)] - 4*x^3*Log[x^(-1)]^3)*Log[x] + (6*x^3 + 6*x^3*Log[x^(-1)]^2)*Log[ x]^2 - 4*x^3*Log[x^(-1)]*Log[x]^3 + x^3*Log[x]^4),x]
Output:
$Aborted
Time = 1.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(\frac {x^{2}+8 x +16}{x^{2} \left (4 \ln \left (x \right )^{2}+3\right )}\) | \(23\) |
| parallelrisch | \(\frac {6 x^{2}+48 x +96}{6 x^{2} \left (\ln \left (x \right )^{2}-2 \ln \left (x \right ) \ln \left (\frac {1}{x}\right )+\ln \left (\frac {1}{x}\right )^{2}+3\right )}\) | \(38\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(936\) |
| default | \(\text {Expression too large to display}\) | \(936\) |
Input:
int(((-8*x-32)*ln(x)^2+((16*x+64)*ln(1/x)-4*x^2-32*x-64)*ln(x)+(-8*x-32)*l n(1/x)^2+(4*x^2+32*x+64)*ln(1/x)-24*x-96)/(x^3*ln(x)^4-4*x^3*ln(1/x)*ln(x) ^3+(6*x^3*ln(1/x)^2+6*x^3)*ln(x)^2+(-4*x^3*ln(1/x)^3-12*x^3*ln(1/x))*ln(x) +x^3*ln(1/x)^4+6*x^3*ln(1/x)^2+9*x^3),x,method=_RETURNVERBOSE)
Output:
(x^2+8*x+16)/x^2/(4*ln(x)^2+3)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {x^{2} + 8 \, x + 16}{4 \, x^{2} \log \left (\frac {1}{x}\right )^{2} + 3 \, x^{2}} \] Input:
integrate(((-8*x-32)*log(x)^2+((16*x+64)*log(1/x)-4*x^2-32*x-64)*log(x)+(- 8*x-32)*log(1/x)^2+(4*x^2+32*x+64)*log(1/x)-24*x-96)/(x^3*log(x)^4-4*x^3*l og(1/x)*log(x)^3+(6*x^3*log(1/x)^2+6*x^3)*log(x)^2+(-4*x^3*log(1/x)^3-12*x ^3*log(1/x))*log(x)+x^3*log(1/x)^4+6*x^3*log(1/x)^2+9*x^3),x, algorithm="f ricas")
Output:
(x^2 + 8*x + 16)/(4*x^2*log(1/x)^2 + 3*x^2)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {x^{2} + 8 x + 16}{4 x^{2} \log {\left (x \right )}^{2} + 3 x^{2}} \] Input:
integrate(((-8*x-32)*ln(x)**2+((16*x+64)*ln(1/x)-4*x**2-32*x-64)*ln(x)+(-8 *x-32)*ln(1/x)**2+(4*x**2+32*x+64)*ln(1/x)-24*x-96)/(x**3*ln(x)**4-4*x**3* ln(1/x)*ln(x)**3+(6*x**3*ln(1/x)**2+6*x**3)*ln(x)**2+(-4*x**3*ln(1/x)**3-1 2*x**3*ln(1/x))*ln(x)+x**3*ln(1/x)**4+6*x**3*ln(1/x)**2+9*x**3),x)
Output:
(x**2 + 8*x + 16)/(4*x**2*log(x)**2 + 3*x**2)
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {x^{2} + 8 \, x + 16}{4 \, x^{2} \log \left (x\right )^{2} + 3 \, x^{2}} \] Input:
integrate(((-8*x-32)*log(x)^2+((16*x+64)*log(1/x)-4*x^2-32*x-64)*log(x)+(- 8*x-32)*log(1/x)^2+(4*x^2+32*x+64)*log(1/x)-24*x-96)/(x^3*log(x)^4-4*x^3*l og(1/x)*log(x)^3+(6*x^3*log(1/x)^2+6*x^3)*log(x)^2+(-4*x^3*log(1/x)^3-12*x ^3*log(1/x))*log(x)+x^3*log(1/x)^4+6*x^3*log(1/x)^2+9*x^3),x, algorithm="m axima")
Output:
(x^2 + 8*x + 16)/(4*x^2*log(x)^2 + 3*x^2)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {x^{2} + 8 \, x + 16}{4 \, x^{2} \log \left (x\right )^{2} + 3 \, x^{2}} \] Input:
integrate(((-8*x-32)*log(x)^2+((16*x+64)*log(1/x)-4*x^2-32*x-64)*log(x)+(- 8*x-32)*log(1/x)^2+(4*x^2+32*x+64)*log(1/x)-24*x-96)/(x^3*log(x)^4-4*x^3*l og(1/x)*log(x)^3+(6*x^3*log(1/x)^2+6*x^3)*log(x)^2+(-4*x^3*log(1/x)^3-12*x ^3*log(1/x))*log(x)+x^3*log(1/x)^4+6*x^3*log(1/x)^2+9*x^3),x, algorithm="g iac")
Output:
(x^2 + 8*x + 16)/(4*x^2*log(x)^2 + 3*x^2)
Time = 2.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {x^2+8\,x+16}{x^2\,{\ln \left (\frac {1}{x}\right )}^2-2\,x^2\,\ln \left (\frac {1}{x}\right )\,\ln \left (x\right )+x^2\,{\ln \left (x\right )}^2+3\,x^2} \] Input:
int(-(24*x - log(1/x)*(32*x + 4*x^2 + 64) + log(1/x)^2*(8*x + 32) + log(x) *(32*x + 4*x^2 - log(1/x)*(16*x + 64) + 64) + log(x)^2*(8*x + 32) + 96)/(l og(x)^2*(6*x^3 + 6*x^3*log(1/x)^2) + x^3*log(x)^4 + 9*x^3 + 6*x^3*log(1/x) ^2 + x^3*log(1/x)^4 - log(x)*(12*x^3*log(1/x) + 4*x^3*log(1/x)^3) - 4*x^3* log(1/x)*log(x)^3),x)
Output:
(8*x + x^2 + 16)/(x^2*log(x)^2 + 3*x^2 + x^2*log(1/x)^2 - 2*x^2*log(1/x)*l og(x))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-96-24 x+\left (64+32 x+4 x^2\right ) \log \left (\frac {1}{x}\right )+(-32-8 x) \log ^2\left (\frac {1}{x}\right )+\left (-64-32 x-4 x^2+(64+16 x) \log \left (\frac {1}{x}\right )\right ) \log (x)+(-32-8 x) \log ^2(x)}{9 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )+x^3 \log ^4\left (\frac {1}{x}\right )+\left (-12 x^3 \log \left (\frac {1}{x}\right )-4 x^3 \log ^3\left (\frac {1}{x}\right )\right ) \log (x)+\left (6 x^3+6 x^3 \log ^2\left (\frac {1}{x}\right )\right ) \log ^2(x)-4 x^3 \log \left (\frac {1}{x}\right ) \log ^3(x)+x^3 \log ^4(x)} \, dx=\frac {-\frac {4 \mathrm {log}\left (x \right )^{2} x^{2}}{3}+8 x +16}{x^{2} \left (4 \mathrm {log}\left (x \right )^{2}+3\right )} \] Input:
int(((-8*x-32)*log(x)^2+((16*x+64)*log(1/x)-4*x^2-32*x-64)*log(x)+(-8*x-32 )*log(1/x)^2+(4*x^2+32*x+64)*log(1/x)-24*x-96)/(x^3*log(x)^4-4*x^3*log(1/x )*log(x)^3+(6*x^3*log(1/x)^2+6*x^3)*log(x)^2+(-4*x^3*log(1/x)^3-12*x^3*log (1/x))*log(x)+x^3*log(1/x)^4+6*x^3*log(1/x)^2+9*x^3),x)
Output:
(4*( - log(x)**2*x**2 + 6*x + 12))/(3*x**2*(4*log(x)**2 + 3))