Integrand size = 89, antiderivative size = 24 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} e^{\frac {x^2}{x+\frac {2}{x^2+\log (5)}}} \]
\[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx \]
Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*L og[5] + 5*x^2*Log[5]^2),x]
Integrate[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*L og[5] + 5*x^2*Log[5]^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.
\(\displaystyle \int \frac {e^{\frac {x^4+x^2 \log (5)}{x^3+x \log (5)+2}} \left (x^6+\left (2 x^4+4 x\right ) \log (5)+8 x^3+x^2 \log ^2(5)\right )}{5 x^6+\left (10 x^4+20 x\right ) \log (5)+20 x^3+5 x^2 \log ^2(5)+20} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {e^{\frac {x^4+x^2 \log (5)}{x^3+x \log (5)+2}} \left (x^6+\left (2 x^4+4 x\right ) \log (5)+8 x^3+x^2 \log ^2(5)\right )}{5 \left (x^3+x \log (5)+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} \left (x^6+8 x^3+\log ^2(5) x^2+2 \left (x^4+2 x\right ) \log (5)\right )}{\left (x^3+\log (5) x+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (-\frac {4\ 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} (\log (5) x+3)}{\left (x^3+\log (5) x+2\right )^2}+5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}+\frac {4\ 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{x^3+\log (5) x+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\int 5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}dx-12 \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{\left (x^3+\log (5) x+2\right )^2}dx-4 \log (5) \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}} x}{\left (x^3+\log (5) x+2\right )^2}dx+4 \int \frac {5^{\frac {x^2}{x^3+\log (5) x+2}} e^{\frac {x^4}{x^3+\log (5) x+2}}}{x^3+\log (5) x+2}dx\right )\) |
Int[(E^((x^4 + x^2*Log[5])/(2 + x^3 + x*Log[5]))*(8*x^3 + x^6 + (4*x + 2*x ^4)*Log[5] + x^2*Log[5]^2))/(20 + 20*x^3 + 5*x^6 + (20*x + 10*x^4)*Log[5] + 5*x^2*Log[5]^2),x]
3.14.3.3.1 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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 2.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
risch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {x^{2} \left (x^{2}+\ln \left (5\right )\right )}{x \ln \left (5\right )+x^{3}+2}}}{5}\) | \(25\) |
norman | \(\frac {\frac {x^{3} {\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}+\frac {x \ln \left (5\right ) {\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}+\frac {2 \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )+x^{4}}{x \ln \left (5\right )+x^{3}+2}}}{5}}{x \ln \left (5\right )+x^{3}+2}\) | \(95\) |
int((x^2*ln(5)^2+(2*x^4+4*x)*ln(5)+x^6+8*x^3)*exp((x^2*ln(5)+x^4)/(x*ln(5) +x^3+2))/(5*x^2*ln(5)^2+(10*x^4+20*x)*ln(5)+5*x^6+20*x^3+20),x,method=_RET URNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4} + x^{2} \log \left (5\right )}{x^{3} + x \log \left (5\right ) + 2}\right )} \]
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm=\
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {e^{\frac {x^{4} + x^{2} \log {\left (5 \right )}}{x^{3} + x \log {\left (5 \right )} + 2}}}{5} \]
integrate((x**2*ln(5)**2+(2*x**4+4*x)*ln(5)+x**6+8*x**3)*exp((x**2*ln(5)+x **4)/(x*ln(5)+x**3+2))/(5*x**2*ln(5)**2+(10*x**4+20*x)*ln(5)+5*x**6+20*x** 3+20),x)
Time = 0.41 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (x - \frac {2 \, x}{x^{3} + x \log \left (5\right ) + 2}\right )} \]
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm=\
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4}}{x^{3} + x \log \left (5\right ) + 2} + \frac {x^{2} \log \left (5\right )}{x^{3} + x \log \left (5\right ) + 2}\right )} \]
integrate((x^2*log(5)^2+(2*x^4+4*x)*log(5)+x^6+8*x^3)*exp((x^2*log(5)+x^4) /(x*log(5)+x^3+2))/(5*x^2*log(5)^2+(10*x^4+20*x)*log(5)+5*x^6+20*x^3+20),x , algorithm=\
Time = 11.98 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {x^4+x^2 \log (5)}{2+x^3+x \log (5)}} \left (8 x^3+x^6+\left (4 x+2 x^4\right ) \log (5)+x^2 \log ^2(5)\right )}{20+20 x^3+5 x^6+\left (20 x+10 x^4\right ) \log (5)+5 x^2 \log ^2(5)} \, dx=\frac {{\mathrm {e}}^{\frac {x^4}{x^3+\ln \left (5\right )\,x+2}}\,{\mathrm {e}}^{\frac {x^2\,\ln \left (5\right )}{x^3+\ln \left (5\right )\,x+2}}}{5} \]