Integrand size = 116, antiderivative size = 28 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\frac {3}{\left (-13-x \left (4+e^5 \log (x)\right )\right ) \log ((-1+2 x) \log (2))} \] Output:
3/(-13-(exp(5)*ln(x)+4)*x)/ln((-1+2*x)*ln(2))
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\frac {3}{\left (-13-4 x-e^5 x \log (x)\right ) \log ((-1+2 x) \log (2))} \] Input:
Integrate[(78 + 24*x + 6*E^5*x*Log[x] + (-12 + 24*x + E^5*(-3 + 6*x) + E^5 *(-3 + 6*x)*Log[x])*Log[(-1 + 2*x)*Log[2]])/((-169 + 234*x + 192*x^2 + 32* x^3 + E^5*(-26*x + 44*x^2 + 16*x^3)*Log[x] + E^10*(-x^2 + 2*x^3)*Log[x]^2) *Log[(-1 + 2*x)*Log[2]]^2),x]
Output:
3/((-13 - 4*x - E^5*x*Log[x])*Log[(-1 + 2*x)*Log[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.
\(\displaystyle \int \frac {24 x+6 e^5 x \log (x)+\left (24 x+e^5 (6 x-3)+e^5 (6 x-3) \log (x)-12\right ) \log ((2 x-1) \log (2))+78}{\left (32 x^3+192 x^2+e^{10} \left (2 x^3-x^2\right ) \log ^2(x)+e^5 \left (16 x^3+44 x^2-26 x\right ) \log (x)+234 x-169\right ) \log ^2((2 x-1) \log (2))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-24 x-6 e^5 x \log (x)-\left (24 x+e^5 (6 x-3)+e^5 (6 x-3) \log (x)-12\right ) \log ((2 x-1) \log (2))-78}{(1-2 x) \left (4 x+e^5 x \log (x)+13\right )^2 \log ^2(2 x \log (2)-\log (2))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {6 e^5 x \log (x)}{(2 x-1) \left (4 x+e^5 x \log (x)+13\right )^2 \log ^2(2 x \log (2)-\log (2))}+\frac {24 x}{(2 x-1) \left (4 x+e^5 x \log (x)+13\right )^2 \log ^2(2 x \log (2)-\log (2))}+\frac {78}{(2 x-1) \left (4 x+e^5 x \log (x)+13\right )^2 \log ^2(2 x \log (2)-\log (2))}+\frac {6 e^5 x \log (x)}{(2 x-1) \left (4 x+e^5 x \log (x)+13\right )^2 \log (2 x \log (2)-\log (2))}+\frac {24 \left (1+\frac {e^5}{4}\right ) x}{(2 x-1) \left (4 x+e^5 x \log (x)+13\right )^2 \log (2 x \log (2)-\log (2))}+\frac {3 e^5 \log (x)}{(1-2 x) \left (4 x+e^5 x \log (x)+13\right )^2 \log (2 x \log (2)-\log (2))}+\frac {12 \left (1+\frac {e^5}{4}\right )}{(1-2 x) \left (4 x+e^5 x \log (x)+13\right )^2 \log (2 x \log (2)-\log (2))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 12 \int \frac {1}{\left (e^5 \log (x) x+4 x+13\right )^2 \log ^2(2 x \log (2)-\log (2))}dx+90 \int \frac {1}{(2 x-1) \left (e^5 \log (x) x+4 x+13\right )^2 \log ^2(2 x \log (2)-\log (2))}dx+3 e^5 \int \frac {\log (x)}{\left (e^5 \log (x) x+4 x+13\right )^2 \log ^2(2 x \log (2)-\log (2))}dx+3 e^5 \int \frac {\log (x)}{(2 x-1) \left (e^5 \log (x) x+4 x+13\right )^2 \log ^2(2 x \log (2)-\log (2))}dx+3 \left (4+e^5\right ) \int \frac {1}{\left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx+3 \left (4+e^5\right ) \int \frac {1}{(1-2 x) \left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx+3 \left (4+e^5\right ) \int \frac {1}{(2 x-1) \left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx+3 e^5 \int \frac {\log (x)}{\left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx+3 e^5 \int \frac {\log (x)}{(1-2 x) \left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx+3 e^5 \int \frac {\log (x)}{(2 x-1) \left (e^5 \log (x) x+4 x+13\right )^2 \log (2 x \log (2)-\log (2))}dx\) |
Input:
Int[(78 + 24*x + 6*E^5*x*Log[x] + (-12 + 24*x + E^5*(-3 + 6*x) + E^5*(-3 + 6*x)*Log[x])*Log[(-1 + 2*x)*Log[2]])/((-169 + 234*x + 192*x^2 + 32*x^3 + E^5*(-26*x + 44*x^2 + 16*x^3)*Log[x] + E^10*(-x^2 + 2*x^3)*Log[x]^2)*Log[( -1 + 2*x)*Log[2]]^2),x]
Output:
$Aborted
Time = 3.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {3}{\left (x \,{\mathrm e}^{5} \ln \left (x \right )+4 x +13\right ) \ln \left (\left (-1+2 x \right ) \ln \left (2\right )\right )}\) | \(27\) |
parallelrisch | \(-\frac {3}{\left (x \,{\mathrm e}^{5} \ln \left (x \right )+4 x +13\right ) \ln \left (\left (-1+2 x \right ) \ln \left (2\right )\right )}\) | \(27\) |
default | \(-\frac {3}{\left (\ln \left (x \right ) {\mathrm e}^{5+\ln \left (x \right )}+4 x +13\right ) \left (\ln \left (\ln \left (2\right )\right )+\ln \left (-1+2 x \right )\right )}\) | \(30\) |
Input:
int((((-3+6*x)*exp(5)*ln(x)+(-3+6*x)*exp(5)+24*x-12)*ln((-1+2*x)*ln(2))+6* x*exp(5)*ln(x)+24*x+78)/((2*x^3-x^2)*exp(5)^2*ln(x)^2+(16*x^3+44*x^2-26*x) *exp(5)*ln(x)+32*x^3+192*x^2+234*x-169)/ln((-1+2*x)*ln(2))^2,x,method=_RET URNVERBOSE)
Output:
-3/(x*exp(5)*ln(x)+4*x+13)/ln((-1+2*x)*ln(2))
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{{\left (x e^{5} \log \left (x\right ) + 4 \, x + 13\right )} \log \left ({\left (2 \, x - 1\right )} \log \left (2\right )\right )} \] Input:
integrate((((-3+6*x)*exp(5)*log(x)+(-3+6*x)*exp(5)+24*x-12)*log((-1+2*x)*l og(2))+6*x*exp(5)*log(x)+24*x+78)/((2*x^3-x^2)*exp(5)^2*log(x)^2+(16*x^3+4 4*x^2-26*x)*exp(5)*log(x)+32*x^3+192*x^2+234*x-169)/log((-1+2*x)*log(2))^2 ,x, algorithm="fricas")
Output:
-3/((x*e^5*log(x) + 4*x + 13)*log((2*x - 1)*log(2)))
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=- \frac {3}{\left (x e^{5} \log {\left (x \right )} + 4 x + 13\right ) \log {\left (\left (2 x - 1\right ) \log {\left (2 \right )} \right )}} \] Input:
integrate((((-3+6*x)*exp(5)*ln(x)+(-3+6*x)*exp(5)+24*x-12)*ln((-1+2*x)*ln( 2))+6*x*exp(5)*ln(x)+24*x+78)/((2*x**3-x**2)*exp(5)**2*ln(x)**2+(16*x**3+4 4*x**2-26*x)*exp(5)*ln(x)+32*x**3+192*x**2+234*x-169)/ln((-1+2*x)*ln(2))** 2,x)
Output:
-3/((x*exp(5)*log(x) + 4*x + 13)*log((2*x - 1)*log(2)))
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{x e^{5} \log \left (x\right ) \log \left (\log \left (2\right )\right ) + {\left (x e^{5} \log \left (x\right ) + 4 \, x + 13\right )} \log \left (2 \, x - 1\right ) + 4 \, x \log \left (\log \left (2\right )\right ) + 13 \, \log \left (\log \left (2\right )\right )} \] Input:
integrate((((-3+6*x)*exp(5)*log(x)+(-3+6*x)*exp(5)+24*x-12)*log((-1+2*x)*l og(2))+6*x*exp(5)*log(x)+24*x+78)/((2*x^3-x^2)*exp(5)^2*log(x)^2+(16*x^3+4 4*x^2-26*x)*exp(5)*log(x)+32*x^3+192*x^2+234*x-169)/log((-1+2*x)*log(2))^2 ,x, algorithm="maxima")
Output:
-3/(x*e^5*log(x)*log(log(2)) + (x*e^5*log(x) + 4*x + 13)*log(2*x - 1) + 4* x*log(log(2)) + 13*log(log(2)))
Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{x e^{5} \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right ) \log \left (x\right ) + 4 \, x \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right ) + 13 \, \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right )} \] Input:
integrate((((-3+6*x)*exp(5)*log(x)+(-3+6*x)*exp(5)+24*x-12)*log((-1+2*x)*l og(2))+6*x*exp(5)*log(x)+24*x+78)/((2*x^3-x^2)*exp(5)^2*log(x)^2+(16*x^3+4 4*x^2-26*x)*exp(5)*log(x)+32*x^3+192*x^2+234*x-169)/log((-1+2*x)*log(2))^2 ,x, algorithm="giac")
Output:
-3/(x*e^5*log(2*x*log(2) - log(2))*log(x) + 4*x*log(2*x*log(2) - log(2)) + 13*log(2*x*log(2) - log(2)))
Timed out. \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\int \frac {24\,x+\ln \left (\ln \left (2\right )\,\left (2\,x-1\right )\right )\,\left (24\,x+{\mathrm {e}}^5\,\left (6\,x-3\right )+{\mathrm {e}}^5\,\ln \left (x\right )\,\left (6\,x-3\right )-12\right )+6\,x\,{\mathrm {e}}^5\,\ln \left (x\right )+78}{{\ln \left (\ln \left (2\right )\,\left (2\,x-1\right )\right )}^2\,\left (234\,x+192\,x^2+32\,x^3+{\mathrm {e}}^5\,\ln \left (x\right )\,\left (16\,x^3+44\,x^2-26\,x\right )-{\mathrm {e}}^{10}\,{\ln \left (x\right )}^2\,\left (x^2-2\,x^3\right )-169\right )} \,d x \] Input:
int((24*x + log(log(2)*(2*x - 1))*(24*x + exp(5)*(6*x - 3) + exp(5)*log(x) *(6*x - 3) - 12) + 6*x*exp(5)*log(x) + 78)/(log(log(2)*(2*x - 1))^2*(234*x + 192*x^2 + 32*x^3 + exp(5)*log(x)*(44*x^2 - 26*x + 16*x^3) - exp(10)*log (x)^2*(x^2 - 2*x^3) - 169)),x)
Output:
int((24*x + log(log(2)*(2*x - 1))*(24*x + exp(5)*(6*x - 3) + exp(5)*log(x) *(6*x - 3) - 12) + 6*x*exp(5)*log(x) + 78)/(log(log(2)*(2*x - 1))^2*(234*x + 192*x^2 + 32*x^3 + exp(5)*log(x)*(44*x^2 - 26*x + 16*x^3) - exp(10)*log (x)^2*(x^2 - 2*x^3) - 169)), x)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{\mathrm {log}\left (2 \,\mathrm {log}\left (2\right ) x -\mathrm {log}\left (2\right )\right ) \left (\mathrm {log}\left (x \right ) e^{5} x +4 x +13\right )} \] Input:
int((((-3+6*x)*exp(5)*log(x)+(-3+6*x)*exp(5)+24*x-12)*log((-1+2*x)*log(2)) +6*x*exp(5)*log(x)+24*x+78)/((2*x^3-x^2)*exp(5)^2*log(x)^2+(16*x^3+44*x^2- 26*x)*exp(5)*log(x)+32*x^3+192*x^2+234*x-169)/log((-1+2*x)*log(2))^2,x)
Output:
( - 3)/(log(2*log(2)*x - log(2))*(log(x)*e**5*x + 4*x + 13))