Integrand size = 72, antiderivative size = 29 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {1}{5 x}+\frac {e^{e^{x^3-\frac {4}{x \log (2)}}}}{x} \]
\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx \]
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]
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])))/x^3, x]/(5*Log[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.
| \(\displaystyle \int \frac {e^{e^{\frac {x^4 \log (2)-4}{x \log (2)}}} \left (e^{\frac {x^4 \log (2)-4}{x \log (2)}} \left (15 x^4 \log (2)+20\right )-5 x \log (2)\right )-x \log (2)}{5 x^3 \log (2)} \, dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle -\frac {-20 \int \frac {e^{x^3+e^{x^3-\frac {4}{x \log (2)}}-\frac {4}{\log (2) x}}}{x^3}dx-5 \log (8) \int e^{x^3+e^{x^3-\frac {4}{x \log (2)}}-\frac {4}{\log (2) x}} xdx+\log (32) \int \frac {e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2}dx-\frac {\log (2)}{x}}{5 \log (2)}\) |
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]
3.7.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 1.24 (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 | \(\frac {5 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}+\ln \left (2\right )}{5 \ln \left (2\right ) x}\) | \(35\) |
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)
Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {5 \, e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} + 1}{5 \, x} \]
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=\
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {e^{e^{\frac {x^{4} \log {\left (2 \right )} - 4}{x \log {\left (2 \right )}}}}}{x} + \frac {1}{5 x} \]
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)
\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \]
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=\
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)
\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \]
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=\
Time = 11.62 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (2\right )}}}+\frac {1}{5}}{x} \]