Optimal. Leaf size=25 \[ \frac {58 \left (e^{e^x}+x\right )}{x (-5-x+\log (5)) \log (x)} \]
[Out]
________________________________________________________________________________________
Rubi [F]
time = 5.27, 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 {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\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 {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \left (25+\log ^2(5)\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {58 x^2+x (290-58 \log (5))+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \left (25+\log ^2(5)\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {58 \left (e^{e^x}+x\right ) (5+x-\log (5))-58 \left (-x^2+e^{e^x+x} x (5+x-\log (5))+e^{e^x} (-5-2 x+\log (5))\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {58 e^{e^x+x}}{x (5+x-\log (5)) \log (x)}+\frac {58 \left (e^{e^x} x+x^2+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )+5 x \left (1-\frac {\log (5)}{5}\right )+2 e^{e^x} x \log (x)+x^2 \log (x)+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)}\right ) \, dx\\ &=-\left (58 \int \frac {e^{e^x+x}}{x (5+x-\log (5)) \log (x)} \, dx\right )+58 \int \frac {e^{e^x} x+x^2+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )+5 x \left (1-\frac {\log (5)}{5}\right )+2 e^{e^x} x \log (x)+x^2 \log (x)+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=-\left (58 \int \left (-\frac {e^{e^x+x}}{x (-5+\log (5)) \log (x)}+\frac {e^{e^x+x}}{(5+x-\log (5)) (-5+\log (5)) \log (x)}\right ) \, dx\right )+58 \int \frac {\left (e^{e^x}+x\right ) (5+x-\log (5))+\left (x^2+e^{e^x} (5+2 x-\log (5))\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=58 \int \left (\frac {x+5 \left (1-\frac {\log (5)}{5}\right )+x \log (x)}{x (5+x-\log (5))^2 \log ^2(x)}+\frac {e^{e^x} \left (x+5 \left (1-\frac {\log (5)}{5}\right )+2 x \log (x)+5 \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)}\right ) \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {x+5 \left (1-\frac {\log (5)}{5}\right )+x \log (x)}{x (5+x-\log (5))^2 \log ^2(x)} \, dx+58 \int \frac {e^{e^x} \left (x+5 \left (1-\frac {\log (5)}{5}\right )+2 x \log (x)+5 \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \left (\frac {1}{x (5+x-\log (5)) \log ^2(x)}+\frac {1}{(5+x-\log (5))^2 \log (x)}\right ) \, dx+58 \int \left (\frac {e^{e^x}}{x^2 (5+x-\log (5)) \log ^2(x)}+\frac {e^{e^x} (5+2 x-\log (5))}{x^2 (5+x-\log (5))^2 \log (x)}\right ) \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {e^{e^x}}{x^2 (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx+58 \int \frac {e^{e^x} (5+2 x-\log (5))}{x^2 (5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \left (-\frac {e^{e^x}}{x (-5+\log (5))^2 \log ^2(x)}+\frac {e^{e^x}}{(5+x-\log (5)) (-5+\log (5))^2 \log ^2(x)}-\frac {e^{e^x}}{x^2 (-5+\log (5)) \log ^2(x)}\right ) \, dx+58 \int \left (-\frac {e^{e^x}}{x^2 (-5+\log (5)) \log (x)}+\frac {e^{e^x}}{(5+x-\log (5))^2 (-5+\log (5)) \log (x)}\right ) \, dx+58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x}}{x \log ^2(x)} \, dx}{(5-\log (5))^2}+\frac {58 \int \frac {e^{e^x}}{(5+x-\log (5)) \log ^2(x)} \, dx}{(5-\log (5))^2}+\frac {58 \int \frac {e^{e^x}}{x^2 \log ^2(x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x}}{x^2 \log (x)} \, dx}{5-\log (5)}-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}-\frac {58 \int \frac {e^{e^x}}{(5+x-\log (5))^2 \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A]
time = 0.29, size = 25, normalized size = 1.00 \begin {gather*} -\frac {58 \left (e^{e^x}+x\right )}{x (5+x-\log (5)) \log (x)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A]
time = 1.70, size = 38, normalized size = 1.52
method | result | size |
risch | \(\frac {58}{\ln \left (x \right ) \left (\ln \left (5\right )-x -5\right )}+\frac {58 \,{\mathrm e}^{{\mathrm e}^{x}}}{x \left (\ln \left (5\right )-x -5\right ) \ln \left (x \right )}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A]
time = 0.50, size = 24, normalized size = 0.96 \begin {gather*} -\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x {\left (\log \left (5\right ) - 5\right )}\right )} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A]
time = 0.35, size = 25, normalized size = 1.00 \begin {gather*} -\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x \log \left (5\right ) + 5 \, x\right )} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 41 vs.
\(2 (20) = 40\).
time = 0.19, size = 41, normalized size = 1.64 \begin {gather*} - \frac {58 e^{e^{x}}}{x^{2} \log {\left (x \right )} - x \log {\left (5 \right )} \log {\left (x \right )} + 5 x \log {\left (x \right )}} - \frac {58}{\left (x - \log {\left (5 \right )} + 5\right ) \log {\left (x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A]
time = 0.39, size = 39, normalized size = 1.56 \begin {gather*} -\frac {58 \, {\left (x e^{x} + e^{\left (x + e^{x}\right )}\right )}}{x^{2} e^{x} \log \left (x\right ) - x e^{x} \log \left (5\right ) \log \left (x\right ) + 5 \, x e^{x} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {290\,x+58\,x^2\,\ln \left (x\right )-58\,x\,\ln \left (5\right )+58\,x^2+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (58\,x-58\,\ln \left (5\right )+\ln \left (x\right )\,\left (116\,x-58\,\ln \left (5\right )-{\mathrm {e}}^x\,\left (290\,x-58\,x\,\ln \left (5\right )+58\,x^2\right )+290\right )+290\right )}{{\ln \left (x\right )}^2\,\left (x^2\,{\ln \left (5\right )}^2-\ln \left (5\right )\,\left (2\,x^3+10\,x^2\right )+25\,x^2+10\,x^3+x^4\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________