Integrand size = 93, antiderivative size = 26 \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-4+2 x-e^{-e^x} \log (x) \log ^2\left (2-5 x^2\right ) \end {dmath*}
Time = 0.70 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=2 x-e^{-e^x} \log (x) \log ^2\left (2-5 x^2\right ) \end {dmath*}
Integrate[(E^E^x*(-4*x + 10*x^3) + (2 - 5*x^2)*Log[2 - 5*x^2]^2 + Log[x]*( -20*x^2*Log[2 - 5*x^2] + E^x*(-2*x + 5*x^3)*Log[2 - 5*x^2]^2))/(E^E^x*(-2* x + 5*x^3)),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.
\(\displaystyle \int \frac {e^{-e^x} \left (e^{e^x} \left (10 x^3-4 x\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (e^x \left (5 x^3-2 x\right ) \log ^2\left (2-5 x^2\right )-20 x^2 \log \left (2-5 x^2\right )\right )\right )}{5 x^3-2 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-e^x} \left (e^{e^x} \left (10 x^3-4 x\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (e^x \left (5 x^3-2 x\right ) \log ^2\left (2-5 x^2\right )-20 x^2 \log \left (2-5 x^2\right )\right )\right )}{x \left (5 x^2-2\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (e^{x-e^x} \log (x) \log ^2\left (2-5 x^2\right )+\frac {e^{-e^x} \left (10 e^{e^x} x^3-5 x^2 \log ^2\left (2-5 x^2\right )+2 \log ^2\left (2-5 x^2\right )-20 x^2 \log (x) \log \left (2-5 x^2\right )-4 e^{e^x} x\right )}{x \left (5 x^2-2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 x+2 \sqrt {5} \log (x) \log \left (2-5 x^2\right ) \int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx-2 \sqrt {5} \log (x) \log \left (2-5 x^2\right ) \int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx-\int \frac {e^{-e^x} \log ^2\left (2-5 x^2\right )}{x}dx+\int e^{x-e^x} \log (x) \log ^2\left (2-5 x^2\right )dx-2 \sqrt {5} \log \left (2-5 x^2\right ) \int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{x}dx+10 \log (x) \int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{\sqrt {2}-\sqrt {5} x}dx-10 \log (x) \int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{\sqrt {5} x+\sqrt {2}}dx+2 \sqrt {5} \log \left (2-5 x^2\right ) \int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{x}dx-10 \log (x) \int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{\sqrt {2}-\sqrt {5} x}dx+10 \log (x) \int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{\sqrt {5} x+\sqrt {2}}dx-10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{x}dx}{\sqrt {2}-\sqrt {5} x}dx+10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{x}dx}{\sqrt {5} x+\sqrt {2}}dx-10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{\sqrt {2}-\sqrt {5} x}dx}{x}dx+10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {2}-\sqrt {5} x}dx}{\sqrt {5} x+\sqrt {2}}dx}{x}dx+10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{x}dx}{\sqrt {2}-\sqrt {5} x}dx-10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{x}dx}{\sqrt {5} x+\sqrt {2}}dx+10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{\sqrt {2}-\sqrt {5} x}dx}{x}dx-10 \int \frac {\int \frac {\int \frac {e^{-e^x}}{\sqrt {5} x+\sqrt {2}}dx}{\sqrt {5} x+\sqrt {2}}dx}{x}dx\) |
Int[(E^E^x*(-4*x + 10*x^3) + (2 - 5*x^2)*Log[2 - 5*x^2]^2 + Log[x]*(-20*x^ 2*Log[2 - 5*x^2] + E^x*(-2*x + 5*x^3)*Log[2 - 5*x^2]^2))/(E^E^x*(-2*x + 5* x^3)),x]
3.7.50.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 17.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(2 x -\ln \left (x \right ) \ln \left (-5 x^{2}+2\right )^{2} {\mathrm e}^{-{\mathrm e}^{x}}\) | \(24\) |
parallelrisch | \(\frac {\left (-100 \ln \left (-5 x^{2}+2\right )^{2} \ln \left (x \right )+200 x \,{\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-{\mathrm e}^{x}}}{100}\) | \(29\) |
int(((10*x^3-4*x)*exp(exp(x))+((5*x^3-2*x)*exp(x)*ln(-5*x^2+2)^2-20*x^2*ln (-5*x^2+2))*ln(x)+(-5*x^2+2)*ln(-5*x^2+2)^2)/(5*x^3-2*x)/exp(exp(x)),x,met hod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-{\left (\log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) - 2 \, x e^{\left (e^{x}\right )}\right )} e^{\left (-e^{x}\right )} \end {dmath*}
integrate(((10*x^3-4*x)*exp(exp(x))+((5*x^3-2*x)*exp(x)*log(-5*x^2+2)^2-20 *x^2*log(-5*x^2+2))*log(x)+(-5*x^2+2)*log(-5*x^2+2)^2)/(5*x^3-2*x)/exp(exp (x)),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=\text {Timed out} \end {dmath*}
integrate(((10*x**3-4*x)*exp(exp(x))+((5*x**3-2*x)*exp(x)*ln(-5*x**2+2)**2 -20*x**2*ln(-5*x**2+2))*ln(x)+(-5*x**2+2)*ln(-5*x**2+2)**2)/(5*x**3-2*x)/e xp(exp(x)),x)
Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-e^{\left (-e^{x}\right )} \log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) + 2 \, x \end {dmath*}
integrate(((10*x^3-4*x)*exp(exp(x))+((5*x^3-2*x)*exp(x)*log(-5*x^2+2)^2-20 *x^2*log(-5*x^2+2))*log(x)+(-5*x^2+2)*log(-5*x^2+2)^2)/(5*x^3-2*x)/exp(exp (x)),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-{\left (e^{\left (x - e^{x}\right )} \log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) - 2 \, x e^{x}\right )} e^{\left (-x\right )} \end {dmath*}
integrate(((10*x^3-4*x)*exp(exp(x))+((5*x^3-2*x)*exp(x)*log(-5*x^2+2)^2-20 *x^2*log(-5*x^2+2))*log(x)+(-5*x^2+2)*log(-5*x^2+2)^2)/(5*x^3-2*x)/exp(exp (x)),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=\int \frac {{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x-10\,x^3\right )+{\ln \left (2-5\,x^2\right )}^2\,\left (5\,x^2-2\right )+\ln \left (x\right )\,\left (20\,x^2\,\ln \left (2-5\,x^2\right )+{\ln \left (2-5\,x^2\right )}^2\,{\mathrm {e}}^x\,\left (2\,x-5\,x^3\right )\right )\right )}{2\,x-5\,x^3} \,d x \end {dmath*}
int((exp(-exp(x))*(exp(exp(x))*(4*x - 10*x^3) + log(2 - 5*x^2)^2*(5*x^2 - 2) + log(x)*(20*x^2*log(2 - 5*x^2) + log(2 - 5*x^2)^2*exp(x)*(2*x - 5*x^3) )))/(2*x - 5*x^3),x)