Integrand size = 106, antiderivative size = 17 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=x+\frac {x}{8+e^{(3+x)^2}+\log (x)} \]
[Out]
\[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {71+e^{2 (3+x)^2}+e^{(3+x)^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{(3+x)^2}\right ) \log (x)+\log ^2(x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx \\ & = \int \left (1-\frac {-1+6 x+2 x^2}{8+e^{(3+x)^2}+\log (x)}+\frac {-1+48 x+16 x^2+6 x \log (x)+2 x^2 \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}\right ) \, dx \\ & = x-\int \frac {-1+6 x+2 x^2}{8+e^{(3+x)^2}+\log (x)} \, dx+\int \frac {-1+48 x+16 x^2+6 x \log (x)+2 x^2 \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx \\ & = x+\int \left (-\frac {1}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}+\frac {48 x}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}+\frac {16 x^2}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}+\frac {6 x \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}+\frac {2 x^2 \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2}\right ) \, dx-\int \left (-\frac {1}{8+e^{(3+x)^2}+\log (x)}+\frac {6 x}{8+e^{(3+x)^2}+\log (x)}+\frac {2 x^2}{8+e^{(3+x)^2}+\log (x)}\right ) \, dx \\ & = x+2 \int \frac {x^2 \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx-2 \int \frac {x^2}{8+e^{(3+x)^2}+\log (x)} \, dx+6 \int \frac {x \log (x)}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx-6 \int \frac {x}{8+e^{(3+x)^2}+\log (x)} \, dx+16 \int \frac {x^2}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx+48 \int \frac {x}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx-\int \frac {1}{\left (8+e^{(3+x)^2}+\log (x)\right )^2} \, dx+\int \frac {1}{8+e^{(3+x)^2}+\log (x)} \, dx \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=x \left (1+\frac {1}{8+e^{(3+x)^2}+\log (x)}\right ) \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x +\frac {x}{\ln \left (x \right )+8+{\mathrm e}^{\left (3+x \right )^{2}}}\) | \(17\) |
parallelrisch | \(\frac {x \ln \left (x \right )+{\mathrm e}^{x^{2}+6 x +9} x +9 x}{8+\ln \left (x \right )+{\mathrm e}^{x^{2}+6 x +9}}\) | \(36\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=\frac {x e^{\left (x^{2} + 6 \, x + 9\right )} + x \log \left (x\right ) + 9 \, x}{e^{\left (x^{2} + 6 \, x + 9\right )} + \log \left (x\right ) + 8} \]
[In]
[Out]
Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=x + \frac {x}{e^{x^{2} + 6 x + 9} + \log {\left (x \right )} + 8} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=\frac {x e^{\left (x^{2} + 6 \, x + 9\right )} + x \log \left (x\right ) + 9 \, x}{e^{\left (x^{2} + 6 \, x + 9\right )} + \log \left (x\right ) + 8} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=\frac {x e^{\left (x^{2} + 6 \, x + 9\right )} + x \log \left (x\right ) + 9 \, x}{e^{\left (x^{2} + 6 \, x + 9\right )} + \log \left (x\right ) + 8} \]
[In]
[Out]
Time = 16.99 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int \frac {71+e^{18+12 x+2 x^2}+e^{9+6 x+x^2} \left (17-6 x-2 x^2\right )+\left (17+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)}{64+16 e^{9+6 x+x^2}+e^{18+12 x+2 x^2}+\left (16+2 e^{9+6 x+x^2}\right ) \log (x)+\log ^2(x)} \, dx=\frac {x\,\left ({\mathrm {e}}^{x^2+6\,x+9}+\ln \left (x\right )+9\right )}{{\mathrm {e}}^{x^2+6\,x+9}+\ln \left (x\right )+8} \]
[In]
[Out]