Integrand size = 72, antiderivative size = 29 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {1}{5 x}+\frac {e^{e^{x^3-\frac {4}{x \log (2)}}}}{x} \] Output:
exp(exp(x^3-4/x/ln(2)))/x+1/5/x
Time = 2.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {1+5 e^{e^{x^3-\frac {4}{x \log (2)}}}}{5 x} \] Input:
Integrate[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((-4 + x^4*Log[2])/(x*Log[2]))*(20 + 15*x^4*Log[2])))/(5*x^3*Log[2]),x]
Output:
(1 + 5*E^E^(x^3 - 4/(x*Log[2])))/(5*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.
\(\displaystyle \int \frac {e^{e^{\frac {x^4 \log (2)-4}{x \log (2)}}} \left (e^{\frac {x^4 \log (2)-4}{x \log (2)}} \left (15 x^4 \log (2)+20\right )-5 x \log (2)\right )-x \log (2)}{5 x^3 \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\log (2) x+5 e^{2^{\frac {x^3}{\log (2)}} e^{-\frac {4}{x \log (2)}}} \left (x \log (2)-2^{\frac {x^3}{\log (2)}} e^{-\frac {4}{x \log (2)}} \left (3 \log (2) x^4+4\right )\right )}{x^3}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\log (2) x+5 e^{2^{\frac {x^3}{\log (2)}} e^{-\frac {4}{x \log (2)}}} \left (x \log (2)-2^{\frac {x^3}{\log (2)}} e^{-\frac {4}{x \log (2)}} \left (3 \log (2) x^4+4\right )\right )}{x^3}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (\frac {\log (2)+e^{e^{x^3-\frac {4}{x \log (2)}}} \log (32)}{x^2}-\frac {5 e^{x^3+e^{x^3-\frac {4}{x \log (2)}}-\frac {4}{\log (2) x}} \left (\log (8) x^4+4\right )}{x^3}\right )dx}{5 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-20 \int \frac {e^{x^3+e^{x^3-\frac {4}{x \log (2)}}-\frac {4}{\log (2) x}}}{x^3}dx-5 \log (8) \int e^{x^3+e^{x^3-\frac {4}{x \log (2)}}-\frac {4}{\log (2) x}} xdx+\log (32) \int \frac {e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2}dx-\frac {\log (2)}{x}}{5 \log (2)}\) |
Input:
Int[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((- 4 + x^4*Log[2])/(x*Log[2]))*(20 + 15*x^4*Log[2])))/(5*x^3*Log[2]),x]
Output:
$Aborted
Time = 1.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}+\frac {x}{5}}{x^{2}}\) | \(29\) |
risch | \(\frac {1}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}}{x}\) | \(29\) |
parallelrisch | \(\frac {5 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}+\ln \left (2\right )}{5 \ln \left (2\right ) x}\) | \(35\) |
Input:
int(1/5*(((15*x^4*ln(2)+20)*exp((x^4*ln(2)-4)/x/ln(2))-5*x*ln(2))*exp(exp( (x^4*ln(2)-4)/x/ln(2)))-x*ln(2))/x^3/ln(2),x,method=_RETURNVERBOSE)
Output:
(x*exp(exp((x^4*ln(2)-4)/x/ln(2)))+1/5*x)/x^2
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {5 \, e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} + 1}{5 \, x} \] Input:
integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2) )*exp(exp((x^4*log(2)-4)/x/log(2)))-x*log(2))/x^3/log(2),x, algorithm="fri cas")
Output:
1/5*(5*e^(e^((x^4*log(2) - 4)/(x*log(2)))) + 1)/x
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {e^{e^{\frac {x^{4} \log {\left (2 \right )} - 4}{x \log {\left (2 \right )}}}}}{x} + \frac {1}{5 x} \] Input:
integrate(1/5*(((15*x**4*ln(2)+20)*exp((x**4*ln(2)-4)/x/ln(2))-5*x*ln(2))* exp(exp((x**4*ln(2)-4)/x/ln(2)))-x*ln(2))/x**3/ln(2),x)
Output:
exp(exp((x**4*log(2) - 4)/(x*log(2))))/x + 1/(5*x)
\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \] Input:
integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2) )*exp(exp((x^4*log(2)-4)/x/log(2)))-x*log(2))/x^3/log(2),x, algorithm="max ima")
Output:
1/5*(log(2)/x - integrate(5*(x*e^(4/(x*log(2)))*log(2) - (3*x^4*log(2) + 4 )*e^(x^3))*e^(-4/(x*log(2)) + e^(x^3 - 4/(x*log(2))))/x^3, x))/log(2)
\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \] Input:
integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2) )*exp(exp((x^4*log(2)-4)/x/log(2)))-x*log(2))/x^3/log(2),x, algorithm="gia c")
Output:
undef
Time = 2.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (2\right )}}}+\frac {1}{5}}{x} \] Input:
int(-((x*log(2))/5 + (exp(exp((x^4*log(2) - 4)/(x*log(2))))*(5*x*log(2) - exp((x^4*log(2) - 4)/(x*log(2)))*(15*x^4*log(2) + 20)))/5)/(x^3*log(2)),x)
Output:
(exp(exp(x^3)*exp(-4/(x*log(2)))) + 1/5)/x
Time = 0.64 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {5 e^{\frac {e^{x^{3}}}{e^{\frac {4}{\mathrm {log}\left (2\right ) x}}}}+1}{5 x} \] Input:
int(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp( exp((x^4*log(2)-4)/x/log(2)))-x*log(2))/x^3/log(2),x)
Output:
(5*e**(e**(x**3)/e**(4/(log(2)*x))) + 1)/(5*x)