Integrand size = 121, antiderivative size = 28 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=1+\frac {4}{5+\frac {1}{2} \left (e^8-x\right )+\left (e^x-x\right )^2} \]
[Out]
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 12, 6818} \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 x^2-4 e^x x-x+2 e^{2 x}+e^8+10} \]
[In]
[Out]
Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {8 \left (1-4 e^{2 x}-4 x+4 e^x (1+x)\right )}{\left (2 e^{2 x}+10 \left (1+\frac {e^8}{10}\right )-x-4 e^x x+2 x^2\right )^2} \, dx \\ & = 8 \int \frac {1-4 e^{2 x}-4 x+4 e^x (1+x)}{\left (2 e^{2 x}+10 \left (1+\frac {e^8}{10}\right )-x-4 e^x x+2 x^2\right )^2} \, dx \\ & = \frac {8}{10+e^8+2 e^{2 x}-x-4 e^x x+2 x^2} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{10+e^8+2 e^{2 x}-x-4 e^x x+2 x^2} \]
[In]
[Out]
Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\) | \(28\) |
risch | \(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\) | \(28\) |
parallelrisch | \(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\) | \(28\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]
[In]
[Out]
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {4}{x^{2} - 2 x e^{x} - \frac {x}{2} + e^{2 x} + 5 + \frac {e^{8}}{2}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]
[In]
[Out]
none
Time = 0.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {16}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]
[In]
[Out]
Time = 12.50 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=-\frac {\frac {16\,x^2}{{\mathrm {e}}^8+10}+\frac {16\,{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^8+10}-\frac {8\,x}{{\mathrm {e}}^8+10}-\frac {32\,x\,{\mathrm {e}}^x}{{\mathrm {e}}^8+10}}{2\,{\mathrm {e}}^{2\,x}-x+{\mathrm {e}}^8-4\,x\,{\mathrm {e}}^x+2\,x^2+10} \]
[In]
[Out]