Integrand size = 102, antiderivative size = 27 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\frac {3 x}{1+x-x^2 \log ^2(4 x)+4 \log \left (e^{10} x\right )} \] Output:
3/(x+1+4*ln(x*exp(5)^2)-x^2*ln(4*x)^2)*x
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\frac {3 x}{41+x+4 \log (x)-x^2 \log ^2(4 x)} \] Input:
Integrate[(-9 + 6*x^2*Log[4*x] + 3*x^2*Log[4*x]^2 + 12*Log[E^10*x])/(1 + 2 *x + x^2 + (-2*x^2 - 2*x^3)*Log[4*x]^2 + x^4*Log[4*x]^4 + (8 + 8*x - 8*x^2 *Log[4*x]^2)*Log[E^10*x] + 16*Log[E^10*x]^2),x]
Output:
(3*x)/(41 + x + 4*Log[x] - x^2*Log[4*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.
\(\displaystyle \int \frac {3 x^2 \log ^2(4 x)+6 x^2 \log (4 x)+12 \log \left (e^{10} x\right )-9}{x^4 \log ^4(4 x)+x^2+\left (-8 x^2 \log ^2(4 x)+8 x+8\right ) \log \left (e^{10} x\right )+\left (-2 x^3-2 x^2\right ) \log ^2(4 x)+2 x+16 \log ^2\left (e^{10} x\right )+1} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 \left (x^2 \log ^2(4 x)+2 x^2 \log (4 x)+4 \log (x)+37\right )}{\left (-x^2 \log ^2(4 x)+x+4 \log (x)+41\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int \frac {\log ^2(4 x) x^2+2 \log (4 x) x^2+4 \log (x)+37}{\left (-x^2 \log ^2(4 x)+x+4 \log (x)+41\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 3 \int \left (\frac {2 \log (4 x) x^2+x+8 \log (x)+78}{\left (x^2 \log ^2(4 x)-x-4 \log (x)-41\right )^2}+\frac {1}{x^2 \log ^2(4 x)-x-4 \log (x)-41}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (8 \int \frac {\log (x)}{\left (-x^2 \log ^2(4 x)+x+4 \log (x)+41\right )^2}dx+78 \int \frac {1}{\left (x^2 \log ^2(4 x)-x-4 \log (x)-41\right )^2}dx+\int \frac {x}{\left (x^2 \log ^2(4 x)-x-4 \log (x)-41\right )^2}dx+2 \int \frac {x^2 \log (4 x)}{\left (x^2 \log ^2(4 x)-x-4 \log (x)-41\right )^2}dx+\int \frac {1}{x^2 \log ^2(4 x)-x-4 \log (x)-41}dx\right )\) |
Input:
Int[(-9 + 6*x^2*Log[4*x] + 3*x^2*Log[4*x]^2 + 12*Log[E^10*x])/(1 + 2*x + x ^2 + (-2*x^2 - 2*x^3)*Log[4*x]^2 + x^4*Log[4*x]^4 + (8 + 8*x - 8*x^2*Log[4 *x]^2)*Log[E^10*x] + 16*Log[E^10*x]^2),x]
Output:
$Aborted
Time = 3.90 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(-\frac {3 x}{x^{2} \ln \left (4 x \right )^{2}-4 \ln \left (x \,{\mathrm e}^{10}\right )-x -1}\) | \(30\) |
default | \(-\frac {3 x}{4 x^{2} \ln \left (2\right )^{2}+4 x^{2} \ln \left (2\right ) \ln \left (x \right )+x^{2} \ln \left (x \right )^{2}-x -4 \ln \left (x \right )-41}\) | \(41\) |
risch | \(\frac {12 x}{164+4 x -16 x^{2} \ln \left (2\right ) \ln \left (x \right )-4 x^{2} \ln \left (x \right )^{2}-16 x^{2} \ln \left (2\right )^{2}+16 \ln \left (x \right )}\) | \(42\) |
Input:
int((12*ln(x*exp(5)^2)+3*x^2*ln(4*x)^2+6*x^2*ln(4*x)-9)/(16*ln(x*exp(5)^2) ^2+(-8*x^2*ln(4*x)^2+8*x+8)*ln(x*exp(5)^2)+x^4*ln(4*x)^4+(-2*x^3-2*x^2)*ln (4*x)^2+x^2+2*x+1),x,method=_RETURNVERBOSE)
Output:
-3*x/(x^2*ln(4*x)^2-4*ln(x*exp(5)^2)-x-1)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{x^{2} \log \left (4 \, x\right )^{2} - x + 8 \, \log \left (2\right ) - 4 \, \log \left (4 \, x\right ) - 41} \] Input:
integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x *exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2* x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="fricas")
Output:
-3*x/(x^2*log(4*x)^2 - x + 8*log(2) - 4*log(4*x) - 41)
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=- \frac {3 x}{x^{2} \log {\left (4 x \right )}^{2} - x - 4 \log {\left (4 x \right )} - 41 + 8 \log {\left (2 \right )}} \] Input:
integrate((12*ln(x*exp(5)**2)+3*x**2*ln(4*x)**2+6*x**2*ln(4*x)-9)/(16*ln(x *exp(5)**2)**2+(-8*x**2*ln(4*x)**2+8*x+8)*ln(x*exp(5)**2)+x**4*ln(4*x)**4+ (-2*x**3-2*x**2)*ln(4*x)**2+x**2+2*x+1),x)
Output:
-3*x/(x**2*log(4*x)**2 - x - 4*log(4*x) - 41 + 8*log(2))
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{4 \, x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x\right )^{2} + 4 \, {\left (x^{2} \log \left (2\right ) - 1\right )} \log \left (x\right ) - x - 41} \] Input:
integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x *exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2* x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="maxima")
Output:
-3*x/(4*x^2*log(2)^2 + x^2*log(x)^2 + 4*(x^2*log(2) - 1)*log(x) - x - 41)
Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=-\frac {3 \, x}{4 \, x^{2} \log \left (2\right )^{2} + 4 \, x^{2} \log \left (2\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2} - x - 4 \, \log \left (x\right ) - 41} \] Input:
integrate((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x *exp(5)^2)^2+(-8*x^2*log(4*x)^2+8*x+8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2* x^3-2*x^2)*log(4*x)^2+x^2+2*x+1),x, algorithm="giac")
Output:
-3*x/(4*x^2*log(2)^2 + 4*x^2*log(2)*log(x) + x^2*log(x)^2 - x - 4*log(x) - 41)
Timed out. \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\int \frac {12\,\ln \left (x\,{\mathrm {e}}^{10}\right )+6\,x^2\,\ln \left (4\,x\right )+3\,x^2\,{\ln \left (4\,x\right )}^2-9}{2\,x+16\,{\ln \left (x\,{\mathrm {e}}^{10}\right )}^2+\ln \left (x\,{\mathrm {e}}^{10}\right )\,\left (-8\,x^2\,{\ln \left (4\,x\right )}^2+8\,x+8\right )-{\ln \left (4\,x\right )}^2\,\left (2\,x^3+2\,x^2\right )+x^2+x^4\,{\ln \left (4\,x\right )}^4+1} \,d x \] Input:
int((12*log(x*exp(10)) + 6*x^2*log(4*x) + 3*x^2*log(4*x)^2 - 9)/(2*x + 16* log(x*exp(10))^2 + log(x*exp(10))*(8*x - 8*x^2*log(4*x)^2 + 8) - log(4*x)^ 2*(2*x^2 + 2*x^3) + x^2 + x^4*log(4*x)^4 + 1),x)
Output:
int((12*log(x*exp(10)) + 6*x^2*log(4*x) + 3*x^2*log(4*x)^2 - 9)/(2*x + 16* log(x*exp(10))^2 + log(x*exp(10))*(8*x - 8*x^2*log(4*x)^2 + 8) - log(4*x)^ 2*(2*x^2 + 2*x^3) + x^2 + x^4*log(4*x)^4 + 1), x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-9+6 x^2 \log (4 x)+3 x^2 \log ^2(4 x)+12 \log \left (e^{10} x\right )}{1+2 x+x^2+\left (-2 x^2-2 x^3\right ) \log ^2(4 x)+x^4 \log ^4(4 x)+\left (8+8 x-8 x^2 \log ^2(4 x)\right ) \log \left (e^{10} x\right )+16 \log ^2\left (e^{10} x\right )} \, dx=\frac {3 x}{4 \,\mathrm {log}\left (e^{10} x \right )-\mathrm {log}\left (4 x \right )^{2} x^{2}+x +1} \] Input:
int((12*log(x*exp(5)^2)+3*x^2*log(4*x)^2+6*x^2*log(4*x)-9)/(16*log(x*exp(5 )^2)^2+(-8*x^2*log(4*x)^2+8*x+8)*log(x*exp(5)^2)+x^4*log(4*x)^4+(-2*x^3-2* x^2)*log(4*x)^2+x^2+2*x+1),x)
Output:
(3*x)/(4*log(e**10*x) - log(4*x)**2*x**2 + x + 1)