Integrand size = 134, antiderivative size = 31 \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=4 x^2 \log ^4\left (\frac {x^2 \log (3)}{2 (1-x) \log (-3+\log (x))}\right ) \] Output:
4*ln(1/2*x^2/(1-x)*ln(3)/ln(ln(x)-3))^4*x^2
\[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=\int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx \] Input:
Integrate[((16*x - 16*x^2 + (96*x - 48*x^2 + (-32*x + 16*x^2)*Log[x])*Log[ -3 + Log[x]])*Log[-((x^2*Log[3])/((-2 + 2*x)*Log[-3 + Log[x]]))]^3 + (24*x - 24*x^2 + (-8*x + 8*x^2)*Log[x])*Log[-3 + Log[x]]*Log[-((x^2*Log[3])/((- 2 + 2*x)*Log[-3 + Log[x]]))]^4)/((3 - 3*x + (-1 + x)*Log[x])*Log[-3 + Log[ x]]),x]
Output:
Integrate[((16*x - 16*x^2 + (96*x - 48*x^2 + (-32*x + 16*x^2)*Log[x])*Log[ -3 + Log[x]])*Log[-((x^2*Log[3])/((-2 + 2*x)*Log[-3 + Log[x]]))]^3 + (24*x - 24*x^2 + (-8*x + 8*x^2)*Log[x])*Log[-3 + Log[x]]*Log[-((x^2*Log[3])/((- 2 + 2*x)*Log[-3 + Log[x]]))]^4)/((3 - 3*x + (-1 + x)*Log[x])*Log[-3 + Log[ x]]), 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.
| \(\displaystyle \int \frac {\left (-24 x^2+\left (8 x^2-8 x\right ) \log (x)+24 x\right ) \log (\log (x)-3) \log ^4\left (-\frac {x^2 \log (3)}{(2 x-2) \log (\log (x)-3)}\right )+\left (-16 x^2+\left (-48 x^2+\left (16 x^2-32 x\right ) \log (x)+96 x\right ) \log (\log (x)-3)+16 x\right ) \log ^3\left (-\frac {x^2 \log (3)}{(2 x-2) \log (\log (x)-3)}\right )}{(-3 x+(x-1) \log (x)+3) \log (\log (x)-3)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-24 x^2+\left (8 x^2-8 x\right ) \log (x)+24 x\right ) \log (\log (x)-3) \log ^4\left (-\frac {x^2 \log (3)}{(2 x-2) \log (\log (x)-3)}\right )+\left (-16 x^2+\left (-48 x^2+\left (16 x^2-32 x\right ) \log (x)+96 x\right ) \log (\log (x)-3)+16 x\right ) \log ^3\left (-\frac {x^2 \log (3)}{(2 x-2) \log (\log (x)-3)}\right )}{(1-x) (3-\log (x)) \log (\log (x)-3)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (8 x \log ^4\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )+\frac {16 x (-x+x \log (x) \log (\log (x)-3)-3 x \log (\log (x)-3)-2 \log (x) \log (\log (x)-3)+6 \log (\log (x)-3)+1) \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{(x-1) (\log (x)-3) \log (\log (x)-3)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 8 \int x \log ^4\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )dx+48 \int \frac {\log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{\log (x)-3}dx+48 \int \frac {\log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{(x-1) (\log (x)-3)}dx-48 \int \frac {x \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{\log (x)-3}dx-16 \int \frac {\log (x) \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{\log (x)-3}dx-16 \int \frac {\log (x) \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{(x-1) (\log (x)-3)}dx+16 \int \frac {x \log (x) \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{\log (x)-3}dx-16 \int \frac {x \log ^3\left (-\frac {x^2 \log (3)}{2 (x-1) \log (\log (x)-3)}\right )}{(\log (x)-3) \log (\log (x)-3)}dx\) |
Input:
Int[((16*x - 16*x^2 + (96*x - 48*x^2 + (-32*x + 16*x^2)*Log[x])*Log[-3 + L og[x]])*Log[-((x^2*Log[3])/((-2 + 2*x)*Log[-3 + Log[x]]))]^3 + (24*x - 24* x^2 + (-8*x + 8*x^2)*Log[x])*Log[-3 + Log[x]]*Log[-((x^2*Log[3])/((-2 + 2* x)*Log[-3 + Log[x]]))]^4)/((3 - 3*x + (-1 + x)*Log[x])*Log[-3 + Log[x]]),x ]
Output:
$Aborted
Time = 54.92 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
| method | result | size |
| parallelrisch | \(4 \ln \left (-\frac {x^{2} \ln \left (3\right )}{\left (-2+2 x \right ) \ln \left (\ln \left (x \right )-3\right )}\right )^{4} x^{2}\) | \(30\) |
| risch | \(\text {Expression too large to display}\) | \(278332\) |
Input:
int((((8*x^2-8*x)*ln(x)-24*x^2+24*x)*ln(ln(x)-3)*ln(-x^2*ln(3)/(-2+2*x)/ln (ln(x)-3))^4+(((16*x^2-32*x)*ln(x)-48*x^2+96*x)*ln(ln(x)-3)-16*x^2+16*x)*l n(-x^2*ln(3)/(-2+2*x)/ln(ln(x)-3))^3)/((-1+x)*ln(x)-3*x+3)/ln(ln(x)-3),x,m ethod=_RETURNVERBOSE)
Output:
4*ln(-x^2*ln(3)/(-2+2*x)/ln(ln(x)-3))^4*x^2
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=4 \, x^{2} \log \left (-\frac {x^{2} \log \left (3\right )}{2 \, {\left (x - 1\right )} \log \left (\log \left (x\right ) - 3\right )}\right )^{4} \] Input:
integrate((((8*x^2-8*x)*log(x)-24*x^2+24*x)*log(log(x)-3)*log(-x^2*log(3)/ (2*x-2)/log(log(x)-3))^4+(((16*x^2-32*x)*log(x)-48*x^2+96*x)*log(log(x)-3) -16*x^2+16*x)*log(-x^2*log(3)/(2*x-2)/log(log(x)-3))^3)/((-1+x)*log(x)-3*x +3)/log(log(x)-3),x, algorithm="fricas")
Output:
4*x^2*log(-1/2*x^2*log(3)/((x - 1)*log(log(x) - 3)))^4
Exception generated. \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((8*x**2-8*x)*ln(x)-24*x**2+24*x)*ln(ln(x)-3)*ln(-x**2*ln(3)/(2 *x-2)/ln(ln(x)-3))**4+(((16*x**2-32*x)*ln(x)-48*x**2+96*x)*ln(ln(x)-3)-16* x**2+16*x)*ln(-x**2*ln(3)/(2*x-2)/ln(ln(x)-3))**3)/((-1+x)*ln(x)-3*x+3)/ln (ln(x)-3),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Leaf count of result is larger than twice the leaf count of optimal. 668 vs. \(2 (27) = 54\).
Time = 0.20 (sec) , antiderivative size = 668, normalized size of antiderivative = 21.55 \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=\text {Too large to display} \] Input:
integrate((((8*x^2-8*x)*log(x)-24*x^2+24*x)*log(log(x)-3)*log(-x^2*log(3)/ (2*x-2)/log(log(x)-3))^4+(((16*x^2-32*x)*log(x)-48*x^2+96*x)*log(log(x)-3) -16*x^2+16*x)*log(-x^2*log(3)/(2*x-2)/log(log(x)-3))^3)/((-1+x)*log(x)-3*x +3)/log(log(x)-3),x, algorithm="maxima")
Output:
-128*x^2*(log(2) - log(log(3)))*log(x)^3 + 64*x^2*log(x)^4 + 4*x^2*log(-x + 1)^4 + 4*x^2*log(log(log(x) - 3))^4 + 96*(log(2)^2 - 2*log(2)*log(log(3) ) + log(log(3))^2)*x^2*log(x)^2 - 32*(log(2)^3 - 3*log(2)^2*log(log(3)) + 3*log(2)*log(log(3))^2 - log(log(3))^3)*x^2*log(x) + 16*(x^2*(log(2) - log (log(3))) - 2*x^2*log(x))*log(-x + 1)^3 + 16*(x^2*(log(2) - log(log(3))) - 2*x^2*log(x) + x^2*log(-x + 1))*log(log(log(x) - 3))^3 + 4*(log(2)^4 - 4* log(2)^3*log(log(3)) + 6*log(2)^2*log(log(3))^2 - 4*log(2)*log(log(3))^3 + log(log(3))^4)*x^2 - 24*(4*x^2*(log(2) - log(log(3)))*log(x) - 4*x^2*log( x)^2 - (log(2)^2 - 2*log(2)*log(log(3)) + log(log(3))^2)*x^2)*log(-x + 1)^ 2 - 24*(4*x^2*(log(2) - log(log(3)))*log(x) - 4*x^2*log(x)^2 - x^2*log(-x + 1)^2 - (log(2)^2 - 2*log(2)*log(log(3)) + log(log(3))^2)*x^2 - 2*(x^2*(l og(2) - log(log(3))) - 2*x^2*log(x))*log(-x + 1))*log(log(log(x) - 3))^2 + 16*(12*x^2*(log(2) - log(log(3)))*log(x)^2 - 8*x^2*log(x)^3 - 6*(log(2)^2 - 2*log(2)*log(log(3)) + log(log(3))^2)*x^2*log(x) + (log(2)^3 - 3*log(2) ^2*log(log(3)) + 3*log(2)*log(log(3))^2 - log(log(3))^3)*x^2)*log(-x + 1) + 16*(12*x^2*(log(2) - log(log(3)))*log(x)^2 - 8*x^2*log(x)^3 + x^2*log(-x + 1)^3 - 6*(log(2)^2 - 2*log(2)*log(log(3)) + log(log(3))^2)*x^2*log(x) + (log(2)^3 - 3*log(2)^2*log(log(3)) + 3*log(2)*log(log(3))^2 - log(log(3)) ^3)*x^2 + 3*(x^2*(log(2) - log(log(3))) - 2*x^2*log(x))*log(-x + 1)^2 - 3* (4*x^2*(log(2) - log(log(3)))*log(x) - 4*x^2*log(x)^2 - (log(2)^2 - 2*l...
Timed out. \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=\text {Timed out} \] Input:
integrate((((8*x^2-8*x)*log(x)-24*x^2+24*x)*log(log(x)-3)*log(-x^2*log(3)/ (2*x-2)/log(log(x)-3))^4+(((16*x^2-32*x)*log(x)-48*x^2+96*x)*log(log(x)-3) -16*x^2+16*x)*log(-x^2*log(3)/(2*x-2)/log(log(x)-3))^3)/((-1+x)*log(x)-3*x +3)/log(log(x)-3),x, algorithm="giac")
Output:
Timed out
Time = 3.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=4\,x^2\,{\ln \left (-\frac {x^2\,\ln \left (3\right )}{2\,\ln \left (\ln \left (x\right )-3\right )\,\left (x-1\right )}\right )}^4 \] Input:
int(-(log(-(x^2*log(3))/(log(log(x) - 3)*(2*x - 2)))^3*(log(log(x) - 3)*(l og(x)*(32*x - 16*x^2) - 96*x + 48*x^2) - 16*x + 16*x^2) + log(log(x) - 3)* log(-(x^2*log(3))/(log(log(x) - 3)*(2*x - 2)))^4*(log(x)*(8*x - 8*x^2) - 2 4*x + 24*x^2))/(log(log(x) - 3)*(log(x)*(x - 1) - 3*x + 3)),x)
Output:
4*x^2*log(-(x^2*log(3))/(2*log(log(x) - 3)*(x - 1)))^4
Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {\left (16 x-16 x^2+\left (96 x-48 x^2+\left (-32 x+16 x^2\right ) \log (x)\right ) \log (-3+\log (x))\right ) \log ^3\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )+\left (24 x-24 x^2+\left (-8 x+8 x^2\right ) \log (x)\right ) \log (-3+\log (x)) \log ^4\left (-\frac {x^2 \log (3)}{(-2+2 x) \log (-3+\log (x))}\right )}{(3-3 x+(-1+x) \log (x)) \log (-3+\log (x))} \, dx=4 \mathrm {log}\left (-\frac {\mathrm {log}\left (3\right ) x^{2}}{2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )-3\right ) x -2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )-3\right )}\right )^{4} x^{2} \] Input:
int((((8*x^2-8*x)*log(x)-24*x^2+24*x)*log(log(x)-3)*log(-x^2*log(3)/(2*x-2 )/log(log(x)-3))^4+(((16*x^2-32*x)*log(x)-48*x^2+96*x)*log(log(x)-3)-16*x^ 2+16*x)*log(-x^2*log(3)/(2*x-2)/log(log(x)-3))^3)/((-1+x)*log(x)-3*x+3)/lo g(log(x)-3),x)
Output:
4*log(( - log(3)*x**2)/(2*log(log(x) - 3)*x - 2*log(log(x) - 3)))**4*x**2