Integrand size = 151, antiderivative size = 26 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=9+\log \left (e^{2+\frac {4}{4-2 x-\log (x)}}+\frac {3}{x}\right ) \] Output:
9+ln(exp(4/(4-2*x-ln(x))+2)+3/x)
\[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx \] Input:
Integrate[(-48 + 48*x - 12*x^2 + E^((-12 + 4*x + 2*Log[x])/(-4 + 2*x + Log [x]))*(4*x + 8*x^2) + (24 - 12*x)*Log[x] - 3*Log[x]^2)/(48*x - 48*x^2 + 12 *x^3 + (-24*x + 12*x^2)*Log[x] + 3*x*Log[x]^2 + E^((-12 + 4*x + 2*Log[x])/ (-4 + 2*x + Log[x]))*(16*x^2 - 16*x^3 + 4*x^4 + (-8*x^2 + 4*x^3)*Log[x] + x^2*Log[x]^2)),x]
Output:
Integrate[(-48 + 48*x - 12*x^2 + E^((-12 + 4*x + 2*Log[x])/(-4 + 2*x + Log [x]))*(4*x + 8*x^2) + (24 - 12*x)*Log[x] - 3*Log[x]^2)/(48*x - 48*x^2 + 12 *x^3 + (-24*x + 12*x^2)*Log[x] + 3*x*Log[x]^2 + E^((-12 + 4*x + 2*Log[x])/ (-4 + 2*x + Log[x]))*(16*x^2 - 16*x^3 + 4*x^4 + (-8*x^2 + 4*x^3)*Log[x] + x^2*Log[x]^2)), 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.
| \(\displaystyle \int \frac {-12 x^2+\left (8 x^2+4 x\right ) e^{\frac {4 x+2 \log (x)-12}{2 x+\log (x)-4}}+48 x-3 \log ^2(x)+(24-12 x) \log (x)-48}{12 x^3-48 x^2+\left (12 x^2-24 x\right ) \log (x)+e^{\frac {4 x+2 \log (x)-12}{2 x+\log (x)-4}} \left (4 x^4-16 x^3+16 x^2+x^2 \log ^2(x)+\left (4 x^3-8 x^2\right ) \log (x)\right )+48 x+3 x \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {12}{2 x+\log (x)-4}} \left (-12 x^2+\left (8 x^2+4 x\right ) e^{\frac {4 x+2 \log (x)-12}{2 x+\log (x)-4}}+48 x-3 \log ^2(x)+(24-12 x) \log (x)-48\right )}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (-2 x-\log (x)+4)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {8 x^{\frac {2}{2 x+\log (x)-4}+1} \exp \left (\frac {4 (x-3)}{2 x+\log (x)-4}+\frac {12}{2 x+\log (x)-4}\right )}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (-2 x-\log (x)+4)^2}+\frac {4 x^{\frac {2}{2 x+\log (x)-4}} \exp \left (\frac {4 (x-3)}{2 x+\log (x)-4}+\frac {12}{2 x+\log (x)-4}\right )}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (-2 x-\log (x)+4)^2}-\frac {3 e^{\frac {12}{2 x+\log (x)-4}} \log ^2(x)}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}-\frac {12 x e^{\frac {12}{2 x+\log (x)-4}}}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}-\frac {12 e^{\frac {12}{2 x+\log (x)-4}} \log (x)}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}+\frac {48 e^{\frac {12}{2 x+\log (x)-4}}}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}+\frac {24 e^{\frac {12}{2 x+\log (x)-4}} \log (x)}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}-\frac {48 e^{\frac {12}{2 x+\log (x)-4}}}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}+1}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -3 \int \frac {e^{\frac {12}{2 x+\log (x)-4}} \log ^2(x)}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx+8 \int \frac {e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (-2 x-\log (x)+4)^2}dx+48 \int \frac {e^{\frac {12}{2 x+\log (x)-4}}}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx-48 \int \frac {e^{\frac {12}{2 x+\log (x)-4}}}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx-12 \int \frac {e^{\frac {12}{2 x+\log (x)-4}} x}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx-12 \int \frac {e^{\frac {12}{2 x+\log (x)-4}} \log (x)}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx+24 \int \frac {e^{\frac {12}{2 x+\log (x)-4}} \log (x)}{x \left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (2 x+\log (x)-4)^2}dx+4 \int \frac {e^{\frac {4 x}{2 x+\log (x)-4}} x^{\frac {2}{2 x+\log (x)-4}}}{\left (e^{\frac {4 x}{2 x+\log (x)-4}} x^{1+\frac {2}{2 x+\log (x)-4}}+3 e^{\frac {12}{2 x+\log (x)-4}}\right ) (-2 x-\log (x)+4)^2}dx\) |
Input:
Int[(-48 + 48*x - 12*x^2 + E^((-12 + 4*x + 2*Log[x])/(-4 + 2*x + Log[x]))* (4*x + 8*x^2) + (24 - 12*x)*Log[x] - 3*Log[x]^2)/(48*x - 48*x^2 + 12*x^3 + (-24*x + 12*x^2)*Log[x] + 3*x*Log[x]^2 + E^((-12 + 4*x + 2*Log[x])/(-4 + 2*x + Log[x]))*(16*x^2 - 16*x^3 + 4*x^4 + (-8*x^2 + 4*x^3)*Log[x] + x^2*Lo g[x]^2)),x]
Output:
$Aborted
Time = 0.78 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
| method | result | size |
| parallelrisch | \(\ln \left ({\mathrm e}^{\frac {2 \ln \left (x \right )+4 x -12}{\ln \left (x \right )+2 x -4}} x +3\right )-\ln \left (x \right )-16\) | \(31\) |
| risch | \(-\frac {4}{\ln \left (x \right )+2 x -4}-\frac {2 \ln \left (x \right )+4 x -12}{\ln \left (x \right )+2 x -4}+\ln \left ({\mathrm e}^{\frac {2 \ln \left (x \right )+4 x -12}{\ln \left (x \right )+2 x -4}}+\frac {3}{x}\right )\) | \(59\) |
Input:
int(((8*x^2+4*x)*exp((2*ln(x)+4*x-12)/(ln(x)+2*x-4))-3*ln(x)^2+(-12*x+24)* ln(x)-12*x^2+48*x-48)/((x^2*ln(x)^2+(4*x^3-8*x^2)*ln(x)+4*x^4-16*x^3+16*x^ 2)*exp((2*ln(x)+4*x-12)/(ln(x)+2*x-4))+3*x*ln(x)^2+(12*x^2-24*x)*ln(x)+12* x^3-48*x^2+48*x),x,method=_RETURNVERBOSE)
Output:
ln(exp(2*(ln(x)+2*x-6)/(ln(x)+2*x-4))*x+3)-ln(x)-16
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\log \left (\frac {x e^{\left (\frac {2 \, {\left (2 \, x + \log \left (x\right ) - 6\right )}}{2 \, x + \log \left (x\right ) - 4}\right )} + 3}{x}\right ) \] Input:
integrate(((8*x^2+4*x)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))-3*log(x)^2+(- 12*x+24)*log(x)-12*x^2+48*x-48)/((x^2*log(x)^2+(4*x^3-8*x^2)*log(x)+4*x^4- 16*x^3+16*x^2)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))+3*x*log(x)^2+(12*x^2- 24*x)*log(x)+12*x^3-48*x^2+48*x),x, algorithm="fricas")
Output:
log((x*e^(2*(2*x + log(x) - 6)/(2*x + log(x) - 4)) + 3)/x)
Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\log {\left (e^{\frac {4 x + 2 \log {\left (x \right )} - 12}{2 x + \log {\left (x \right )} - 4}} + \frac {3}{x} \right )} \] Input:
integrate(((8*x**2+4*x)*exp((2*ln(x)+4*x-12)/(ln(x)+2*x-4))-3*ln(x)**2+(-1 2*x+24)*ln(x)-12*x**2+48*x-48)/((x**2*ln(x)**2+(4*x**3-8*x**2)*ln(x)+4*x** 4-16*x**3+16*x**2)*exp((2*ln(x)+4*x-12)/(ln(x)+2*x-4))+3*x*ln(x)**2+(12*x* *2-24*x)*ln(x)+12*x**3-48*x**2+48*x),x)
Output:
log(exp((4*x + 2*log(x) - 12)/(2*x + log(x) - 4)) + 3/x)
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (23) = 46\).
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.54 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\frac {4 \, {\left (x - 2\right )}}{2 \, x + \log \left (x\right ) - 4} + \log \left (\frac {1}{3} \, {\left (x e^{\left (\frac {2 \, \log \left (x\right )}{2 \, x + \log \left (x\right ) - 4} + \frac {8}{2 \, x + \log \left (x\right ) - 4} + 2\right )} + 3 \, e^{\left (\frac {2 \, \log \left (x\right )}{2 \, x + \log \left (x\right ) - 4} + \frac {12}{2 \, x + \log \left (x\right ) - 4}\right )}\right )} e^{\left (-\frac {12}{2 \, x + \log \left (x\right ) - 4}\right )}\right ) - \log \left (x\right ) \] Input:
integrate(((8*x^2+4*x)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))-3*log(x)^2+(- 12*x+24)*log(x)-12*x^2+48*x-48)/((x^2*log(x)^2+(4*x^3-8*x^2)*log(x)+4*x^4- 16*x^3+16*x^2)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))+3*x*log(x)^2+(12*x^2- 24*x)*log(x)+12*x^3-48*x^2+48*x),x, algorithm="maxima")
Output:
4*(x - 2)/(2*x + log(x) - 4) + log(1/3*(x*e^(2*log(x)/(2*x + log(x) - 4) + 8/(2*x + log(x) - 4) + 2) + 3*e^(2*log(x)/(2*x + log(x) - 4) + 12/(2*x + log(x) - 4)))*e^(-12/(2*x + log(x) - 4))) - log(x)
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\log \left (x e^{\left (-\frac {2 \, x + \log \left (x\right )}{2 \, x + \log \left (x\right ) - 4} + 3\right )} + 3\right ) - \log \left (x\right ) \] Input:
integrate(((8*x^2+4*x)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))-3*log(x)^2+(- 12*x+24)*log(x)-12*x^2+48*x-48)/((x^2*log(x)^2+(4*x^3-8*x^2)*log(x)+4*x^4- 16*x^3+16*x^2)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))+3*x*log(x)^2+(12*x^2- 24*x)*log(x)+12*x^3-48*x^2+48*x),x, algorithm="giac")
Output:
log(x*e^(-(2*x + log(x))/(2*x + log(x) - 4) + 3) + 3) - log(x)
Time = 3.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\ln \left (\frac {3}{x}+x^{\frac {2}{2\,x+\ln \left (x\right )-4}}\,{\mathrm {e}}^{-\frac {12}{2\,x+\ln \left (x\right )-4}}\,{\mathrm {e}}^{\frac {4\,x}{2\,x+\ln \left (x\right )-4}}\right ) \] Input:
int(-(3*log(x)^2 - 48*x + log(x)*(12*x - 24) - exp((4*x + 2*log(x) - 12)/( 2*x + log(x) - 4))*(4*x + 8*x^2) + 12*x^2 + 48)/(48*x + 3*x*log(x)^2 + exp ((4*x + 2*log(x) - 12)/(2*x + log(x) - 4))*(x^2*log(x)^2 - log(x)*(8*x^2 - 4*x^3) + 16*x^2 - 16*x^3 + 4*x^4) - log(x)*(24*x - 12*x^2) - 48*x^2 + 12* x^3),x)
Output:
log(3/x + x^(2/(2*x + log(x) - 4))*exp(-12/(2*x + log(x) - 4))*exp((4*x)/( 2*x + log(x) - 4)))
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.96 \[ \int \frac {-48+48 x-12 x^2+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (4 x+8 x^2\right )+(24-12 x) \log (x)-3 \log ^2(x)}{48 x-48 x^2+12 x^3+\left (-24 x+12 x^2\right ) \log (x)+3 x \log ^2(x)+e^{\frac {-12+4 x+2 \log (x)}{-4+2 x+\log (x)}} \left (16 x^2-16 x^3+4 x^4+\left (-8 x^2+4 x^3\right ) \log (x)+x^2 \log ^2(x)\right )} \, dx=\frac {\mathrm {log}\left (3 e^{\frac {4}{\mathrm {log}\left (x \right )+2 x -4}}+e^{2} x \right ) \mathrm {log}\left (x \right )+2 \,\mathrm {log}\left (3 e^{\frac {4}{\mathrm {log}\left (x \right )+2 x -4}}+e^{2} x \right ) x -4 \,\mathrm {log}\left (3 e^{\frac {4}{\mathrm {log}\left (x \right )+2 x -4}}+e^{2} x \right )-\mathrm {log}\left (x \right )^{2}-2 \,\mathrm {log}\left (x \right ) x +3 \,\mathrm {log}\left (x \right )-2 x}{\mathrm {log}\left (x \right )+2 x -4} \] Input:
int(((8*x^2+4*x)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))-3*log(x)^2+(-12*x+2 4)*log(x)-12*x^2+48*x-48)/((x^2*log(x)^2+(4*x^3-8*x^2)*log(x)+4*x^4-16*x^3 +16*x^2)*exp((2*log(x)+4*x-12)/(log(x)+2*x-4))+3*x*log(x)^2+(12*x^2-24*x)* log(x)+12*x^3-48*x^2+48*x),x)
Output:
(log(3*e**(4/(log(x) + 2*x - 4)) + e**2*x)*log(x) + 2*log(3*e**(4/(log(x) + 2*x - 4)) + e**2*x)*x - 4*log(3*e**(4/(log(x) + 2*x - 4)) + e**2*x) - lo g(x)**2 - 2*log(x)*x + 3*log(x) - 2*x)/(log(x) + 2*x - 4)