Integrand size = 101, antiderivative size = 24 \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=4+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x}+x \]
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x}+x \]
Integrate[(E^x*x + 5*x*Log[x^2] + E^(-E^Log[E^x + 5*Log[x^2]]^4 + x)*(E^x* x + 5*x*Log[x^2] + E^Log[E^x + 5*Log[x^2]]^4*(-40 - 4*E^x*x)*Log[E^x + 5*L og[x^2]]^3))/(E^x*x + 5*x*Log[x^2]),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.
\(\displaystyle \int \frac {e^{x-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}} \left (\left (-4 e^x x-40\right ) e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )} \log ^3\left (5 \log \left (x^2\right )+e^x\right )+5 x \log \left (x^2\right )+e^x x\right )+5 x \log \left (x^2\right )+e^x x}{5 x \log \left (x^2\right )+e^x x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}} \left (e^{e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}}+e^x\right )-\frac {4 \left (e^x x+10\right ) \log ^3\left (5 \log \left (x^2\right )+e^x\right ) \exp \left (\log ^4\left (5 \log \left (x^2\right )+e^x\right )-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}+x\right )}{x \left (5 \log \left (x^2\right )+e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \exp \left (\log ^4\left (5 \log \left (x^2\right )+e^x\right )-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}+x\right ) \log ^3\left (5 \log \left (x^2\right )+e^x\right )dx-40 \int \frac {\exp \left (\log ^4\left (5 \log \left (x^2\right )+e^x\right )-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}+x\right ) \log ^3\left (5 \log \left (x^2\right )+e^x\right )}{x \left (5 \log \left (x^2\right )+e^x\right )}dx+20 \int \frac {\exp \left (\log ^4\left (5 \log \left (x^2\right )+e^x\right )-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}+x\right ) \log \left (x^2\right ) \log ^3\left (5 \log \left (x^2\right )+e^x\right )}{5 \log \left (x^2\right )+e^x}dx+\int e^{x-e^{\log ^4\left (5 \log \left (x^2\right )+e^x\right )}}dx+x\) |
Int[(E^x*x + 5*x*Log[x^2] + E^(-E^Log[E^x + 5*Log[x^2]]^4 + x)*(E^x*x + 5* x*Log[x^2] + E^Log[E^x + 5*Log[x^2]]^4*(-40 - 4*E^x*x)*Log[E^x + 5*Log[x^2 ]]^3))/(E^x*x + 5*x*Log[x^2]),x]
3.25.100.3.1 Defintions of rubi rules used
Time = 79.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(x +{\mathrm e}^{-{\mathrm e}^{{\ln \left (5 \ln \left (x^{2}\right )+{\mathrm e}^{x}\right )}^{4}}+x}\) | \(21\) |
risch | \(x +{\mathrm e}^{x} {\mathrm e}^{-{\mathrm e}^{{\ln \left (10 \ln \left (x \right )-\frac {5 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}+{\mathrm e}^{x}\right )}^{4}}}\) | \(48\) |
int((((-4*exp(x)*x-40)*ln(5*ln(x^2)+exp(x))^3*exp(ln(5*ln(x^2)+exp(x))^4)+ 5*x*ln(x^2)+exp(x)*x)*exp(-exp(ln(5*ln(x^2)+exp(x))^4)+x)+5*x*ln(x^2)+exp( x)*x)/(5*x*ln(x^2)+exp(x)*x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=x + e^{\left (x - e^{\left (\log \left (e^{x} + 5 \, \log \left (x^{2}\right )\right )^{4}\right )}\right )} \]
integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+e xp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x *log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm=\
Timed out. \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=\text {Timed out} \]
integrate((((-4*exp(x)*x-40)*ln(5*ln(x**2)+exp(x))**3*exp(ln(5*ln(x**2)+ex p(x))**4)+5*x*ln(x**2)+exp(x)*x)*exp(-exp(ln(5*ln(x**2)+exp(x))**4)+x)+5*x *ln(x**2)+exp(x)*x)/(5*x*ln(x**2)+exp(x)*x),x)
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=x + e^{\left (x - e^{\left (\log \left (e^{x} + 10 \, \log \left (x\right )\right )^{4}\right )}\right )} \]
integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+e xp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x *log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm=\
\[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=\int { -\frac {{\left (4 \, {\left (x e^{x} + 10\right )} e^{\left (\log \left (e^{x} + 5 \, \log \left (x^{2}\right )\right )^{4}\right )} \log \left (e^{x} + 5 \, \log \left (x^{2}\right )\right )^{3} - x e^{x} - 5 \, x \log \left (x^{2}\right )\right )} e^{\left (x - e^{\left (\log \left (e^{x} + 5 \, \log \left (x^{2}\right )\right )^{4}\right )}\right )} - x e^{x} - 5 \, x \log \left (x^{2}\right )}{x e^{x} + 5 \, x \log \left (x^{2}\right )} \,d x } \]
integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+e xp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x *log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm=\
integrate(-((4*(x*e^x + 10)*e^(log(e^x + 5*log(x^2))^4)*log(e^x + 5*log(x^ 2))^3 - x*e^x - 5*x*log(x^2))*e^(x - e^(log(e^x + 5*log(x^2))^4)) - x*e^x - 5*x*log(x^2))/(x*e^x + 5*x*log(x^2)), x)
Time = 11.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx=x+{\mathrm {e}}^{-{\mathrm {e}}^{{\ln \left (\ln \left (x^{10}\right )+{\mathrm {e}}^x\right )}^4}}\,{\mathrm {e}}^x \]