Optimal. Leaf size=33 \[ 2-\frac {3 x}{4-x-\frac {\left (5+e^{3 e^3}\right ) x}{9+x-\log (x)}} \]
________________________________________________________________________________________
Rubi [F] time = 3.25, 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 {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \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 {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+\left (28+e^{6 e^3}\right ) x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {3 \left (-324-\left (77+e^{3 e^3}\right ) x+\left (1+e^{3 e^3}\right ) x^2+8 (9+x) \log (x)-4 \log ^2(x)\right )}{\left (36-\left (10+e^{3 e^3}\right ) x-x^2+(-4+x) \log (x)\right )^2} \, dx\\ &=3 \int \frac {-324-\left (77+e^{3 e^3}\right ) x+\left (1+e^{3 e^3}\right ) x^2+8 (9+x) \log (x)-4 \log ^2(x)}{\left (36-\left (10+e^{3 e^3}\right ) x-x^2+(-4+x) \log (x)\right )^2} \, dx\\ &=3 \int \left (-\frac {4}{(-4+x)^2}+\frac {x \left (-16 \left (5+e^{3 e^3}\right )+4 \left (5-4 e^{3 e^3}-e^{6 e^3}\right ) x-9 \left (5+e^{3 e^3}\right ) x^2+\left (5+e^{3 e^3}\right ) x^3\right )}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}+\frac {8 \left (-5-e^{3 e^3}\right ) x}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )}\right ) \, dx\\ &=-\frac {12}{4-x}+3 \int \frac {x \left (-16 \left (5+e^{3 e^3}\right )+4 \left (5-4 e^{3 e^3}-e^{6 e^3}\right ) x-9 \left (5+e^{3 e^3}\right ) x^2+\left (5+e^{3 e^3}\right ) x^3\right )}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx-\left (24 \left (5+e^{3 e^3}\right )\right ) \int \frac {x}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )} \, dx\\ &=-\frac {12}{4-x}+3 \int \left (-\frac {4 \left (5+e^{3 e^3}\right )^2}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}-\frac {64 \left (5+e^{3 e^3}\right )^2}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}+\frac {32 \left (5+e^{3 e^3}\right )^2}{(4-x) \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}+\frac {\left (-5-e^{3 e^3}\right ) x}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}+\frac {\left (5+e^{3 e^3}\right ) x^2}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2}\right ) \, dx-\left (24 \left (5+e^{3 e^3}\right )\right ) \int \left (\frac {4}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )}+\frac {1}{(-4+x) \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )}\right ) \, dx\\ &=-\frac {12}{4-x}-\left (3 \left (5+e^{3 e^3}\right )\right ) \int \frac {x}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx+\left (3 \left (5+e^{3 e^3}\right )\right ) \int \frac {x^2}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx-\left (24 \left (5+e^{3 e^3}\right )\right ) \int \frac {1}{(-4+x) \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )} \, dx-\left (96 \left (5+e^{3 e^3}\right )\right ) \int \frac {1}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )} \, dx-\left (12 \left (5+e^{3 e^3}\right )^2\right ) \int \frac {1}{\left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx+\left (96 \left (5+e^{3 e^3}\right )^2\right ) \int \frac {1}{(4-x) \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx-\left (192 \left (5+e^{3 e^3}\right )^2\right ) \int \frac {1}{(4-x)^2 \left (36-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-x^2-4 \log (x)+x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.73, size = 45, normalized size = 1.36 \begin {gather*} \frac {3 \left (36-\left (1+e^{3 e^3}\right ) x-4 \log (x)\right )}{-36+\left (10+e^{3 e^3}\right ) x+x^2-(-4+x) \log (x)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.52, size = 40, normalized size = 1.21 \begin {gather*} -\frac {3 \, {\left (x e^{\left (3 \, e^{3}\right )} + x + 4 \, \log \relax (x) - 36\right )}}{x^{2} + x e^{\left (3 \, e^{3}\right )} - {\left (x - 4\right )} \log \relax (x) + 10 \, x - 36} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.16, size = 42, normalized size = 1.27 \begin {gather*} -\frac {3 \, {\left (x e^{\left (3 \, e^{3}\right )} + x + 4 \, \log \relax (x) - 36\right )}}{x^{2} + x e^{\left (3 \, e^{3}\right )} - x \log \relax (x) + 10 \, x + 4 \, \log \relax (x) - 36} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.22, size = 45, normalized size = 1.36
method | result | size |
norman | \(\frac {-12 \ln \relax (x )+\left (-3 \,{\mathrm e}^{3 \,{\mathrm e}^{3}}-3\right ) x +108}{{\mathrm e}^{3 \,{\mathrm e}^{3}} x +x^{2}-x \ln \relax (x )+10 x +4 \ln \relax (x )-36}\) | \(45\) |
risch | \(\frac {12}{x -4}-\frac {3 \left (5+{\mathrm e}^{3 \,{\mathrm e}^{3}}\right ) x^{2}}{\left (x -4\right ) \left ({\mathrm e}^{3 \,{\mathrm e}^{3}} x +x^{2}-x \ln \relax (x )+10 x +4 \ln \relax (x )-36\right )}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.39, size = 40, normalized size = 1.21 \begin {gather*} -\frac {3 \, {\left (x {\left (e^{\left (3 \, e^{3}\right )} + 1\right )} + 4 \, \log \relax (x) - 36\right )}}{x^{2} + x {\left (e^{\left (3 \, e^{3}\right )} + 10\right )} - {\left (x - 4\right )} \log \relax (x) - 36} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 8.23, size = 45, normalized size = 1.36 \begin {gather*} -\frac {12\,\ln \relax (x)+x\,\left (3\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}+3\right )-108}{10\,x+4\,\ln \relax (x)+x\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}-x\,\ln \relax (x)+x^2-36} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 0.38, size = 66, normalized size = 2.00 \begin {gather*} \frac {15 x^{2} + 3 x^{2} e^{3 e^{3}}}{- x^{3} - x^{2} e^{3 e^{3}} - 6 x^{2} + 76 x + 4 x e^{3 e^{3}} + \left (x^{2} - 8 x + 16\right ) \log {\relax (x )} - 144} + \frac {12}{x - 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________