Optimal. Leaf size=20 \[ \frac {64}{\left (-5+e^{\frac {5 e}{4 x \log (3)}}\right )^2} \]
________________________________________________________________________________________
Rubi [A] time = 0.35, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 6688, 6686} \begin {gather*} \frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 6686
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=160 \int \frac {e^{1+\frac {5 e}{4 x \log (3)}}}{-125 x^2 \log (3)+75 e^{\frac {5 e}{4 x \log (3)}} x^2 \log (3)-15 e^{\frac {5 e}{2 x \log (3)}} x^2 \log (3)+e^{\frac {15 e}{4 x \log (3)}} x^2 \log (3)} \, dx\\ &=160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2 \log (3)} \, dx\\ &=\frac {160 \int \frac {e^{1+\frac {5 e}{x \log (81)}}}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^3 x^2} \, dx}{\log (3)}\\ &=\frac {64}{\left (5-e^{\frac {5 e}{x \log (81)}}\right )^2}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.04, size = 18, normalized size = 0.90 \begin {gather*} \frac {64}{\left (-5+e^{\frac {5 e}{x \log (81)}}\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.77, size = 85, normalized size = 4.25 \begin {gather*} -\frac {64 \, e^{\left (2 \, \log \relax (5) + 2\right )}}{10 \, e^{\left (\frac {4 \, x \log \relax (5) \log \relax (3) + 4 \, x \log \relax (3) + e^{\left (\log \relax (5) + 1\right )}}{4 \, x \log \relax (3)} + \log \relax (5) + 1\right )} - e^{\left (\frac {4 \, x \log \relax (5) \log \relax (3) + 4 \, x \log \relax (3) + e^{\left (\log \relax (5) + 1\right )}}{2 \, x \log \relax (3)}\right )} - 25 \, e^{\left (2 \, \log \relax (5) + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.55, size = 41, normalized size = 2.05 \begin {gather*} \frac {64 \, e}{25 \, e + e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)} + 1\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.18, size = 19, normalized size = 0.95
method | result | size |
risch | \(\frac {64}{\left ({\mathrm e}^{\frac {5 \,{\mathrm e}}{4 x \ln \relax (3)}}-5\right )^{2}}\) | \(19\) |
norman | \(\frac {64}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\) | \(22\) |
derivativedivides | \(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\) | \(33\) |
default | \(\frac {64 \,{\mathrm e}^{2} {\mathrm e}^{-2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{\ln \relax (5)+1}}{4 x \ln \relax (3)}}-5\right )^{2}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.54, size = 59, normalized size = 2.95 \begin {gather*} -\frac {64 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)}\right )}\right )}}{25 \, {\left (e^{\left (\frac {5 \, e}{2 \, x \log \relax (3)}\right )} - 10 \, e^{\left (\frac {5 \, e}{4 \, x \log \relax (3)}\right )} + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.28, size = 18, normalized size = 0.90 \begin {gather*} \frac {64}{{\left ({\mathrm {e}}^{\frac {5\,\mathrm {e}}{4\,x\,\ln \relax (3)}}-5\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.14, size = 31, normalized size = 1.55 \begin {gather*} \frac {64}{- 10 e^{\frac {5 e}{4 x \log {\relax (3 )}}} + e^{\frac {5 e}{2 x \log {\relax (3 )}}} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________