Integrand size = 66, antiderivative size = 20 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=e^{\frac {1}{72} (2-x)^2 (-1-2 x+\log (x))} \end {dmath*}
Time = 1.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=e^{-\frac {1}{72} (-2+x)^2 (1+2 x)} x^{\frac {1}{72} (-2+x)^2} \end {dmath*}
Integrate[(E^((-4 - 4*x + 7*x^2 - 2*x^3 + (4 - 4*x + x^2)*Log[x])/72)*(4 - 8*x + 15*x^2 - 6*x^3 + (-4*x + 2*x^2)*Log[x]))/(72*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.
\(\displaystyle \int \frac {\left (-6 x^3+15 x^2+\left (2 x^2-4 x\right ) \log (x)-8 x+4\right ) \exp \left (\frac {1}{72} \left (-2 x^3+7 x^2+\left (x^2-4 x+4\right ) \log (x)-4 x-4\right )\right )}{72 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{72} \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {1}{72} \left (x^2-4 x+4\right )-1} \left (-6 x^3+15 x^2-8 x-2 \left (2 x-x^2\right ) \log (x)+4\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{72} \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} (2-x) x^{\frac {x^2}{72}-\frac {x}{18}-\frac {17}{18}} \left (6 x^2-2 \log (x) x-3 x+2\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{72} \int \left (4 e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}-\frac {17}{18}}-8 e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {1}{18}}+2 e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} (x-2) \log (x) x^{\frac {x^2}{72}-\frac {x}{18}+\frac {1}{18}}+15 e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {19}{18}}-6 e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {37}{18}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{72} \left (4 \int \frac {\int e^{-\frac {1}{72} (x-2)^2 (2 x+1)} x^{\frac {1}{72} (x-2)^2}dx}{x}dx-2 \int \frac {\int e^{-\frac {1}{72} (x-2)^2 (2 x+1)} x^{\frac {1}{72} \left (x^2-4 x+76\right )}dx}{x}dx+4 \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}-\frac {17}{18}}dx-8 \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {1}{18}}dx+15 \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {19}{18}}dx-6 \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {37}{18}}dx-4 \log (x) \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {1}{18}}dx+2 \log (x) \int e^{\frac {1}{72} \left (-2 x^3+7 x^2-4 x-4\right )} x^{\frac {x^2}{72}-\frac {x}{18}+\frac {19}{18}}dx\right )\) |
Int[(E^((-4 - 4*x + 7*x^2 - 2*x^3 + (4 - 4*x + x^2)*Log[x])/72)*(4 - 8*x + 15*x^2 - 6*x^3 + (-4*x + 2*x^2)*Log[x]))/(72*x),x]
3.1.16.3.1 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]]
Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20
method | result | size |
risch | \(x^{\frac {\left (-2+x \right )^{2}}{72}} {\mathrm e}^{-\frac {\left (1+2 x \right ) \left (-2+x \right )^{2}}{72}}\) | \(24\) |
norman | \({\mathrm e}^{\frac {\left (x^{2}-4 x +4\right ) \ln \left (x \right )}{72}-\frac {x^{3}}{36}+\frac {7 x^{2}}{72}-\frac {x}{18}-\frac {1}{18}}\) | \(29\) |
parallelrisch | \({\mathrm e}^{\frac {\left (x^{2}-4 x +4\right ) \ln \left (x \right )}{72}-\frac {x^{3}}{36}+\frac {7 x^{2}}{72}-\frac {x}{18}-\frac {1}{18}}\) | \(29\) |
int(1/72*((2*x^2-4*x)*ln(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4)*ln(x) -1/36*x^3+7/72*x^2-1/18*x-1/18)/x,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=e^{\left (-\frac {1}{36} \, x^{3} + \frac {7}{72} \, x^{2} + \frac {1}{72} \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (x\right ) - \frac {1}{18} \, x - \frac {1}{18}\right )} \end {dmath*}
integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4 )*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/18)/x,x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=e^{- \frac {x^{3}}{36} + \frac {7 x^{2}}{72} - \frac {x}{18} + \left (\frac {x^{2}}{72} - \frac {x}{18} + \frac {1}{18}\right ) \log {\left (x \right )} - \frac {1}{18}} \end {dmath*}
integrate(1/72*((2*x**2-4*x)*ln(x)-6*x**3+15*x**2-8*x+4)*exp(1/72*(x**2-4* x+4)*ln(x)-1/36*x**3+7/72*x**2-1/18*x-1/18)/x,x)
\begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=\int { -\frac {{\left (6 \, x^{3} - 15 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} \log \left (x\right ) + 8 \, x - 4\right )} e^{\left (-\frac {1}{36} \, x^{3} + \frac {7}{72} \, x^{2} + \frac {1}{72} \, {\left (x^{2} - 4 \, x + 4\right )} \log \left (x\right ) - \frac {1}{18} \, x - \frac {1}{18}\right )}}{72 \, x} \,d x } \end {dmath*}
integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4 )*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/18)/x,x, algorithm=\
-1/72*integrate((6*x^3 - 15*x^2 - 2*(x^2 - 2*x)*log(x) + 8*x - 4)*e^(-1/36 *x^3 + 7/72*x^2 + 1/72*(x^2 - 4*x + 4)*log(x) - 1/18*x - 1/18)/x, x)
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=e^{\left (-\frac {1}{36} \, x^{3} + \frac {1}{72} \, x^{2} \log \left (x\right ) + \frac {7}{72} \, x^{2} - \frac {1}{18} \, x \log \left (x\right ) - \frac {1}{18} \, x + \frac {1}{18} \, \log \left (x\right ) - \frac {1}{18}\right )} \end {dmath*}
integrate(1/72*((2*x^2-4*x)*log(x)-6*x^3+15*x^2-8*x+4)*exp(1/72*(x^2-4*x+4 )*log(x)-1/36*x^3+7/72*x^2-1/18*x-1/18)/x,x, algorithm=\
Time = 13.73 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \begin {dmath*} \int \frac {e^{\frac {1}{72} \left (-4-4 x+7 x^2-2 x^3+\left (4-4 x+x^2\right ) \log (x)\right )} \left (4-8 x+15 x^2-6 x^3+\left (-4 x+2 x^2\right ) \log (x)\right )}{72 x} \, dx=x^{\frac {x^2}{72}+\frac {1}{18}}\,{\mathrm {e}}^{-\frac {x\,\ln \left (x\right )}{18}}\,{\mathrm {e}}^{-\frac {x}{18}}\,{\mathrm {e}}^{-\frac {1}{18}}\,{\mathrm {e}}^{-\frac {x^3}{36}}\,{\mathrm {e}}^{\frac {7\,x^2}{72}} \end {dmath*}