Integrand size = 62, antiderivative size = 27 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{e^x-x^3+\frac {1}{8} e^{x^2} x^3} \]
[Out]
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 2306, 6838} \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{\frac {1}{8} \left (-8 x^3+e^{x^2} x^3+8 e^x\right )} \]
[In]
[Out]
Rule 12
Rule 2306
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{8} \int \frac {1}{4} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{32} \int e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{4} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \\ \end{align*}
Time = 0.65 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{e^x-x^3+\frac {1}{8} e^{x^2} x^3} \]
[In]
[Out]
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {{\mathrm e}^{\frac {x^{3} {\mathrm e}^{x^{2}}}{8}+{\mathrm e}^{x}-x^{3}}}{4}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {x^{3} {\mathrm e}^{x^{2}}}{8}+{\mathrm e}^{x}-2 \ln \left (2\right )-x^{3}}\) | \(23\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x} - 2 \, \log \left (2\right )\right )} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {e^{\frac {x^{3} e^{x^{2}}}{8} - x^{3} + e^{x}}}{4} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} \, e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x}\right )} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x} - 2 \, \log \left (2\right )\right )} \]
[In]
[Out]
Time = 8.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{x^2}}{8}}}{4} \]
[In]
[Out]