Integrand size = 71, antiderivative size = 27 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=25 \left (4-e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}}\right ) \]
[Out]
\[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {100 e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)}+\frac {125 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)}\right ) \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {2 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)}\right ) \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)} \, dx+250 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ & = 100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log (-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{2 \log (-3+2 x)}\right ) \, dx+250 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log ^2(-3+2 x)}+\frac {3 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 (-3+2 x) \log ^2(-3+2 x)}\right ) \, dx \\ & = \frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log (-3+2 x)} \, dx+\frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{\log (-3+2 x)} \, dx+100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log ^2(-3+2 x)} \, dx+375 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \\ \end{align*}
Time = 1.77 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}} \]
[In]
[Out]
Time = 1.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (-3+2 x \right )}}\) | \(25\) |
parallelrisch | \(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (-3+2 x \right )}}\) | \(25\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25 \, e^{\left (x - \frac {{\left (x e^{x} \log \left (2 \, x - 3\right ) - 5 \, x - 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )} \]
[In]
[Out]
Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=- 25 e^{\frac {\left (5 x + 2 e^{x}\right ) e^{- x}}{\log {\left (2 x - 3 \right )}}} \]
[In]
[Out]
Exception generated. \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
\[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=\int { \frac {25 \, {\left (5 \, {\left (2 \, x^{2} - 5 \, x + 3\right )} \log \left (2 \, x - 3\right ) + 10 \, x + 4 \, e^{x}\right )} e^{\left (-x + \frac {{\left (5 \, x + 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )}}{{\left (2 \, x - 3\right )} \log \left (2 \, x - 3\right )^{2}} \,d x } \]
[In]
[Out]
Time = 12.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}} \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx=-25\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{-x}+2}{\ln \left (2\,x-3\right )}} \]
[In]
[Out]