Integrand size = 75, antiderivative size = 25 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^2} \] Output:
exp(exp(exp(x)+3*x+5)*ln(ln(x)^2)+4+x)/x^2
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {e^{4+x} \log ^2(x)^{e^{5+e^x+3 x}}}{x^2} \] Input:
Integrate[(E^(4 + x + E^(5 + E^x + 3*x)*Log[Log[x]^2])*(2*E^(5 + E^x + 3*x ) + (-2 + x)*Log[x] + E^(5 + E^x + 3*x)*(3*x + E^x*x)*Log[x]*Log[Log[x]^2] ))/(x^3*Log[x]),x]
Output:
(E^(4 + x)*(Log[x]^2)^E^(5 + E^x + 3*x))/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.
| \(\displaystyle \int \frac {e^{x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+4} \left (2 e^{3 x+e^x+5}+e^{3 x+e^x+5} \left (e^x x+3 x\right ) \log \left (\log ^2(x)\right ) \log (x)+(x-2) \log (x)\right )}{x^3 \log (x)} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {(x-2) e^{x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+4}}{x^3}+\frac {e^{4 x+e^x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+9} \left (3 x \log (x) \log \left (\log ^2(x)\right )+2\right )}{x^3 \log (x)}+\frac {e^{5 x+e^x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+9} \log \left (\log ^2(x)\right )}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {e^{x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+4}}{x^3}dx+2 \int \frac {e^{4 x+e^x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+9}}{x^3 \log (x)}dx+\int \frac {e^{x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+4}}{x^2}dx+3 \int \frac {e^{4 x+e^x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+9} \log \left (\log ^2(x)\right )}{x^2}dx+\int \frac {e^{5 x+e^x+e^{3 x+e^x+5} \log \left (\log ^2(x)\right )+9} \log \left (\log ^2(x)\right )}{x^2}dx\) |
Input:
Int[(E^(4 + x + E^(5 + E^x + 3*x)*Log[Log[x]^2])*(2*E^(5 + E^x + 3*x) + (- 2 + x)*Log[x] + E^(5 + E^x + 3*x)*(3*x + E^x*x)*Log[x]*Log[Log[x]^2]))/(x^ 3*Log[x]),x]
Output:
$Aborted
Time = 10.48 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
| method | result | size |
| parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}+3 x +5} \ln \left (\ln \left (x \right )^{2}\right )+4+x}}{x^{2}}\) | \(23\) |
| risch | \(\frac {\ln \left (x \right )^{2 \,{\mathrm e}^{{\mathrm e}^{x}+3 x +5}} {\mathrm e}^{4-\frac {i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{2}+i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )-\frac {i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2}}{2}+x}}{x^{2}}\) | \(100\) |
Input:
int(((exp(x)*x+3*x)*ln(x)*exp(exp(x)+3*x+5)*ln(ln(x)^2)+2*exp(exp(x)+3*x+5 )+(-2+x)*ln(x))*exp(exp(exp(x)+3*x+5)*ln(ln(x)^2)+4+x)/x^3/ln(x),x,method= _RETURNVERBOSE)
Output:
exp(exp(exp(x)+3*x+5)*ln(ln(x)^2)+4+x)/x^2
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {e^{\left (e^{\left (3 \, x + e^{x} + 5\right )} \log \left (\log \left (x\right )^{2}\right ) + x + 4\right )}}{x^{2}} \] Input:
integrate(((exp(x)*x+3*x)*log(x)*exp(exp(x)+3*x+5)*log(log(x)^2)+2*exp(exp (x)+3*x+5)+(-2+x)*log(x))*exp(exp(exp(x)+3*x+5)*log(log(x)^2)+4+x)/x^3/log (x),x, algorithm="fricas")
Output:
e^(e^(3*x + e^x + 5)*log(log(x)^2) + x + 4)/x^2
Time = 3.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {e^{x + e^{3 x + e^{x} + 5} \log {\left (\log {\left (x \right )}^{2} \right )} + 4}}{x^{2}} \] Input:
integrate(((exp(x)*x+3*x)*ln(x)*exp(exp(x)+3*x+5)*ln(ln(x)**2)+2*exp(exp(x )+3*x+5)+(-2+x)*ln(x))*exp(exp(exp(x)+3*x+5)*ln(ln(x)**2)+4+x)/x**3/ln(x), x)
Output:
exp(x + exp(3*x + exp(x) + 5)*log(log(x)**2) + 4)/x**2
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {e^{\left (2 \, e^{\left (3 \, x + e^{x} + 5\right )} \log \left (\log \left (x\right )\right ) + x + 4\right )}}{x^{2}} \] Input:
integrate(((exp(x)*x+3*x)*log(x)*exp(exp(x)+3*x+5)*log(log(x)^2)+2*exp(exp (x)+3*x+5)+(-2+x)*log(x))*exp(exp(exp(x)+3*x+5)*log(log(x)^2)+4+x)/x^3/log (x),x, algorithm="maxima")
Output:
e^(2*e^(3*x + e^x + 5)*log(log(x)) + x + 4)/x^2
\[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\int { \frac {{\left ({\left (x e^{x} + 3 \, x\right )} e^{\left (3 \, x + e^{x} + 5\right )} \log \left (x\right ) \log \left (\log \left (x\right )^{2}\right ) + {\left (x - 2\right )} \log \left (x\right ) + 2 \, e^{\left (3 \, x + e^{x} + 5\right )}\right )} e^{\left (e^{\left (3 \, x + e^{x} + 5\right )} \log \left (\log \left (x\right )^{2}\right ) + x + 4\right )}}{x^{3} \log \left (x\right )} \,d x } \] Input:
integrate(((exp(x)*x+3*x)*log(x)*exp(exp(x)+3*x+5)*log(log(x)^2)+2*exp(exp (x)+3*x+5)+(-2+x)*log(x))*exp(exp(exp(x)+3*x+5)*log(log(x)^2)+4+x)/x^3/log (x),x, algorithm="giac")
Output:
undef
Time = 3.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\frac {{\mathrm {e}}^4\,{\mathrm {e}}^x\,{\left ({\ln \left (x\right )}^2\right )}^{{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^5}}{x^2} \] Input:
int((exp(x + exp(3*x + exp(x) + 5)*log(log(x)^2) + 4)*(2*exp(3*x + exp(x) + 5) + log(x)*(x - 2) + exp(3*x + exp(x) + 5)*log(log(x)^2)*log(x)*(3*x + x*exp(x))))/(x^3*log(x)),x)
Output:
(exp(4)*exp(x)*(log(x)^2)^(exp(3*x)*exp(exp(x))*exp(5)))/x^2
\[ \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx=\int \frac {\left (\left ({\mathrm e}^{x} x +3 x \right ) \mathrm {log}\left (x \right ) {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )+2 \,{\mathrm e}^{{\mathrm e}^{x}+3 x +5}+\left (x -2\right ) \mathrm {log}\left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}+3 x +5} \mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )+4+x}}{x^{3} \mathrm {log}\left (x \right )}d x \] Input:
int(((exp(x)*x+3*x)*log(x)*exp(exp(x)+3*x+5)*log(log(x)^2)+2*exp(exp(x)+3* x+5)+(-2+x)*log(x))*exp(exp(exp(x)+3*x+5)*log(log(x)^2)+4+x)/x^3/log(x),x)
Output:
int(((exp(x)*x+3*x)*log(x)*exp(exp(x)+3*x+5)*log(log(x)^2)+2*exp(exp(x)+3* x+5)+(-2+x)*log(x))*exp(exp(exp(x)+3*x+5)*log(log(x)^2)+4+x)/x^3/log(x),x)