Integrand size = 91, antiderivative size = 26 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]
[Out]
\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int \frac {\exp \left (\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}\right ) \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\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}\right ) \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{x^4 (-25+25 x) \log \left (x^2\right )} \, dx \\ & = \int \frac {e^{1+\frac {81}{25 x^2}} (1-x) \left (50 (-1+x) x^2-\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{25 x^4 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \frac {e^{1+\frac {81}{25 x^2}} (1-x) \left (50 (-1+x) x^2-\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{x^4 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \left (-\frac {50 e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x^2 \log ^2\left (x^2\right )}+\frac {e^{1+\frac {81}{25 x^2}} \left (-162+324 x-187 x^2+25 x^4\right )}{x^4 \log \left (x^2\right )}\right ) \, dx \\ & = \frac {1}{25} \int \frac {e^{1+\frac {81}{25 x^2}} \left (-162+324 x-187 x^2+25 x^4\right )}{x^4 \log \left (x^2\right )} \, dx-2 \int \frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x^2 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \left (\frac {25 e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )}-\frac {162 e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )}+\frac {324 e^{1+\frac {81}{25 x^2}}}{x^3 \log \left (x^2\right )}-\frac {187 e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )}\right ) \, dx-2 \int \left (\frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )}+\frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )}-\frac {2 e^{1+\frac {81}{25 x^2}}}{x \log ^2\left (x^2\right )}\right ) \, dx \\ & = -\left (2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )} \, dx\right )-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )} \, dx+4 \int \frac {e^{1+\frac {81}{25 x^2}}}{x \log ^2\left (x^2\right )} \, dx-\frac {162}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )} \, dx-\frac {187}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )} \, dx+\frac {324}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^3 \log \left (x^2\right )} \, dx+\int \frac {e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )} \, dx \\ & = -\left (2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )} \, dx\right )-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )} \, dx+2 \text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{x \log ^2(x)} \, dx,x,x^2\right )-\frac {162}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )} \, dx+\frac {162}{25} \text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{x^2 \log (x)} \, dx,x,x^2\right )-\frac {187}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )} \, dx+\int \frac {e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )} \, dx \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]
[In]
[Out]
Time = 35.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {25 x^{2} \ln \left (\ln \left (x^{2}\right )\right )-25 x^{2} \ln \left (x^{2}-2 x +1\right )-25 x^{2}-81}{25 x^{2}}}}{x}\) | \(42\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{x} \]
[In]
[Out]
Timed out. \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]
[In]
[Out]
\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]
[In]
[Out]
Time = 7.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{\frac {81}{25\,x^2}+1}\,{\left (x-1\right )}^2}{x\,\ln \left (x^2\right )} \]
[In]
[Out]