Integrand size = 91, antiderivative size = 26 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]
Integrate[(E^((81 + 25*x^2 + 25*x^2*Log[1 - 2*x + x^2] - 25*x^2*Log[Log[x^ 2]])/(25*x^2))*(50*x^2 - 50*x^3 + (162 - 162*x + 25*x^2 + 25*x^3)*Log[x^2] ))/((-25*x^4 + 25*x^5)*Log[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.
| \(\displaystyle \int \frac {\left (-50 x^3+50 x^2+\left (25 x^3+25 x^2-162 x+162\right ) \log \left (x^2\right )\right ) \exp \left (\frac {25 x^2+25 x^2 \log \left (x^2-2 x+1\right )-25 x^2 \log \left (\log \left (x^2\right )\right )+81}{25 x^2}\right )}{\left (25 x^5-25 x^4\right ) \log \left (x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (-50 x^3+50 x^2+\left (25 x^3+25 x^2-162 x+162\right ) \log \left (x^2\right )\right ) \exp \left (\frac {25 x^2+25 x^2 \log \left (x^2-2 x+1\right )-25 x^2 \log \left (\log \left (x^2\right )\right )+81}{25 x^2}\right )}{x^4 (25 x-25) \log \left (x^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\left (25 x^3+25 x^2-162 x+162\right ) \exp \left (\frac {25 x^2+25 x^2 \log \left (x^2-2 x+1\right )-25 x^2 \log \left (\log \left (x^2\right )\right )+81}{25 x^2}\right )}{25 (x-1) x^4}-\frac {2 \exp \left (\frac {25 x^2+25 x^2 \log \left (x^2-2 x+1\right )-25 x^2 \log \left (\log \left (x^2\right )\right )+81}{25 x^2}\right )}{x^2 \log \left (x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {\exp \left (\frac {25 \log \left (x^2-2 x+1\right ) x^2-25 \log \left (\log \left (x^2\right )\right ) x^2+25 x^2+81}{25 x^2}\right )}{x^2}dx-2 \int \frac {\exp \left (\frac {25 \log \left (x^2-2 x+1\right ) x^2-25 \log \left (\log \left (x^2\right )\right ) x^2+25 x^2+81}{25 x^2}\right )}{x}dx-\frac {162}{25} \int \frac {\exp \left (\frac {25 \log \left (x^2-2 x+1\right ) x^2-25 \log \left (\log \left (x^2\right )\right ) x^2+25 x^2+81}{25 x^2}\right )}{x^4}dx+2 \text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{x \log ^2(x)}dx,x,x^2\right )+\text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{\log (x)}dx,x,x^2\right )-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )}dx-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )}dx-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )}dx\) |
Int[(E^((81 + 25*x^2 + 25*x^2*Log[1 - 2*x + x^2] - 25*x^2*Log[Log[x^2]])/( 25*x^2))*(50*x^2 - 50*x^3 + (162 - 162*x + 25*x^2 + 25*x^3)*Log[x^2]))/((- 25*x^4 + 25*x^5)*Log[x^2]),x]
3.4.95.3.1 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])
Time = 35.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62
| method | result | size |
| parallelrisch | \(\frac {{\mathrm e}^{-\frac {25 x^{2} \ln \left (\ln \left (x^{2}\right )\right )-25 x^{2} \ln \left (x^{2}-2 x +1\right )-25 x^{2}-81}{25 x^{2}}}}{x}\) | \(42\) |
int(((25*x^3+25*x^2-162*x+162)*ln(x^2)-50*x^3+50*x^2)*exp(1/25*(-25*x^2*ln (ln(x^2))+25*x^2*ln(x^2-2*x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/ln(x^2),x,m ethod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{x} \]
integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25 *x^2*log(log(x^2))+25*x^2*log(x^2-2*x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/l og(x^2),x, algorithm=\
Timed out. \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\text {Timed out} \]
integrate(((25*x**3+25*x**2-162*x+162)*ln(x**2)-50*x**3+50*x**2)*exp(1/25* (-25*x**2*ln(ln(x**2))+25*x**2*ln(x**2-2*x+1)+25*x**2+81)/x**2)/(25*x**5-2 5*x**4)/ln(x**2),x)
\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]
integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25 *x^2*log(log(x^2))+25*x^2*log(x^2-2*x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/l og(x^2),x, algorithm=\
-1/25*integrate((50*x^3 - 50*x^2 - (25*x^3 + 25*x^2 - 162*x + 162)*log(x^2 ))*e^(1/25*(25*x^2*log(x^2 - 2*x + 1) - 25*x^2*log(log(x^2)) + 25*x^2 + 81 )/x^2)/((x^5 - x^4)*log(x^2)), x)
\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]
integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25 *x^2*log(log(x^2))+25*x^2*log(x^2-2*x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/l og(x^2),x, algorithm=\
integrate(-1/25*(50*x^3 - 50*x^2 - (25*x^3 + 25*x^2 - 162*x + 162)*log(x^2 ))*e^(1/25*(25*x^2*log(x^2 - 2*x + 1) - 25*x^2*log(log(x^2)) + 25*x^2 + 81 )/x^2)/((x^5 - x^4)*log(x^2)), x)
Time = 7.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{\frac {81}{25\,x^2}+1}\,{\left (x-1\right )}^2}{x\,\ln \left (x^2\right )} \]