Integrand size = 95, antiderivative size = 22 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=\left (15+x^2\right ) \log \left (5 \left (e^{1-\frac {1}{x^2}}+x^2\right )\right ) \]
Time = 0.50 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=1-\frac {15}{x^2}+x^2 \log \left (5 \left (e^{1-\frac {1}{x^2}}+x^2\right )\right )+15 \log \left (e+e^{\frac {1}{x^2}} x^2\right ) \]
Integrate[(30*x^4 + 2*x^6 + E^((-1 + x^2)/x^2)*(30 + 2*x^2) + (2*E^((-1 + x^2)/x^2)*x^4 + 2*x^6)*Log[5*E^((-1 + x^2)/x^2) + 5*x^2])/(E^((-1 + x^2)/x ^2)*x^3 + x^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.
\(\displaystyle \int \frac {2 x^6+30 x^4+e^{\frac {x^2-1}{x^2}} \left (2 x^2+30\right )+\left (2 x^6+2 e^{\frac {x^2-1}{x^2}} x^4\right ) \log \left (5 x^2+5 e^{\frac {x^2-1}{x^2}}\right )}{x^5+e^{\frac {x^2-1}{x^2}} x^3} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (2 x \log \left (5 \left (x^2+e^{1-\frac {1}{x^2}}\right )\right )+\frac {2 \left (x^2+15\right ) \left (e^{\frac {1}{x^2}} x^4+e\right )}{x^3 \left (e^{\frac {1}{x^2}} x^2+e\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 15 e \text {Subst}\left (\int \frac {1}{x^2 \left (e^{\frac {1}{x}} x+e\right )}dx,x,x^2\right )-15 e \text {Subst}\left (\int \frac {1}{x \left (e^{\frac {1}{x}} x+e\right )}dx,x,x^2\right )+x^2 \log \left (5 \left (x^2+e^{1-\frac {1}{x^2}}\right )\right )+30 \log (x)\) |
Int[(30*x^4 + 2*x^6 + E^((-1 + x^2)/x^2)*(30 + 2*x^2) + (2*E^((-1 + x^2)/x ^2)*x^4 + 2*x^6)*Log[5*E^((-1 + x^2)/x^2) + 5*x^2])/(E^((-1 + x^2)/x^2)*x^ 3 + x^5),x]
3.13.70.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.57 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05
method | result | size |
risch | \(x^{2} \ln \left (5 \,{\mathrm e}^{\frac {\left (-1+x \right ) \left (1+x \right )}{x^{2}}}+5 x^{2}\right )+15 \ln \left ({\mathrm e}^{\frac {\left (-1+x \right ) \left (1+x \right )}{x^{2}}}+x^{2}\right )-15\) | \(45\) |
parallelrisch | \(x^{2} \ln \left (5 \,{\mathrm e}^{\frac {x^{2}-1}{x^{2}}}+5 x^{2}\right )+15 \ln \left (5 \,{\mathrm e}^{\frac {x^{2}-1}{x^{2}}}+5 x^{2}\right )\) | \(46\) |
int(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*ln(5*exp((x^2-1)/x^2)+5*x^2)+(2*x^2+30 )*exp((x^2-1)/x^2)+2*x^6+30*x^4)/(x^3*exp((x^2-1)/x^2)+x^5),x,method=_RETU RNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx={\left (x^{2} + 15\right )} \log \left (5 \, x^{2} + 5 \, e^{\left (\frac {x^{2} - 1}{x^{2}}\right )}\right ) \]
integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2 *x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algo rithm=\
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=x^{2} \log {\left (5 x^{2} + 5 e^{\frac {x^{2} - 1}{x^{2}}} \right )} + 15 \log {\left (x^{2} + e^{\frac {x^{2} - 1}{x^{2}}} \right )} \]
integrate(((2*x**4*exp((x**2-1)/x**2)+2*x**6)*ln(5*exp((x**2-1)/x**2)+5*x* *2)+(2*x**2+30)*exp((x**2-1)/x**2)+2*x**6+30*x**4)/(x**3*exp((x**2-1)/x**2 )+x**5),x)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (23) = 46\).
Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=\frac {x^{4} \log \left (5\right ) + x^{4} \log \left (x^{2} e^{\left (\frac {1}{x^{2}}\right )} + e\right ) - 15}{x^{2}} + 30 \, \log \left (x\right ) + 15 \, \log \left (\frac {x^{2} e^{\left (\frac {1}{x^{2}}\right )} + e}{x^{2}}\right ) \]
integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2 *x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algo rithm=\
(x^4*log(5) + x^4*log(x^2*e^(x^(-2)) + e) - 15)/x^2 + 30*log(x) + 15*log(( x^2*e^(x^(-2)) + e)/x^2)
Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=x^{2} \log \left (5 \, x^{2} + 5 \, e^{\left (\frac {x^{2} - 1}{x^{2}}\right )}\right ) + 15 \, \log \left (x^{2} + e^{\left (\frac {x^{2} - 1}{x^{2}}\right )}\right ) \]
integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2 *x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algo rithm=\
Time = 11.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {30 x^4+2 x^6+e^{\frac {-1+x^2}{x^2}} \left (30+2 x^2\right )+\left (2 e^{\frac {-1+x^2}{x^2}} x^4+2 x^6\right ) \log \left (5 e^{\frac {-1+x^2}{x^2}}+5 x^2\right )}{e^{\frac {-1+x^2}{x^2}} x^3+x^5} \, dx=15\,\ln \left (\mathrm {e}\,{\mathrm {e}}^{-\frac {1}{x^2}}+x^2\right )+x^2\,\ln \left (5\,\mathrm {e}\,{\mathrm {e}}^{-\frac {1}{x^2}}+5\,x^2\right ) \]