Optimal. Leaf size=20 \[ \frac {\log (x)}{-1+e^3-4 \left (25+e^x-2 x\right )} \]
________________________________________________________________________________________
Rubi [F] time = 1.36, 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 {-101+e^3-4 e^x+8 x+\left (-8 x+4 e^x x\right ) \log (x)}{10201 x+e^6 x+16 e^{2 x} x-1616 x^2+64 x^3+e^x \left (808 x-8 e^3 x-64 x^2\right )+e^3 \left (-202 x+16 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-101+e^3-4 e^x+8 x+\left (-8 x+4 e^x x\right ) \log (x)}{16 e^{2 x} x+\left (10201+e^6\right ) x-1616 x^2+64 x^3+e^x \left (808 x-8 e^3 x-64 x^2\right )+e^3 \left (-202 x+16 x^2\right )} \, dx\\ &=\int \frac {-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x+4 \left (-2+e^x\right ) x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2 x} \, dx\\ &=\int \left (\frac {\left (-109+e^3+8 x\right ) \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2}+\frac {-1+x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right ) x}\right ) \, dx\\ &=\int \frac {\left (-109+e^3+8 x\right ) \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {-1+x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right ) x} \, dx\\ &=(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \left (\frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )}+\frac {\log (x)}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x}\right ) \, dx-\int \frac {\left (-109+e^3\right ) \int \frac {1}{\left (-101+e^3-4 e^x+8 x\right )^2} \, dx+8 \int \frac {x}{\left (-101+e^3-4 e^x+8 x\right )^2} \, dx}{x} \, dx\\ &=(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )} \, dx+\int \frac {\log (x)}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx-\int \left (\frac {\left (-109+e^3\right ) \int \frac {1}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x}+\frac {8 \int \frac {x}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x}\right ) \, dx\\ &=-\left (8 \int \frac {\int \frac {x}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x} \, dx\right )-\left (-109+e^3\right ) \int \frac {\int \frac {1}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x} \, dx+\log (x) \int \frac {1}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx+(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )} \, dx-\int \frac {\int \frac {1}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.35, size = 21, normalized size = 1.05 \begin {gather*} -\frac {\log (x)}{101-e^3+4 e^x-8 x} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.43, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.33, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.03, size = 17, normalized size = 0.85
method | result | size |
risch | \(\frac {\ln \relax (x )}{-4 \,{\mathrm e}^{x}+{\mathrm e}^{3}+8 x -101}\) | \(17\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.74, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 2.21, size = 16, normalized size = 0.80 \begin {gather*} \frac {\ln \relax (x)}{8\,x+{\mathrm {e}}^3-4\,{\mathrm {e}}^x-101} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.27, size = 17, normalized size = 0.85 \begin {gather*} - \frac {\log {\relax (x )}}{- 8 x + 4 e^{x} - e^{3} + 101} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________