Integrand size = 88, antiderivative size = 25 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=-\frac {\left (-4+\frac {1}{\log (3)}\right )^2}{x^2}+\frac {\log (x)}{\log (4-x)} \] Output:
1/ln(4-x)*ln(x)-(1/ln(3)-4)^2/x^2
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=-\frac {(-1+\log (81))^2}{x^2 \log ^2(3)}+\frac {\log (x)}{\log (4-x)} \] Input:
Integrate[((-4*x^2 + x^3)*Log[3]^2*Log[4 - x] + (-8 + 2*x + (64 - 16*x)*Lo g[3] + (-128 + 32*x)*Log[3]^2)*Log[4 - x]^2 - x^3*Log[3]^2*Log[x])/((-4*x^ 3 + x^4)*Log[3]^2*Log[4 - x]^2),x]
Output:
-((-1 + Log[81])^2/(x^2*Log[3]^2)) + Log[x]/Log[4 - 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.
\(\displaystyle \int \frac {x^3 \left (-\log ^2(3)\right ) \log (x)+\left (x^3-4 x^2\right ) \log ^2(3) \log (4-x)+\left (2 x+(32 x-128) \log ^2(3)+(64-16 x) \log (3)-8\right ) \log ^2(4-x)}{\left (x^4-4 x^3\right ) \log ^2(3) \log ^2(4-x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\log ^2(3) \log (x) x^3+2 \left (16 \log ^2(3) (4-x)-8 \log (3) (4-x)-x+4\right ) \log ^2(4-x)+\left (4 x^2-x^3\right ) \log ^2(3) \log (4-x)}{\left (4 x^3-x^4\right ) \log ^2(4-x)}dx}{\log ^2(3)}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {\log ^2(3) \log (x) x^3+2 \left (16 \log ^2(3) (4-x)-8 \log (3) (4-x)-x+4\right ) \log ^2(4-x)+\left (4 x^2-x^3\right ) \log ^2(3) \log (4-x)}{(4-x) x^3 \log ^2(4-x)}dx}{\log ^2(3)}\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {\int \left (-\frac {\log ^2(3) \log (x)}{(x-4) \log ^2(4-x)}+\frac {\log ^2(3)}{x \log (4-x)}+\frac {2 (-1+\log (81))^2}{x^3}\right )dx}{\log ^2(3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\log ^2(3) \text {Subst}\left (\int \frac {\log (4-x)}{x \log ^2(x)}dx,x,4-x\right )+\log ^2(3) \int \frac {1}{x \log (4-x)}dx-\frac {(1-\log (81))^2}{x^2}}{\log ^2(3)}\) |
Input:
Int[((-4*x^2 + x^3)*Log[3]^2*Log[4 - x] + (-8 + 2*x + (64 - 16*x)*Log[3] + (-128 + 32*x)*Log[3]^2)*Log[4 - x]^2 - x^3*Log[3]^2*Log[x])/((-4*x^3 + x^ 4)*Log[3]^2*Log[4 - x]^2),x]
Output:
$Aborted
Time = 1.56 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
risch | \(-\frac {16}{x^{2}}+\frac {8}{\ln \left (3\right ) x^{2}}-\frac {1}{\ln \left (3\right )^{2} x^{2}}+\frac {\ln \left (x \right )}{\ln \left (-x +4\right )}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (3\right )^{2} \ln \left (x \right ) x^{2}-16 \ln \left (3\right )^{2} \ln \left (-x +4\right )+8 \ln \left (3\right ) \ln \left (-x +4\right )-\ln \left (-x +4\right )}{\ln \left (3\right )^{2} x^{2} \ln \left (-x +4\right )}\) | \(58\) |
Input:
int((-x^3*ln(3)^2*ln(x)+((32*x-128)*ln(3)^2+(-16*x+64)*ln(3)+2*x-8)*ln(-x+ 4)^2+(x^3-4*x^2)*ln(3)^2*ln(-x+4))/(x^4-4*x^3)/ln(3)^2/ln(-x+4)^2,x,method =_RETURNVERBOSE)
Output:
-16/x^2+8/ln(3)/x^2-1/ln(3)^2/x^2+1/ln(-x+4)*ln(x)
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=\frac {x^{2} \log \left (3\right )^{2} \log \left (x\right ) - {\left (16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1\right )} \log \left (-x + 4\right )}{x^{2} \log \left (3\right )^{2} \log \left (-x + 4\right )} \] Input:
integrate((-x^3*log(3)^2*log(x)+((32*x-128)*log(3)^2+(-16*x+64)*log(3)+2*x -8)*log(-x+4)^2+(x^3-4*x^2)*log(3)^2*log(-x+4))/(x^4-4*x^3)/log(3)^2/log(- x+4)^2,x, algorithm="fricas")
Output:
(x^2*log(3)^2*log(x) - (16*log(3)^2 - 8*log(3) + 1)*log(-x + 4))/(x^2*log( 3)^2*log(-x + 4))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=\frac {\log {\left (x \right )}}{\log {\left (4 - x \right )}} - \frac {- 16 \log {\left (3 \right )} + 2 + 32 \log {\left (3 \right )}^{2}}{2 x^{2} \log {\left (3 \right )}^{2}} \] Input:
integrate((-x**3*ln(3)**2*ln(x)+((32*x-128)*ln(3)**2+(-16*x+64)*ln(3)+2*x- 8)*ln(-x+4)**2+(x**3-4*x**2)*ln(3)**2*ln(-x+4))/(x**4-4*x**3)/ln(3)**2/ln( -x+4)**2,x)
Output:
log(x)/log(4 - x) - (-16*log(3) + 2 + 32*log(3)**2)/(2*x**2*log(3)**2)
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 4.68 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=-\frac {4 \, {\left (\frac {4 \, {\left (x + 2\right )}}{x^{2}} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (3\right )^{2} - 4 \, {\left (\frac {4}{x} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (3\right )^{2} - 2 \, {\left (\frac {4 \, {\left (x + 2\right )}}{x^{2}} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (3\right ) + 2 \, {\left (\frac {4}{x} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (3\right ) - \frac {2 \, \log \left (3\right )^{2} \log \left (x\right )}{\log \left (-x + 4\right )} + \frac {x + 2}{x^{2}} - \frac {1}{x}}{2 \, \log \left (3\right )^{2}} \] Input:
integrate((-x^3*log(3)^2*log(x)+((32*x-128)*log(3)^2+(-16*x+64)*log(3)+2*x -8)*log(-x+4)^2+(x^3-4*x^2)*log(3)^2*log(-x+4))/(x^4-4*x^3)/log(3)^2/log(- x+4)^2,x, algorithm="maxima")
Output:
-1/2*(4*(4*(x + 2)/x^2 + log(x - 4) - log(x))*log(3)^2 - 4*(4/x + log(x - 4) - log(x))*log(3)^2 - 2*(4*(x + 2)/x^2 + log(x - 4) - log(x))*log(3) + 2 *(4/x + log(x - 4) - log(x))*log(3) - 2*log(3)^2*log(x)/log(-x + 4) + (x + 2)/x^2 - 1/x)/log(3)^2
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=\frac {\frac {\log \left (3\right )^{2} \log \left (x\right )}{\log \left (-x + 4\right )} - \frac {16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1}{x^{2}}}{\log \left (3\right )^{2}} \] Input:
integrate((-x^3*log(3)^2*log(x)+((32*x-128)*log(3)^2+(-16*x+64)*log(3)+2*x -8)*log(-x+4)^2+(x^3-4*x^2)*log(3)^2*log(-x+4))/(x^4-4*x^3)/log(3)^2/log(- x+4)^2,x, algorithm="giac")
Output:
(log(3)^2*log(x)/log(-x + 4) - (16*log(3)^2 - 8*log(3) + 1)/x^2)/log(3)^2
Time = 2.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=\frac {\ln \left (x\right )-\frac {\ln \left (4-x\right )\,\left (x-4\right )}{x}}{\ln \left (4-x\right )}-\frac {4\,x\,{\ln \left (3\right )}^2-8\,\ln \left (3\right )+16\,{\ln \left (3\right )}^2+1}{x^2\,{\ln \left (3\right )}^2} \] Input:
int((log(3)^2*log(4 - x)*(4*x^2 - x^3) - log(4 - x)^2*(2*x - log(3)*(16*x - 64) + log(3)^2*(32*x - 128) - 8) + x^3*log(3)^2*log(x))/(log(3)^2*log(4 - x)^2*(4*x^3 - x^4)),x)
Output:
(log(x) - (log(4 - x)*(x - 4))/x)/log(4 - x) - (4*x*log(3)^2 - 8*log(3) + 16*log(3)^2 + 1)/(x^2*log(3)^2)
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {\left (-4 x^2+x^3\right ) \log ^2(3) \log (4-x)+\left (-8+2 x+(64-16 x) \log (3)+(-128+32 x) \log ^2(3)\right ) \log ^2(4-x)-x^3 \log ^2(3) \log (x)}{\left (-4 x^3+x^4\right ) \log ^2(3) \log ^2(4-x)} \, dx=\frac {-16 \,\mathrm {log}\left (-x +4\right ) \mathrm {log}\left (3\right )^{2}+8 \,\mathrm {log}\left (-x +4\right ) \mathrm {log}\left (3\right )-\mathrm {log}\left (-x +4\right )+\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )^{2} x^{2}}{\mathrm {log}\left (-x +4\right ) \mathrm {log}\left (3\right )^{2} x^{2}} \] Input:
int((-x^3*log(3)^2*log(x)+((32*x-128)*log(3)^2+(-16*x+64)*log(3)+2*x-8)*lo g(-x+4)^2+(x^3-4*x^2)*log(3)^2*log(-x+4))/(x^4-4*x^3)/log(3)^2/log(-x+4)^2 ,x)
Output:
( - 16*log( - x + 4)*log(3)**2 + 8*log( - x + 4)*log(3) - log( - x + 4) + log(x)*log(3)**2*x**2)/(log( - x + 4)*log(3)**2*x**2)