Integrand size = 56, antiderivative size = 21 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=e^{17+e^{\log ^2(x)}+\frac {x \log (5)}{e^5}}-x \]
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=5^{\frac {x}{e^5}} e^{17+e^{\log ^2(x)}}-x \]
Integrate[(-(E^5*x) + E^((17*E^5 + E^(5 + Log[x]^2) + x*Log[5])/E^5)*(x*Lo g[5] + 2*E^(5 + Log[x]^2)*Log[x]))/(E^5*x),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.
\(\displaystyle \int \frac {e^{\frac {e^{\log ^2(x)+5}+x \log (5)+17 e^5}{e^5}} \left (2 e^{\log ^2(x)+5} \log (x)+x \log (5)\right )-e^5 x}{e^5 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^5 x-5^{\frac {x}{e^5}} e^{\frac {17 e^5+e^{\log ^2(x)+5}}{e^5}} \left (\log (5) x+2 e^{\log ^2(x)+5} \log (x)\right )}{x}dx}{e^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^5 x-5^{\frac {x}{e^5}} e^{\frac {17 e^5+e^{\log ^2(x)+5}}{e^5}} \left (\log (5) x+2 e^{\log ^2(x)+5} \log (x)\right )}{x}dx}{e^5}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (-e^5 \left (-1+5^{\frac {x}{e^5}} e^{12+e^{\log ^2(x)}} \log (5)\right )-\frac {2\ 5^{\frac {x}{e^5}} e^{\log ^2(x)+e^{\log ^2(x)}+22} \log (x)}{x}\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-e^5 \log (5) \int 5^{\frac {x}{e^5}} e^{12+e^{\log ^2(x)}}dx-2 \int \frac {5^{\frac {x}{e^5}} e^{\log ^2(x)+e^{\log ^2(x)}+22} \log (x)}{x}dx+e^5 x}{e^5}\) |
Int[(-(E^5*x) + E^((17*E^5 + E^(5 + Log[x]^2) + x*Log[5])/E^5)*(x*Log[5] + 2*E^(5 + Log[x]^2)*Log[x]))/(E^5*x),x]
3.14.20.3.1 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 = 0.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19
method | result | size |
risch | \(-x +{\mathrm e}^{\left ({\mathrm e}^{5+\ln \left (x \right )^{2}}+x \ln \left (5\right )+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}\) | \(25\) |
norman | \(-x +{\mathrm e}^{\left ({\mathrm e}^{5} {\mathrm e}^{\ln \left (x \right )^{2}}+x \ln \left (5\right )+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}\) | \(28\) |
default | \({\mathrm e}^{-5} \left ({\mathrm e}^{\left ({\mathrm e}^{5+\ln \left (x \right )^{2}}+x \ln \left (5\right )+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}} {\mathrm e}^{5}-x \,{\mathrm e}^{5}\right )\) | \(37\) |
parallelrisch | \({\mathrm e}^{-5} \left (-x \,{\mathrm e}^{5}+{\mathrm e}^{5} {\mathrm e}^{\left ({\mathrm e}^{5} {\mathrm e}^{\ln \left (x \right )^{2}}+x \ln \left (5\right )+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}\right )\) | \(38\) |
int(((2*exp(5)*ln(x)*exp(ln(x)^2)+x*ln(5))*exp((exp(5)*exp(ln(x)^2)+x*ln(5 )+17*exp(5))/exp(5))-x*exp(5))/x/exp(5),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=-x + e^{\left ({\left (x \log \left (5\right ) + 17 \, e^{5} + e^{\left (\log \left (x\right )^{2} + 5\right )}\right )} e^{\left (-5\right )}\right )} \]
integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x) ^2)+x*log(5)+17*exp(5))/exp(5))-x*exp(5))/x/exp(5),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=- x + e^{\frac {x \log {\left (5 \right )} + e^{5} e^{\log {\left (x \right )}^{2}} + 17 e^{5}}{e^{5}}} \]
integrate(((2*exp(5)*ln(x)*exp(ln(x)**2)+x*ln(5))*exp((exp(5)*exp(ln(x)**2 )+x*ln(5)+17*exp(5))/exp(5))-x*exp(5))/x/exp(5),x)
Time = 0.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=-{\left (x e^{5} - e^{\left (x e^{\left (-5\right )} \log \left (5\right ) + e^{\left (\log \left (x\right )^{2}\right )} + 22\right )}\right )} e^{\left (-5\right )} \]
integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x) ^2)+x*log(5)+17*exp(5))/exp(5))-x*exp(5))/x/exp(5),x, algorithm=\
\[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=\int { -\frac {{\left (x e^{5} - {\left (x \log \left (5\right ) + 2 \, e^{\left (\log \left (x\right )^{2} + 5\right )} \log \left (x\right )\right )} e^{\left ({\left (x \log \left (5\right ) + 17 \, e^{5} + e^{\left (\log \left (x\right )^{2} + 5\right )}\right )} e^{\left (-5\right )}\right )}\right )} e^{\left (-5\right )}}{x} \,d x } \]
integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x) ^2)+x*log(5)+17*exp(5))/exp(5))-x*exp(5))/x/exp(5),x, algorithm=\
integrate(-(x*e^5 - (x*log(5) + 2*e^(log(x)^2 + 5)*log(x))*e^((x*log(5) + 17*e^5 + e^(log(x)^2 + 5))*e^(-5)))*e^(-5)/x, x)
Time = 13.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{e^5 x} \, dx=5^{x\,{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{{\mathrm {e}}^{{\ln \left (x\right )}^2}}\,{\mathrm {e}}^{17}-x \]