Optimal. Leaf size=29 \[ x-\log \left (\log \left (3-\frac {e^{2+x}}{3}-x+\frac {x}{5-10 x}\right )\right ) \]
________________________________________________________________________________________
Rubi [F] time = 4.21, 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 {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12-60 x+60 x^2-e^{2+x} \left (-5+20 x-20 x^2\right )-\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{(1-2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ &=\int \left (\frac {3 \left (-19+84 x-98 x^2+20 x^3\right )}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {-1+\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx\\ &=3 \int \frac {-19+84 x-98 x^2+20 x^3}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+\int \frac {-1+\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ &=3 \int \left (\frac {20}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}-\frac {44 x}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {10 x^2}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}+\frac {1}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx+\int \left (1-\frac {1}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}\right ) \, dx\\ &=x+3 \int \frac {1}{(-1+2 x) \left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+30 \int \frac {x^2}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx+60 \int \frac {1}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx-132 \int \frac {x}{\left (45-5 e^{2+x}-102 x+10 e^{2+x} x+30 x^2\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx-\int \frac {1}{\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.14, size = 35, normalized size = 1.21 \begin {gather*} x-\log \left (\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.48, size = 36, normalized size = 1.24 \begin {gather*} x - \log \left (\log \left (-\frac {30 \, x^{2} + 5 \, {\left (2 \, x - 1\right )} e^{\left (x + 2\right )} - 102 \, x + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.69, size = 38, normalized size = 1.31 \begin {gather*} x - \log \left (\log \left (-\frac {30 \, x^{2} + 10 \, x e^{\left (x + 2\right )} - 102 \, x - 5 \, e^{\left (x + 2\right )} + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.22, size = 35, normalized size = 1.21
method | result | size |
norman | \(x -\ln \left (\ln \left (\frac {\left (-10 x +5\right ) {\mathrm e}^{2+x}-30 x^{2}+102 x -45}{30 x -15}\right )\right )\) | \(35\) |
risch | \(x -\ln \left (\ln \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )+\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x -\frac {1}{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x -\frac {1}{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{x -\frac {1}{2}}\right )^{3}+2 \pi +2 i \ln \left (x -\frac {1}{2}\right )\right )}{2}\right )\) | \(281\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 1.47, size = 47, normalized size = 1.62 \begin {gather*} x - \log \left (-\log \relax (5) - \log \relax (3) + \log \left (-30 \, x^{2} - 5 \, {\left (2 \, x e^{2} - e^{2}\right )} e^{x} + 102 \, x - 45\right ) - \log \left (2 \, x - 1\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 6.91, size = 35, normalized size = 1.21 \begin {gather*} x-\ln \left (\ln \left (-\frac {30\,x^2-102\,x+{\mathrm {e}}^2\,{\mathrm {e}}^x\,\left (10\,x-5\right )+45}{30\,x-15}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.77, size = 29, normalized size = 1.00 \begin {gather*} x - \log {\left (\log {\left (\frac {- 30 x^{2} + 102 x + \left (5 - 10 x\right ) e^{x + 2} - 45}{30 x - 15} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________