Integrand size = 80, antiderivative size = 28 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {\log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{4 x^2 (1+\log (x))^2} \] Output:
1/4*ln(ln(3)^2/exp(4)^2/exp(x)^2)^2/x^2/(1+ln(x))^2
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {\log ^2\left (e^{-2 (4+x)} \log ^2(3)\right )}{4 x^2 (1+\log (x))^2} \] Input:
Integrate[((-2*x - 2*x*Log[x])*Log[E^(-8 - 2*x)*Log[3]^2] + (-2 - Log[x])* Log[E^(-8 - 2*x)*Log[3]^2]^2)/(2*x^3 + 6*x^3*Log[x] + 6*x^3*Log[x]^2 + 2*x ^3*Log[x]^3),x]
Output:
Log[Log[3]^2/E^(2*(4 + x))]^2/(4*x^2*(1 + 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.
| \(\displaystyle \int \frac {(-\log (x)-2) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )+(-2 x-2 x \log (x)) \log \left (e^{-2 x-8} \log ^2(3)\right )}{2 x^3+2 x^3 \log ^3(x)+6 x^3 \log ^2(x)+6 x^3 \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {(-\log (x)-2) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )+(-2 x-2 x \log (x)) \log \left (e^{-2 x-8} \log ^2(3)\right )}{2 x^3 (\log (x)+1)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {(\log (x)+2) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )+2 (\log (x) x+x) \log \left (e^{-2 x-8} \log ^2(3)\right )}{x^3 (\log (x)+1)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {(\log (x)+2) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )+2 (\log (x) x+x) \log \left (e^{-2 x-8} \log ^2(3)\right )}{x^3 (\log (x)+1)^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {(\log (x)+2) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )}{x^3 (\log (x)+1)^3}+\frac {2 \log \left (e^{-2 x-8} \log ^2(3)\right )}{x^2 (\log (x)+1)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-2 \int \frac {\log ^2\left (e^{-2 x-8} \log ^2(3)\right )}{x^3 (\log (x)+1)^3}dx-\int \frac {\log (x) \log ^2\left (e^{-2 x-8} \log ^2(3)\right )}{x^3 (\log (x)+1)^3}dx-2 \int \frac {\log \left (e^{-2 x-8} \log ^2(3)\right )}{x^2 (\log (x)+1)^2}dx\right )\) |
Input:
Int[((-2*x - 2*x*Log[x])*Log[E^(-8 - 2*x)*Log[3]^2] + (-2 - Log[x])*Log[E^ (-8 - 2*x)*Log[3]^2]^2)/(2*x^3 + 6*x^3*Log[x] + 6*x^3*Log[x]^2 + 2*x^3*Log [x]^3),x]
Output:
$Aborted
Time = 2.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
| method | result | size |
| parallelrisch | \(\frac {\ln \left (\ln \left (3\right )^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )^{2}}{4 x^{2} \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}\) | \(34\) |
| default | \(\frac {2 \ln \left ({\mathrm e}^{x}\right )-2 x}{x \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}-\frac {\left (\ln \left ({\mathrm e}^{x}\right )-x \right ) \left (\ln \left (\ln \left (3\right )^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )}{x^{2} \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}+\frac {1}{\left (\ln \left (x \right )+1\right )^{2}}+\frac {{\left (\ln \left ({\mathrm e}^{x}\right )-x \right )}^{2}}{x^{2} \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}-\frac {\ln \left (\ln \left (3\right )^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )}{x \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}+\frac {{\left (\ln \left (\ln \left (3\right )^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )}^{2}}{4 x^{2} \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}\) | \(177\) |
| risch | \(\frac {\ln \left ({\mathrm e}^{x}\right )^{2}}{x^{2} \left (\ln \left (x \right )+1\right )^{2}}-\frac {\left (i \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-16+i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+4 \ln \left (\ln \left (3\right )\right )\right ) \ln \left ({\mathrm e}^{x}\right )}{2 x^{2} \left (\ln \left (x \right )+1\right )^{2}}+\frac {256+16 \ln \left (\ln \left (3\right )\right )^{2}-128 \ln \left (\ln \left (3\right )\right )-32 i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-\pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{6}+4 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{5}+4 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{3} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-6 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{4}-\pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{4} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+8 i \ln \left (\ln \left (3\right )\right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )-16 i \ln \left (\ln \left (3\right )\right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-32 i \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )+64 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+8 i \ln \left (\ln \left (3\right )\right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{16 x^{2} \left (\ln \left (x \right )+1\right )^{2}}\) | \(339\) |
Input:
int(((-ln(x)-2)*ln(ln(3)^2/exp(4)^2/exp(x)^2)^2+(-2*x*ln(x)-2*x)*ln(ln(3)^ 2/exp(4)^2/exp(x)^2))/(2*x^3*ln(x)^3+6*x^3*ln(x)^2+6*x^3*ln(x)+2*x^3),x,me thod=_RETURNVERBOSE)
Output:
1/4/x^2*ln(ln(3)^2/exp(4)^2/exp(x)^2)^2/(ln(x)^2+2*ln(x)+1)
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {4 \, x^{2} - 4 \, {\left (x + 4\right )} \log \left (\log \left (3\right )^{2}\right ) + \log \left (\log \left (3\right )^{2}\right )^{2} + 32 \, x + 64}{4 \, {\left (x^{2} \log \left (x\right )^{2} + 2 \, x^{2} \log \left (x\right ) + x^{2}\right )}} \] Input:
integrate(((-log(x)-2)*log(log(3)^2/exp(4)^2/exp(x)^2)^2+(-2*x*log(x)-2*x) *log(log(3)^2/exp(4)^2/exp(x)^2))/(2*x^3*log(x)^3+6*x^3*log(x)^2+6*x^3*log (x)+2*x^3),x, algorithm="fricas")
Output:
1/4*(4*x^2 - 4*(x + 4)*log(log(3)^2) + log(log(3)^2)^2 + 32*x + 64)/(x^2*l og(x)^2 + 2*x^2*log(x) + x^2)
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {x^{2} - 2 x \log {\left (\log {\left (3 \right )} \right )} + 8 x - 8 \log {\left (\log {\left (3 \right )} \right )} + \log {\left (\log {\left (3 \right )} \right )}^{2} + 16}{x^{2} \log {\left (x \right )}^{2} + 2 x^{2} \log {\left (x \right )} + x^{2}} \] Input:
integrate(((-ln(x)-2)*ln(ln(3)**2/exp(4)**2/exp(x)**2)**2+(-2*x*ln(x)-2*x) *ln(ln(3)**2/exp(4)**2/exp(x)**2))/(2*x**3*ln(x)**3+6*x**3*ln(x)**2+6*x**3 *ln(x)+2*x**3),x)
Output:
(x**2 - 2*x*log(log(3)) + 8*x - 8*log(log(3)) + log(log(3))**2 + 16)/(x**2 *log(x)**2 + 2*x**2*log(x) + x**2)
Time = 0.14 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {x^{2} - 2 \, x {\left (\log \left (\log \left (3\right )\right ) - 4\right )} + \log \left (\log \left (3\right )\right )^{2} - 8 \, \log \left (\log \left (3\right )\right ) + 16}{x^{2} \log \left (x\right )^{2} + 2 \, x^{2} \log \left (x\right ) + x^{2}} \] Input:
integrate(((-log(x)-2)*log(log(3)^2/exp(4)^2/exp(x)^2)^2+(-2*x*log(x)-2*x) *log(log(3)^2/exp(4)^2/exp(x)^2))/(2*x^3*log(x)^3+6*x^3*log(x)^2+6*x^3*log (x)+2*x^3),x, algorithm="maxima")
Output:
(x^2 - 2*x*(log(log(3)) - 4) + log(log(3))^2 - 8*log(log(3)) + 16)/(x^2*lo g(x)^2 + 2*x^2*log(x) + x^2)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {x^{2} - 2 \, x \log \left (\log \left (3\right )\right ) + \log \left (\log \left (3\right )\right )^{2} + 8 \, x - 8 \, \log \left (\log \left (3\right )\right ) + 16}{x^{2} \log \left (x\right )^{2} + 2 \, x^{2} \log \left (x\right ) + x^{2}} \] Input:
integrate(((-log(x)-2)*log(log(3)^2/exp(4)^2/exp(x)^2)^2+(-2*x*log(x)-2*x) *log(log(3)^2/exp(4)^2/exp(x)^2))/(2*x^3*log(x)^3+6*x^3*log(x)^2+6*x^3*log (x)+2*x^3),x, algorithm="giac")
Output:
(x^2 - 2*x*log(log(3)) + log(log(3))^2 + 8*x - 8*log(log(3)) + 16)/(x^2*lo g(x)^2 + 2*x^2*log(x) + x^2)
Time = 2.75 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=-\frac {{\ln \left (x\right )}^2+2\,\ln \left (x\right )}{{\left (\ln \left (x\right )+1\right )}^2}-\frac {8\,\ln \left (\ln \left (3\right )\right )-{\ln \left (\ln \left (3\right )\right )}^2+x\,\left (2\,\ln \left (\ln \left (3\right )\right )-8\right )-16}{x^2\,{\left (\ln \left (x\right )+1\right )}^2} \] Input:
int(-(log(exp(-2*x)*exp(-8)*log(3)^2)^2*(log(x) + 2) + log(exp(-2*x)*exp(- 8)*log(3)^2)*(2*x + 2*x*log(x)))/(6*x^3*log(x) + 6*x^3*log(x)^2 + 2*x^3*lo g(x)^3 + 2*x^3),x)
Output:
- (2*log(x) + log(x)^2)/(log(x) + 1)^2 - (8*log(log(3)) - log(log(3))^2 + x*(2*log(log(3)) - 8) - 16)/(x^2*(log(x) + 1)^2)
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (3\right )^{2}}{e^{2 x} e^{8}}\right )^{2}}{4 x^{2} \left (\mathrm {log}\left (x \right )^{2}+2 \,\mathrm {log}\left (x \right )+1\right )} \] Input:
int(((-log(x)-2)*log(log(3)^2/exp(4)^2/exp(x)^2)^2+(-2*x*log(x)-2*x)*log(l og(3)^2/exp(4)^2/exp(x)^2))/(2*x^3*log(x)^3+6*x^3*log(x)^2+6*x^3*log(x)+2* x^3),x)
Output:
log(log(3)**2/(e**(2*x)*e**8))**2/(4*x**2*(log(x)**2 + 2*log(x) + 1))