Optimal. Leaf size=37 \[ \frac {1}{4} \left (\frac {x}{3}+\frac {e^{\left .\frac {2}{5}\right /x}}{-e^{4+9 x}-x+\log (x)}\right ) \]
________________________________________________________________________________________
Rubi [F] time = 4.04, 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 {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{60} \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{60} \int \left (5+\frac {3 e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}-\frac {15 e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx\\ &=\frac {x}{12}+\frac {1}{20} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ &=\frac {x}{12}+\frac {1}{20} \int \left (\frac {45 e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)}+\frac {2 e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}\right ) \, dx-\frac {1}{4} \int \left (-\frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {9 e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2}-\frac {9 e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx\\ &=\frac {x}{12}+\frac {1}{10} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx+\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.12, size = 48, normalized size = 1.30 \begin {gather*} \frac {-3 e^{\left .\frac {2}{5}\right /x}+e^{4+9 x} x+x^2-x \log (x)}{12 \left (e^{4+9 x}+x-\log (x)\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.60, size = 41, normalized size = 1.11 \begin {gather*} \frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \relax (x) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \relax (x)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.09, size = 27, normalized size = 0.73
method | result | size |
risch | \(\frac {x}{12}-\frac {{\mathrm e}^{\frac {2}{5 x}}}{4 \left (x +{\mathrm e}^{9 x +4}-\ln \relax (x )\right )}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.40, size = 41, normalized size = 1.11 \begin {gather*} \frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \relax (x) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \relax (x)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 7.75, size = 30, normalized size = 0.81 \begin {gather*} \frac {x}{12}-\frac {{\mathrm {e}}^{\frac {2}{5\,x}}}{4\,\left (x+{\mathrm {e}}^{9\,x+4}-\ln \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.34, size = 26, normalized size = 0.70 \begin {gather*} \frac {x}{12} - \frac {e^{\frac {2}{5 x}}}{4 x + 4 e^{9 x + 4} - 4 \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________