Integrand size = 15, antiderivative size = 19 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \left (-\frac {e^x}{x}+\log \left (\frac {2 x^2}{3}\right )\right ) \]
[Out]
Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2225, 2332} \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2 x^2}{3}\right )-e^x \]
[In]
[Out]
Rule 2225
Rule 2332
Rubi steps \begin{align*} \text {integral}& = 2 x-\int e^x \, dx+\int \log \left (\frac {2 x^2}{3}\right ) \, dx \\ & = -e^x+x \log \left (\frac {2 x^2}{3}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=-e^x+x \log \left (\frac {2 x^2}{3}\right ) \]
[In]
[Out]
Time = 1.79 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74
method | result | size |
norman | \(x \ln \left (\frac {2 x^{2}}{3}\right )-{\mathrm e}^{x}\) | \(14\) |
default | \(2 x \ln \left (x \right )+x \left (\ln \left (\frac {2 x^{2}}{3}\right )-2 \ln \left (x \right )\right )-{\mathrm e}^{x}\) | \(24\) |
parallelrisch | \(\frac {12 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} {\mathrm e}^{2 x}-24 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} {\mathrm e}^{x} \ln \left (\frac {2 x^{2}}{3}\right ) x +12 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} \ln \left (\frac {2 x^{2}}{3}\right )^{2} x^{2}}{12 {\left (x \ln \left (\frac {2 x^{2}}{3}\right )-{\mathrm e}^{x}\right )}^{2}}\) | \(115\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2}{3}\right ) + 2 \, x \log \left (x\right ) - e^{x} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log {\left (\frac {2 x^{2}}{3} \right )} - e^{x} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2}{3} \, x^{2}\right ) - e^{x} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=-x \log \left (3\right ) + x \log \left (2\right ) + 2 \, x \log \left (x \mathrm {sgn}\left (x\right )\right ) - e^{x} \]
[In]
[Out]
Time = 15.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x\,\ln \left (x^2\right )-{\mathrm {e}}^x+x\,\ln \left (2\right )-x\,\ln \left (3\right ) \]
[In]
[Out]