Optimal. Leaf size=31 \[ \frac {e^{e^3-x} \left (1-e+\frac {5}{x}-\log (2)\right )}{x+x^2} \]
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Rubi [C] time = 1.46, antiderivative size = 116, normalized size of antiderivative = 3.74, number of steps used = 21, number of rules used = 10, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6, 1586, 3, 1593, 1584, 21, 6741, 6742, 2177, 2178} \begin {gather*} -5 e^{e^3} \text {Ei}(-x)+e^{e^3} (4+e+\log (2)) \text {Ei}(-x)+e^{e^3} (1-e-\log (2)) \text {Ei}(-x)+\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}+\frac {e^{e^3-x} (4+e+\log (2))}{x+1}+\frac {e^{e^3-x} (1-e-\log (2))}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 3
Rule 6
Rule 21
Rule 1584
Rule 1586
Rule 1593
Rule 2177
Rule 2178
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^3-x} (5+(1-e) x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+\left (x^2+x^3\right ) (e+\log (2))\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-\frac {25}{(1-e-\log (2))^3}+\frac {25 e}{(1-e-\log (2))^3}+\frac {30}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {5}{1-e-\log (2)}+\frac {25 \log (2)}{(1-e-\log (2))^3}-\frac {5 \log (2)}{(1-e-\log (2))^2}+x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^2 (1+x) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^3 (1+x) \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}+x \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=-\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^3 (1+x)^2} \, dx\\ &=-\int \frac {e^{e^3-x} \left (10+x^3 (1-e-\log (2))+x (21-e-\log (2))+x^2 (8-3 e-\log (8))\right )}{x^3 (1+x)^2} \, dx\\ &=-\int \left (\frac {10 e^{e^3-x}}{x^3}+\frac {e^{e^3-x} (-4-e-\log (2))}{x}+\frac {e^{e^3-x} (1-e-\log (2))}{x^2}+\frac {e^{e^3-x} (4+e+\log (2))}{(1+x)^2}+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}\right ) \, dx\\ &=-\left (10 \int \frac {e^{e^3-x}}{x^3} \, dx\right )-(-4-e-\log (2)) \int \frac {e^{e^3-x}}{x} \, dx-(1-e-\log (2)) \int \frac {e^{e^3-x}}{x^2} \, dx-(4+e+\log (2)) \int \frac {e^{e^3-x}}{(1+x)^2} \, dx-(4+e+\log (2)) \int \frac {e^{e^3-x}}{1+x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}+\frac {e^{e^3-x} (1-e-\log (2))}{x}+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}-e^{1+e^3} \text {Ei}(-1-x) (4+e+\log (2))+e^{e^3} \text {Ei}(-x) (4+e+\log (2))+5 \int \frac {e^{e^3-x}}{x^2} \, dx-(-4-e-\log (2)) \int \frac {e^{e^3-x}}{1+x} \, dx-(-1+e+\log (2)) \int \frac {e^{e^3-x}}{x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}+\frac {e^{e^3-x} (1-e-\log (2))}{x}+e^{e^3} \text {Ei}(-x) (1-e-\log (2))+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}+e^{e^3} \text {Ei}(-x) (4+e+\log (2))-5 \int \frac {e^{e^3-x}}{x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}-5 e^{e^3} \text {Ei}(-x)+\frac {e^{e^3-x} (1-e-\log (2))}{x}+e^{e^3} \text {Ei}(-x) (1-e-\log (2))+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}+e^{e^3} \text {Ei}(-x) (4+e+\log (2))\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.29, size = 28, normalized size = 0.90 \begin {gather*} \frac {e^{e^3-x} (5-x (-1+e+\log (2)))}{x^2 (1+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.04, size = 32, normalized size = 1.03 \begin {gather*} e^{\left (-x + e^{3} + \log \left (-\frac {x e + x \log \relax (2) - x - 5}{x^{3} + x^{2}}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.29, size = 277, normalized size = 8.94 \begin {gather*} -\frac {x e^{\left (-x + e^{3}\right )} \log \relax (2)^{3} + 3 \, x e^{\left (-x + e^{3} + 1\right )} \log \relax (2)^{2} + 7 \, x e^{\left (-x + e^{3}\right )} \log \relax (2)^{2} + 3 \, x e^{\left (-x + e^{3} + 2\right )} \log \relax (2) + 14 \, x e^{\left (-x + e^{3} + 1\right )} \log \relax (2) + 8 \, x e^{\left (-x + e^{3}\right )} \log \relax (2) - 5 \, e^{\left (-x + e^{3}\right )} \log \relax (2)^{2} + x e^{\left (-x + e^{3} + 3\right )} + 7 \, x e^{\left (-x + e^{3} + 2\right )} + 8 \, x e^{\left (-x + e^{3} + 1\right )} - 16 \, x e^{\left (-x + e^{3}\right )} - 10 \, e^{\left (-x + e^{3} + 1\right )} \log \relax (2) - 40 \, e^{\left (-x + e^{3}\right )} \log \relax (2) - 5 \, e^{\left (-x + e^{3} + 2\right )} - 40 \, e^{\left (-x + e^{3} + 1\right )} - 80 \, e^{\left (-x + e^{3}\right )}}{2 \, x^{3} e \log \relax (2) + x^{3} \log \relax (2)^{2} + x^{3} e^{2} + 8 \, x^{3} e + 8 \, x^{3} \log \relax (2) + 2 \, x^{2} e \log \relax (2) + x^{2} \log \relax (2)^{2} + 16 \, x^{3} + x^{2} e^{2} + 8 \, x^{2} e + 8 \, x^{2} \log \relax (2) + 16 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.22, size = 31, normalized size = 1.00
method | result | size |
risch | \(\frac {\left (-x \ln \relax (2)-x \,{\mathrm e}+5+x \right ) {\mathrm e}^{-x +{\mathrm e}^{3}}}{x^{3}+x^{2}}\) | \(31\) |
gosper | \({\mathrm e}^{\ln \left (-\frac {x \,{\mathrm e}+x \ln \relax (2)-x -5}{\left (x +1\right ) x^{2}}\right )-x +{\mathrm e}^{3}}\) | \(32\) |
norman | \({\mathrm e}^{\ln \left (\frac {-x \ln \relax (2)-x \,{\mathrm e}+5+x}{x^{3}+x^{2}}\right )-x +{\mathrm e}^{3}}\) | \(32\) |
default | error in convert/parfrac: cannot convert to partial fraction form\ | N/A |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 37, normalized size = 1.19 \begin {gather*} -\frac {{\left ({\left ({\left (\log \relax (2) - 1\right )} e^{\left (e^{3}\right )} + e^{\left (e^{3} + 1\right )}\right )} x - 5 \, e^{\left (e^{3}\right )}\right )} e^{\left (-x\right )}}{x^{3} + x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.96, size = 78, normalized size = 2.52 \begin {gather*} \frac {5\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x+1}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}\,\ln \relax (2)}{x^3+x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 26, normalized size = 0.84 \begin {gather*} \frac {\left (- e x - x \log {\relax (2 )} + x + 5\right ) e^{- x + e^{3}}}{x^{3} + x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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