Optimal. Leaf size=38 \[ \frac {\frac {x}{5}+\frac {e^3}{\frac {4 x}{5}-\frac {5 e^{-e^x}}{\log (2)}}}{2 x} \]
________________________________________________________________________________________
Rubi [F] time = 3.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{3+e^x} \log (2) \left (25-25 e^x x-8 e^{e^x} x \log (2)\right )}{2 x^2 \left (25-e^{e^x} x \log (16)\right )^2} \, dx\\ &=\frac {1}{2} (5 \log (2)) \int \frac {e^{3+e^x} \left (25-25 e^x x-8 e^{e^x} x \log (2)\right )}{x^2 \left (25-e^{e^x} x \log (16)\right )^2} \, dx\\ &=\frac {1}{2} (5 \log (2)) \int \left (-\frac {25 e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2}-\frac {e^{3+e^x} \left (-25+8 e^{e^x} x \log (2)\right )}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{2} (5 \log (2)) \int \frac {e^{3+e^x} \left (-25+8 e^{e^x} x \log (2)\right )}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\\ &=-\left (\frac {1}{2} (5 \log (2)) \int \left (\frac {25 e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2}+\frac {8 e^{3+e^x} \log (2)}{x^2 \log (16) \left (-25+e^{e^x} x \log (16)\right )}\right ) \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\\ &=-\left (\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx-\frac {\left (20 \log ^2(2)\right ) \int \frac {e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )} \, dx}{\log (16)}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.60, size = 31, normalized size = 0.82 \begin {gather*} -\frac {5 e^{3+e^x} \log (2)}{2 \left (25 x-e^{e^x} x^2 \log (16)\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.80, size = 25, normalized size = 0.66 \begin {gather*} \frac {5 \, e^{\left (e^{x} + 3\right )} \log \relax (2)}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \relax (2) - 25 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.17, size = 25, normalized size = 0.66 \begin {gather*} \frac {5 \, e^{\left (e^{x} + 3\right )} \log \relax (2)}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \relax (2) - 25 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.29, size = 25, normalized size = 0.66
method | result | size |
norman | \(\frac {5 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{3} \ln \relax (2)}{2 x \left (4 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{x}} x -25\right )}\) | \(25\) |
risch | \(\frac {5 \,{\mathrm e}^{3}}{8 x^{2}}+\frac {125 \,{\mathrm e}^{3}}{8 x^{2} \left (4 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{x}} x -25\right )}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.50, size = 25, normalized size = 0.66 \begin {gather*} \frac {5 \, e^{\left (e^{x} + 3\right )} \log \relax (2)}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \relax (2) - 25 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.85, size = 54, normalized size = 1.42 \begin {gather*} \frac {5\,{\mathrm {e}}^3}{8\,x^2}+\frac {125\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^{x+3}\right )}{32\,x\,\left (x^2\,\ln \relax (2)+x^3\,{\mathrm {e}}^x\,\ln \relax (2)\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}-\frac {25}{4\,x\,\ln \relax (2)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.17, size = 32, normalized size = 0.84 \begin {gather*} \frac {125 e^{3}}{32 x^{3} e^{e^{x}} \log {\relax (2 )} - 200 x^{2}} + \frac {5 e^{3}}{8 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________