Integrand size = 100, antiderivative size = 24 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (-5+(3-x+\log (x))^2\right ) \end {dmath*}
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (4-6 x+x^2+(6-2 x) \log (x)+\log ^2(x)\right ) \end {dmath*}
Integrate[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x ^2 + x^3) + (2 - 2*x + E^((1 + 4*x)/4)*(6*x - 2*x^2))*Log[x] + E^((1 + 4*x )/4)*x*Log[x]^2))/(2*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.
\(\displaystyle \int \frac {e^{e^{\frac {1}{4} (4 x+1)}} \left (2 x^2+\left (e^{\frac {1}{4} (4 x+1)} \left (6 x-2 x^2\right )-2 x+2\right ) \log (x)+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )-8 x+e^{\frac {1}{4} (4 x+1)} x \log ^2(x)+6\right )}{2 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{e^{\frac {1}{4} (4 x+1)}} \left (2 x^2+e^{\frac {1}{4} (4 x+1)} \log ^2(x) x-8 x+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )+2 \left (-x+e^{\frac {1}{4} (4 x+1)} \left (3 x-x^2\right )+1\right ) \log (x)+6\right )}{x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{2} \int \frac {e^{e^{x+\frac {1}{4}}} \left (2 x^2+e^{\frac {1}{4} (4 x+1)} \log ^2(x) x-8 x+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )+2 \left (-x+e^{\frac {1}{4} (4 x+1)} \left (3 x-x^2\right )+1\right ) \log (x)+6\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {2 e^{e^{x+\frac {1}{4}}} (x-1) (x-\log (x)-3)}{x}+e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} \left (x^2-2 \log (x) x-6 x+\log ^2(x)+6 \log (x)+4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {\operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right )}{x}dx+\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} x^2dx+2 \int e^{e^{x+\frac {1}{4}}} xdx-6 \int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx-2 \int \frac {\int \frac {e^{e^{x+\frac {1}{4}}}}{x}dx}{x}dx+2 \int \frac {\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx}{x}dx+\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} \log ^2(x)dx+2 \log (x) \int \frac {e^{e^{x+\frac {1}{4}}}}{x}dx-2 \log (x) \int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx-8 \operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right )-2 \operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right ) \log (x)+4 e^{e^{x+\frac {1}{4}}}+6 e^{e^{x+\frac {1}{4}}} \log (x)\right )\) |
Int[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x^2 + x ^3) + (2 - 2*x + E^((1 + 4*x)/4)*(6*x - 2*x^2))*Log[x] + E^((1 + 4*x)/4)*x *Log[x]^2))/(2*x),x]
3.12.9.3.1 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 = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
risch | \(\frac {\left (x^{2}-2 x \ln \left (x \right )+\ln \left (x \right )^{2}-6 x +6 \ln \left (x \right )+4\right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}}{2}\) | \(29\) |
parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} x^{2}}{2}-\ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} x +\frac {{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} \ln \left (x \right )^{2}}{2}-3 x \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}+3 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}+2 \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}\) | \(57\) |
int(1/2*(x*exp(x+1/4)*ln(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*ln(x)+(x^3-6 *x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x,method=_RETURNVERBOS E)
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, {\left (x^{2} - 2 \, {\left (x - 3\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 6 \, x + 4\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \end {dmath*}
integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x )+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=\
Time = 31.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {\left (x^{2} - 2 x \log {\left (x \right )} - 6 x + \log {\left (x \right )}^{2} + 6 \log {\left (x \right )} + 4\right ) e^{e^{x + \frac {1}{4}}}}{2} \end {dmath*}
integrate(1/2*(x*exp(x+1/4)*ln(x)**2+((-2*x**2+6*x)*exp(x+1/4)-2*x+2)*ln(x )+(x**3-6*x**2+4*x)*exp(x+1/4)+2*x**2-8*x+6)*exp(exp(x+1/4))/x,x)
\begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\int { \frac {{\left (x e^{\left (x + \frac {1}{4}\right )} \log \left (x\right )^{2} + 2 \, x^{2} + {\left (x^{3} - 6 \, x^{2} + 4 \, x\right )} e^{\left (x + \frac {1}{4}\right )} - 2 \, {\left ({\left (x^{2} - 3 \, x\right )} e^{\left (x + \frac {1}{4}\right )} + x - 1\right )} \log \left (x\right ) - 8 \, x + 6\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )}}{2 \, x} \,d x } \end {dmath*}
integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x )+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=\
1/2*(x^2 - 2*(x - 3)*log(x) + log(x)^2 - 6*x + 4)*e^(e^(x + 1/4)) - 4*Ei(e ^(x + 1/4)) + 4*integrate(e^(e^(x + 1/4)), x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (18) = 36\).
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, x^{2} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} - x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + \frac {1}{2} \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right )^{2} - 3 \, x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} + 3 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + 2 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \end {dmath*}
integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x )+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=\
1/2*x^2*e^(e^(x + 1/4)) - x*e^(e^(x + 1/4))*log(x) + 1/2*e^(e^(x + 1/4))*l og(x)^2 - 3*x*e^(e^(x + 1/4)) + 3*e^(e^(x + 1/4))*log(x) + 2*e^(e^(x + 1/4 ))
Time = 15.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{1/4}\,{\mathrm {e}}^x}\,\left (\frac {x^2}{2}-x\,\ln \left (x\right )-3\,x+\frac {{\ln \left (x\right )}^2}{2}+3\,\ln \left (x\right )+2\right ) \end {dmath*}