Integrand size = 27, antiderivative size = 16 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=x+\frac {1}{2} e x^2 (-2+e+x) \log (625) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12} \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=\frac {1}{2} e x^3 \log (625)+\frac {1}{2} e^2 x^2 \log (625)-e x^2 \log (625)+x \]
[In]
[Out]
Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx \\ & = x+\frac {1}{2} \log (625) \int \left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \, dx \\ & = x+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} (e \log (625)) \int \left (-4 x+3 x^2\right ) \, dx \\ & = x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625) \]
[In]
[Out]
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81
method | result | size |
parallelrisch | \(2 \ln \left (5\right ) \left (x^{2} {\mathrm e}^{2}+x^{3} {\mathrm e}-2 x^{2} {\mathrm e}\right )+x\) | \(29\) |
gosper | \(x \left (2 x \,{\mathrm e}^{2} \ln \left (5\right )+2 \ln \left (5\right ) {\mathrm e} x^{2}-4 x \,{\mathrm e} \ln \left (5\right )+1\right )\) | \(30\) |
norman | \(2 \ln \left (5\right ) {\mathrm e} x^{3}+\left (2 \,{\mathrm e}^{2} \ln \left (5\right )-4 \,{\mathrm e} \ln \left (5\right )\right ) x^{2}+x\) | \(31\) |
default | \(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\) | \(32\) |
risch | \(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\) | \(32\) |
parts | \(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\) | \(32\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]
[In]
[Out]
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 e x^{3} \log {\left (5 \right )} + x^{2} \left (- 4 e \log {\left (5 \right )} + 2 e^{2} \log {\left (5 \right )}\right ) + x \]
[In]
[Out]
none
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]
[In]
[Out]
Time = 7.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2\,\mathrm {e}\,\ln \left (5\right )\,x^3-\ln \left (5\right )\,\left (4\,\mathrm {e}-2\,{\mathrm {e}}^2\right )\,x^2+x \]
[In]
[Out]