Optimal. Leaf size=34 \[ -5+\frac {1}{3} e^{\frac {\frac {4}{5}+x-x^2-\frac {\log (4)}{5+x}}{9 x}} \]
________________________________________________________________________________________
Rubi [F] time = 2.97, 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 {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{x^2 \left (3375+1350 x+135 x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{135 x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \left (-129 x^2-50 x^3-5 x^4-25 (4-\log (4))-10 x (4-\log (4))\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \left (-5 \exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )+\frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) (-4+\log (4))}{x^2}-\frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \log (4)}{(5+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{27} \int \exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \, dx\right )+\frac {1}{135} (-4+\log (4)) \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )}{x^2} \, dx-\frac {1}{135} \log (4) \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )}{(5+x)^2} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [F] time = 2.23, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.61, size = 33, normalized size = 0.97 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {5 \, x^{3} + 20 \, x^{2} - 29 \, x + 10 \, \log \relax (2) - 20}{45 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 0.14, size = 68, normalized size = 2.00 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {x^{3}}{9 \, {\left (x^{2} + 5 \, x\right )}} - \frac {4 \, x^{2}}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {29 \, x}{45 \, {\left (x^{2} + 5 \, x\right )}} - \frac {2 \, \log \relax (2)}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {4}{9 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.14, size = 33, normalized size = 0.97
method | result | size |
gosper | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \relax (2)-29 x -20}{45 \left (5+x \right ) x}}}{3}\) | \(33\) |
risch | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \relax (2)-29 x -20}{45 \left (5+x \right ) x}}}{3}\) | \(33\) |
norman | \(\frac {\frac {5 x \,{\mathrm e}^{\frac {-10 \ln \relax (2)-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}+\frac {x^{2} {\mathrm e}^{\frac {-10 \ln \relax (2)-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}}{\left (5+x \right ) x}\) | \(83\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.62, size = 29, normalized size = 0.85 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {1}{9} \, x + \frac {2 \, \log \relax (2)}{45 \, {\left (x + 5\right )}} - \frac {2 \, \log \relax (2)}{45 \, x} + \frac {4}{45 \, x} + \frac {1}{9}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 9.54, size = 82, normalized size = 2.41 \begin {gather*} \frac {{\mathrm {e}}^{\frac {29\,x}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {5\,x^3}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {20\,x^2}{45\,x^2+225\,x}}\,{\mathrm {e}}^{\frac {20}{45\,x^2+225\,x}}}{3\,2^{\frac {10}{45\,x^2+225\,x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.53, size = 31, normalized size = 0.91 \begin {gather*} \frac {e^{\frac {- 5 x^{3} - 20 x^{2} + 29 x - 10 \log {\relax (2 )} + 20}{45 x^{2} + 225 x}}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________