Integrand size = 68, antiderivative size = 24 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \] Output:
exp(x)*((4*x-ln(ln(x^2)))*exp(3)-ln(x))
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \] Input:
Integrate[(-2*E^(3 + x) + (E^x*(-1 + E^3*(4*x + 4*x^2)) - E^x*x*Log[x])*Lo g[x^2] - E^(3 + x)*x*Log[x^2]*Log[Log[x^2]])/(x*Log[x^2]),x]
Output:
E^x*(-Log[x] + E^3*(4*x - Log[Log[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.
\(\displaystyle \int \frac {\left (e^x \left (e^3 \left (4 x^2+4 x\right )-1\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{x+3} x \log \left (\log \left (x^2\right )\right ) \log \left (x^2\right )-2 e^{x+3}}{x \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^{x+3} \log \left (\log \left (x^2\right )\right )-\frac {2 e^{x+3}}{x \log \left (x^2\right )}+4 e^{x+3} x+4 e^{x+3}-\frac {e^x}{x}-e^x \log (x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {e^{x+3}}{x \log \left (x^2\right )}dx-\int e^{x+3} \log \left (\log \left (x^2\right )\right )dx+4 e^{x+3} x-e^x \log (x)\) |
Input:
Int[(-2*E^(3 + x) + (E^x*(-1 + E^3*(4*x + 4*x^2)) - E^x*x*Log[x])*Log[x^2] - E^(3 + x)*x*Log[x^2]*Log[Log[x^2]])/(x*Log[x^2]),x]
Output:
$Aborted
Time = 12.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(4 x \,{\mathrm e}^{3} {\mathrm e}^{x}-{\mathrm e}^{3} {\mathrm e}^{x} \ln \left (\ln \left (x^{2}\right )\right )-{\mathrm e}^{x} \ln \left (x \right )\) | \(26\) |
risch | \(4 \,{\mathrm e}^{3+x} x -\ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right ) {\mathrm e}^{3+x}-{\mathrm e}^{x} \ln \left (x \right )\) | \(55\) |
Input:
int((-x*exp(3)*exp(x)*ln(x^2)*ln(ln(x^2))+(-x*exp(x)*ln(x)+((4*x^2+4*x)*ex p(3)-1)*exp(x))*ln(x^2)-2*exp(x)*exp(3))/x/ln(x^2),x,method=_RETURNVERBOSE )
Output:
4*x*exp(3)*exp(x)-exp(3)*exp(x)*ln(ln(x^2))-exp(x)*ln(x)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx={\left (4 \, x e^{\left (x + 6\right )} - e^{\left (x + 3\right )} \log \left (x\right ) - e^{\left (x + 6\right )} \log \left (2 \, \log \left (x\right )\right )\right )} e^{\left (-3\right )} \] Input:
integrate((-x*exp(3)*exp(x)*log(x^2)*log(log(x^2))+(-x*exp(x)*log(x)+((4*x ^2+4*x)*exp(3)-1)*exp(x))*log(x^2)-2*exp(x)*exp(3))/x/log(x^2),x, algorith m="fricas")
Output:
(4*x*e^(x + 6) - e^(x + 3)*log(x) - e^(x + 6)*log(2*log(x)))*e^(-3)
Time = 8.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=\left (4 x e^{3} - \log {\left (x \right )} - e^{3} \log {\left (2 \log {\left (x \right )} \right )}\right ) e^{x} \] Input:
integrate((-x*exp(3)*exp(x)*ln(x**2)*ln(ln(x**2))+(-x*exp(x)*ln(x)+((4*x** 2+4*x)*exp(3)-1)*exp(x))*ln(x**2)-2*exp(x)*exp(3))/x/ln(x**2),x)
Output:
(4*x*exp(3) - log(x) - exp(3)*log(2*log(x)))*exp(x)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4 \, {\left (x e^{3} - e^{3}\right )} e^{x} - e^{\left (x + 3\right )} \log \left (2\right ) - e^{x} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (\log \left (x\right )\right ) + 4 \, e^{\left (x + 3\right )} \] Input:
integrate((-x*exp(3)*exp(x)*log(x^2)*log(log(x^2))+(-x*exp(x)*log(x)+((4*x ^2+4*x)*exp(3)-1)*exp(x))*log(x^2)-2*exp(x)*exp(3))/x/log(x^2),x, algorith m="maxima")
Output:
4*(x*e^3 - e^3)*e^x - e^(x + 3)*log(2) - e^x*log(x) - e^(x + 3)*log(log(x) ) + 4*e^(x + 3)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4 \, x e^{\left (x + 3\right )} - e^{x} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (\log \left (x^{2}\right )\right ) \] Input:
integrate((-x*exp(3)*exp(x)*log(x^2)*log(log(x^2))+(-x*exp(x)*log(x)+((4*x ^2+4*x)*exp(3)-1)*exp(x))*log(x^2)-2*exp(x)*exp(3))/x/log(x^2),x, algorith m="giac")
Output:
4*x*e^(x + 3) - e^x*log(x) - e^(x + 3)*log(log(x^2))
Time = 2.75 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4\,x\,{\mathrm {e}}^3\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (x\right )-{\mathrm {e}}^3\,{\mathrm {e}}^x\,\ln \left (\ln \left (x^2\right )\right ) \] Input:
int(-(2*exp(3)*exp(x) - log(x^2)*(exp(x)*(exp(3)*(4*x + 4*x^2) - 1) - x*ex p(x)*log(x)) + x*log(x^2)*exp(3)*exp(x)*log(log(x^2)))/(x*log(x^2)),x)
Output:
4*x*exp(3)*exp(x) - exp(x)*log(x) - exp(3)*exp(x)*log(log(x^2))
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=e^{x} \left (-\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right ) e^{3}-\mathrm {log}\left (x \right )+4 e^{3} x \right ) \] Input:
int((-x*exp(3)*exp(x)*log(x^2)*log(log(x^2))+(-x*exp(x)*log(x)+((4*x^2+4*x )*exp(3)-1)*exp(x))*log(x^2)-2*exp(x)*exp(3))/x/log(x^2),x)
Output:
e**x*( - log(log(x**2))*e**3 - log(x) + 4*e**3*x)