Integrand size = 51, antiderivative size = 22 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} x \left (x+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}\right ) \]
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} x \left (x+\left (e^{-2 (1+x)} x^4\right )^{\frac {1}{x^2}}\right ) \]
Integrate[(2*x^3 + (E^(-2 - 2*x)*x^4)^x^(-2)*(4 - 2*x + x^2 - 2*Log[E^(-2 - 2*x)*x^4]))/(2*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.
\(\displaystyle \int \frac {2 x^3+\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}} \left (-2 \log \left (e^{-2 x-2} x^4\right )+x^2-2 x+4\right )}{2 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {\left (x^2-2 x-2 \log \left (e^{-2 x-2} x^4\right )+4\right ) \left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}+2 x^3}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (x^2-2 x-2 \log \left (e^{-2 (x+1)} x^4\right )+4\right ) \left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x^2}+2 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\int \left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}dx+4 \int \frac {\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x^2}dx-2 \int \frac {\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x}dx-4 \int \int \frac {\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x^2}dxdx+8 \int \frac {\int \frac {\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x^2}dx}{x}dx-2 \log \left (e^{-2 x-2} x^4\right ) \int \frac {\left (e^{-2 x-2} x^4\right )^{\frac {1}{x^2}}}{x^2}dx+x^2\right )\) |
3.4.59.3.1 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 = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\) | \(26\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\) | \(26\) |
parts | \(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\) | \(26\) |
risch | \(\frac {x^{2}}{2}+\frac {x \,x^{\frac {4}{x^{2}}} \left ({\mathrm e}^{1+x}\right )^{-\frac {2}{x^{2}}} {\mathrm e}^{-\frac {i \pi \left (-\operatorname {csgn}\left (i {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{2}\right )^{3}-2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )-\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\operatorname {csgn}\left (i x^{3}\right )^{3}-\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{3}-\operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{2} \operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{4}\right )^{3}-\operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right )^{3}+2 \operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{1+x}\right )-\operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{1+x}\right )^{2}\right )}{2 x^{2}}}}{2}\) | \(357\) |
int(1/2*((-2*ln(x^4/exp(1+x)^2)+x^2-2*x+4)*exp(ln(x^4/exp(1+x)^2)/x^2)+2*x ^3)/x^2,x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} \, \left (x^{4} e^{\left (-2 \, x - 2\right )}\right )^{\left (\frac {1}{x^{2}}\right )} x + \frac {1}{2} \, x^{2} \]
integrate(1/2*((-2*log(x^4/exp(1+x)^2)+x^2-2*x+4)*exp(log(x^4/exp(1+x)^2)/ x^2)+2*x^3)/x^2,x, algorithm=\
\[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {\int 2 x\, dx + \int e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}\, dx + \int \frac {8 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}}{x^{2}}\, dx + \int \left (- \frac {2 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}}{x}\right )\, dx + \int \left (- \frac {2 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}} \log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}\right )\, dx}{2} \]
integrate(1/2*((-2*ln(x**4/exp(1+x)**2)+x**2-2*x+4)*exp(ln(x**4/exp(1+x)** 2)/x**2)+2*x**3)/x**2,x)
(Integral(2*x, x) + Integral(exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2), x ) + Integral(8*exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2)/x**2, x) + Integ ral(-2*exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2)/x, x) + Integral(-2*exp( -2/x**2)*exp(log(x**4*exp(-2*x))/x**2)*log(x**4*exp(-2*x))/x**2, x))/2
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{\left (-\frac {2}{x} + \frac {4 \, \log \left (x\right )}{x^{2}} - \frac {2}{x^{2}}\right )} \]
integrate(1/2*((-2*log(x^4/exp(1+x)^2)+x^2-2*x+4)*exp(log(x^4/exp(1+x)^2)/ x^2)+2*x^3)/x^2,x, algorithm=\
\[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\int { \frac {2 \, x^{3} + {\left (x^{2} - 2 \, x - 2 \, \log \left (x^{4} e^{\left (-2 \, x - 2\right )}\right ) + 4\right )} \left (x^{4} e^{\left (-2 \, x - 2\right )}\right )^{\left (\frac {1}{x^{2}}\right )}}{2 \, x^{2}} \,d x } \]
integrate(1/2*((-2*log(x^4/exp(1+x)^2)+x^2-2*x+4)*exp(log(x^4/exp(1+x)^2)/ x^2)+2*x^3)/x^2,x, algorithm=\
integrate(1/2*(2*x^3 + (x^2 - 2*x - 2*log(x^4*e^(-2*x - 2)) + 4)*(x^4*e^(- 2*x - 2))^(x^(-2)))/x^2, x)
Time = 7.92 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {x^2}{2}+\frac {x\,{\mathrm {e}}^{-\frac {2}{x}}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\left (x^4\right )}^{\frac {1}{x^2}}}{2} \]