Integrand size = 166, antiderivative size = 27 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-e^x+e^5 \left (4-e^3-\frac {1}{x}+x\right )\right )\right ) \] Output:
x*ln(ln(exp(5)*(x+4-exp(3)-1/x)-exp(x)))
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right ) \] Input:
Integrate[(E^x*x^2 + E^5*(-1 - x^2) + (E^x*x + E^5*(1 - 4*x + E^3*x - x^2) )*Log[(-(E^x*x) + E^5*(-1 + 4*x - E^3*x + x^2))/x]*Log[Log[(-(E^x*x) + E^5 *(-1 + 4*x - E^3*x + x^2))/x]])/((E^x*x + E^5*(1 - 4*x + E^3*x - x^2))*Log [(-(E^x*x) + E^5*(-1 + 4*x - E^3*x + x^2))/x]),x]
Output:
x*Log[Log[-E^8 - E^x + E^5*(4 - x^(-1) + 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.
| \(\displaystyle \int \frac {e^x x^2+e^5 \left (-x^2-1\right )+\left (e^5 \left (-x^2+e^3 x-4 x+1\right )+e^x x\right ) \log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )-e^x x}{x}\right ) \log \left (\log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )-e^x x}{x}\right )\right )}{\left (e^5 \left (-x^2+e^3 x-4 x+1\right )+e^x x\right ) \log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )-e^x x}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^x x^2+e^5 \left (-x^2-1\right )+\left (e^5 \left (-x^2+e^3 x-4 x+1\right )+e^x x\right ) \log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )-e^x x}{x}\right ) \log \left (\log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )-e^x x}{x}\right )\right )}{\left (e^5 \left (-x^2+e^3 x-4 x+1\right )+e^x x\right ) \log \left (\frac {e^5 \left (x^2-e^3 x+4 x-1\right )}{x}-e^x\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^5 \left (x^3+\left (3-e^3\right ) x^2-x-1\right )}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )}+\frac {x+\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right ) \log \left (\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )\right )}{\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^5 \left (3-e^3\right ) \int \frac {x^2}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {1}{\left (e^5 x^2-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {x}{\left (e^5 x^2-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {x^3}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+\int \frac {x}{\log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+\int \log \left (\log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )\right )dx\) |
Input:
Int[(E^x*x^2 + E^5*(-1 - x^2) + (E^x*x + E^5*(1 - 4*x + E^3*x - x^2))*Log[ (-(E^x*x) + E^5*(-1 + 4*x - E^3*x + x^2))/x]*Log[Log[(-(E^x*x) + E^5*(-1 + 4*x - E^3*x + x^2))/x]])/((E^x*x + E^5*(1 - 4*x + E^3*x - x^2))*Log[(-(E^ x*x) + E^5*(-1 + 4*x - E^3*x + x^2))/x]),x]
Output:
$Aborted
Time = 12.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
| method | result | size |
| parallelrisch | \(\ln \left (\ln \left (\frac {-{\mathrm e}^{x} x +\left (-x \,{\mathrm e}^{3}+x^{2}+4 x -1\right ) {\mathrm e}^{5}}{x}\right )\right ) x\) | \(31\) |
| risch | \(x \ln \left (i \pi -\ln \left (x \right )+\ln \left ({\mathrm e}^{8} x +\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}+{\mathrm e}^{x} x \right )+\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )}{x}\right ) \left (\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )}{x}\right )-\operatorname {csgn}\left (i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )}{x}\right )}^{2} \left (-\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{8} x -\left (-x^{2}-4 x +1\right ) {\mathrm e}^{5}-{\mathrm e}^{x} x \right )}{x}\right )-1\right )\right )\) | \(250\) |
Input:
int(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*ln((-exp(x)*x+(-x*exp(3)+x^2+4 *x-1)*exp(5))/x)*ln(ln((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x))+exp(x) *x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/ln((-exp(x)*x +(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x,method=_RETURNVERBOSE)
Output:
ln(ln((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x))*x
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-\frac {x e^{8} - {\left (x^{2} + 4 \, x - 1\right )} e^{5} + x e^{x}}{x}\right )\right ) \] Input:
integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 )+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x ))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( (-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm="fricas")
Output:
x*log(log(-(x*e^8 - (x^2 + 4*x - 1)*e^5 + x*e^x)/x))
Time = 4.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log {\left (\log {\left (\frac {- x e^{x} + \left (x^{2} - x e^{3} + 4 x - 1\right ) e^{5}}{x} \right )} \right )} \] Input:
integrate(((exp(x)*x+(x*exp(3)-x**2-4*x+1)*exp(5))*ln((-exp(x)*x+(-x*exp(3 )+x**2+4*x-1)*exp(5))/x)*ln(ln((-exp(x)*x+(-x*exp(3)+x**2+4*x-1)*exp(5))/x ))+exp(x)*x**2+(-x**2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x**2-4*x+1)*exp(5))/l n((-exp(x)*x+(-x*exp(3)+x**2+4*x-1)*exp(5))/x),x)
Output:
x*log(log((-x*exp(x) + (x**2 - x*exp(3) + 4*x - 1)*exp(5))/x))
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (x^{2} e^{5} - x {\left (e^{8} - 4 \, e^{5}\right )} - x e^{x} - e^{5}\right ) - \log \left (x\right )\right ) \] Input:
integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 )+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x ))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( (-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm="maxima")
Output:
x*log(log(x^2*e^5 - x*(e^8 - 4*e^5) - x*e^x - e^5) - log(x))
\[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=\int { -\frac {x^{2} e^{x} - {\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right ) \log \left (\log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )\right ) - {\left (x^{2} + 1\right )} e^{5}}{{\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )} \,d x } \] Input:
integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 )+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x ))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( (-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm="giac")
Output:
integrate(-(x^2*e^x - ((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2 - x* e^3 + 4*x - 1)*e^5 - x*e^x)/x)*log(log(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^ x)/x)) - (x^2 + 1)*e^5)/(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)/x)), x)
Time = 8.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^5\,\left (4\,x-x\,{\mathrm {e}}^3+x^2-1\right )-x\,{\mathrm {e}}^x}{x}\right )\right ) \] Input:
int((exp(5)*(x^2 + 1) - x^2*exp(x) + log(log((exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x))/x))*log((exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x))/x )*(exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x)))/(log((exp(5)*(4*x - x*ex p(3) + x^2 - 1) - x*exp(x))/x)*(exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp( x))),x)
Output:
x*log(log((exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x))/x))
\[ \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx =\text {Too large to display} \] Input:
int(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3)+x^2+ 4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x))+exp (x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log((-exp( x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x)
Output:
- int(x**2/(e**x*log(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x )*x + log(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*e**8*x - l og(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*e**5*x**2 - 4*log (( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*e**5*x + log(( - e* *x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*e**5),x)*e**5 + int(log(lo g(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x))/(e**x*x + e**8*x - e**5*x**2 - 4*e**5*x + e**5),x)*e**5 + int((e**x*x**2)/(e**x*log(( - e** x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*x + log(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x)*e**8*x - log(( - e**x*x - e**8*x + e**5 *x**2 + 4*e**5*x - e**5)/x)*e**5*x**2 - 4*log(( - e**x*x - e**8*x + e**5*x **2 + 4*e**5*x - e**5)/x)*e**5*x + log(( - e**x*x - e**8*x + e**5*x**2 + 4 *e**5*x - e**5)/x)*e**5),x) + int((e**x*log(log(( - e**x*x - e**8*x + e**5 *x**2 + 4*e**5*x - e**5)/x))*x)/(e**x*x + e**8*x - e**5*x**2 - 4*e**5*x + e**5),x) - int((log(log(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5) /x))*x**2)/(e**x*x + e**8*x - e**5*x**2 - 4*e**5*x + e**5),x)*e**5 + int(( log(log(( - e**x*x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x))*x)/(e**x*x + e**8*x - e**5*x**2 - 4*e**5*x + e**5),x)*e**8 - 4*int((log(log(( - e**x* x - e**8*x + e**5*x**2 + 4*e**5*x - e**5)/x))*x)/(e**x*x + e**8*x - e**5*x **2 - 4*e**5*x + e**5),x)*e**5 - int(1/(e**x*log(( - e**x*x - e**8*x + e** 5*x**2 + 4*e**5*x - e**5)/x)*x + log(( - e**x*x - e**8*x + e**5*x**2 + ...