Optimal. Leaf size=25 \[ \frac {3}{5} x^2 \left (\frac {e^{4+x}}{25}-\frac {3 e^2 x}{5}\right ) \]
________________________________________________________________________________________
Rubi [A] time = 0.06, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {3}{125} e^{x+4} x^2-\frac {9 e^2 x^3}{25} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 1593
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \left (-27 e^2 x^2+\frac {1}{25} e^{4+x} \left (30 x+15 x^2\right )\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int e^{4+x} \left (30 x+15 x^2\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int e^{4+x} x (30+15 x) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {1}{625} \int \left (30 e^{4+x} x+15 e^{4+x} x^2\right ) \, dx\\ &=-\frac {9}{25} e^2 x^3+\frac {3}{125} \int e^{4+x} x^2 \, dx+\frac {6}{125} \int e^{4+x} x \, dx\\ &=\frac {6}{125} e^{4+x} x+\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}-\frac {6}{125} \int e^{4+x} \, dx-\frac {6}{125} \int e^{4+x} x \, dx\\ &=-\frac {6 e^{4+x}}{125}+\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}+\frac {6}{125} \int e^{4+x} \, dx\\ &=\frac {3}{125} e^{4+x} x^2-\frac {9 e^2 x^3}{25}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} \frac {3}{125} e^2 \left (e^{2+x} x^2-15 x^3\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.94, size = 21, normalized size = 0.84 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{5} \, x^{2} e^{\left (x - 2 \, \log \relax (5) + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.16, size = 21, normalized size = 0.84 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{5} \, x^{2} e^{\left (x - 2 \, \log \relax (5) + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.11, size = 18, normalized size = 0.72
method | result | size |
risch | \(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 x^{2} {\mathrm e}^{4+x}}{125}\) | \(18\) |
norman | \(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} x^{2}}{5}\) | \(24\) |
default | \(-\frac {9 x^{3} {\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )^{2}}{5}-\frac {24 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )}{5}+\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x}}{5}-\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)^{2}}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right ) \ln \relax (5)}{5}\) | \(102\) |
derivativedivides | \(-\frac {72 \ln \relax (5)^{3} {\mathrm e}^{2}}{25}-\frac {108 \ln \relax (5)^{2} {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}-\frac {54 \ln \relax (5) {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{2}}{25}-\frac {9 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{3}}{25}+\frac {432 \ln \relax (5)^{2} {\mathrm e}^{2}}{25}+\frac {432 \ln \relax (5) {\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}+\frac {108 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )^{2}}{25}-\frac {864 \,{\mathrm e}^{2} \ln \relax (5)}{25}-\frac {432 \,{\mathrm e}^{2} \left (-2 \ln \relax (5)+4+x \right )}{25}+\frac {576 \,{\mathrm e}^{2}}{25}+\frac {3 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )^{2}}{5}-\frac {24 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right )}{5}+\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x}}{5}-\frac {48 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \ln \relax (5)^{2}}{5}+\frac {12 \,{\mathrm e}^{-2 \ln \relax (5)+4+x} \left (-2 \ln \relax (5)+4+x \right ) \ln \relax (5)}{5}\) | \(219\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.36, size = 41, normalized size = 1.64 \begin {gather*} -\frac {9}{25} \, x^{3} e^{2} + \frac {3}{125} \, {\left (x^{2} e^{4} - 2 \, x e^{4} + 2 \, e^{4}\right )} e^{x} + \frac {6}{125} \, {\left (x e^{4} - e^{4}\right )} e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 0.08, size = 15, normalized size = 0.60 \begin {gather*} \frac {3\,x^2\,\left ({\mathrm {e}}^{x+4}-15\,x\,{\mathrm {e}}^2\right )}{125} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.10, size = 20, normalized size = 0.80 \begin {gather*} - \frac {9 x^{3} e^{2}}{25} + \frac {3 x^{2} e^{x + 4}}{125} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________