Optimal. Leaf size=28 \[ \frac {x \left (5-x+4 \left (x-4 e^2 x^2\right )\right )}{1+e-\log (5)} \]
________________________________________________________________________________________
Rubi [A] time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.57, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12} \begin {gather*} -\frac {16 e^2 x^3}{1+e-\log (5)}+\frac {3 x^2}{1+e-\log (5)}+\frac {5 x}{1+e-\log (5)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-5-6 x+48 e^2 x^2\right ) \, dx}{-1-e+\log (5)}\\ &=\frac {5 x}{1+e-\log (5)}+\frac {3 x^2}{1+e-\log (5)}-\frac {16 e^2 x^3}{1+e-\log (5)}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.00, size = 27, normalized size = 0.96 \begin {gather*} \frac {5 x+3 x^2-16 e^2 x^3}{1+e-\log (5)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.92, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \relax (5) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.15, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \relax (5) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.03, size = 27, normalized size = 0.96
method | result | size |
gosper | \(\frac {x \left (16 x^{2} {\mathrm e}^{2}-3 x -5\right )}{\ln \relax (5)-{\mathrm e}-1}\) | \(27\) |
default | \(\frac {16 x^{3} {\mathrm e}^{2}-3 x^{2}-5 x}{\ln \relax (5)-{\mathrm e}-1}\) | \(30\) |
risch | \(-\frac {5 x}{\ln \relax (5)-{\mathrm e}-1}-\frac {3 x^{2}}{\ln \relax (5)-{\mathrm e}-1}+\frac {16 \,{\mathrm e}^{2} x^{3}}{\ln \relax (5)-{\mathrm e}-1}\) | \(47\) |
norman | \(-\frac {5 x}{\ln \relax (5)-{\mathrm e}-1}-\frac {3 x^{2}}{\ln \relax (5)-{\mathrm e}-1}+\frac {16 \,{\mathrm e}^{2} x^{3}}{\ln \relax (5)-{\mathrm e}-1}\) | \(49\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.35, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \relax (5) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 6.34, size = 24, normalized size = 0.86 \begin {gather*} \frac {x\,\left (-16\,{\mathrm {e}}^2\,x^2+3\,x+5\right )}{\mathrm {e}-\ln \relax (5)+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.06, size = 41, normalized size = 1.46 \begin {gather*} - \frac {16 x^{3} e^{2}}{- \log {\relax (5 )} + 1 + e} + \frac {3 x^{2}}{- \log {\relax (5 )} + 1 + e} + \frac {5 x}{- \log {\relax (5 )} + 1 + e} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________