Optimal. Leaf size=25 \[ \frac {5 x}{-e^{e^{e^2} \left (100-2 e^x+x\right )}+x} \]
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Rubi [A] time = 2.25, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6688, 12, 6711, 32} \begin {gather*} \frac {5}{1-\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 32
Rule 6688
Rule 6711
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-1+e^{e^2} x-2 e^{e^2+x} x\right )}{\left (e^{e^{e^2} \left (100-2 e^x+x\right )}-x\right )^2} \, dx\\ &=5 \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-1+e^{e^2} x-2 e^{e^2+x} x\right )}{\left (e^{e^{e^2} \left (100-2 e^x+x\right )}-x\right )^2} \, dx\\ &=5 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {e^{e^{e^2} \left (100-2 e^x+x\right )}}{x}\right )\\ &=\frac {5}{1-\frac {e^{e^{e^2} \left (100-2 e^x+x\right )}}{x}}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.22, size = 25, normalized size = 1.00 \begin {gather*} -\frac {5 x}{e^{e^{e^2} \left (100-2 e^x+x\right )}-x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 21, normalized size = 0.84 \begin {gather*} \frac {5 \, x}{x - e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 149, normalized size = 5.96 \begin {gather*} \frac {5 \, {\left (2 \, x^{2} e^{\left (x + e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - x^{2} e^{\left (e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (100 \, e^{\left (e^{2}\right )}\right )}\right )}}{2 \, x^{2} e^{\left (x + e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - x^{2} e^{\left (e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - 2 \, x e^{\left (x e^{\left (e^{2}\right )} + x + e^{2} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (x e^{\left (e^{2}\right )} + e^{2} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (100 \, e^{\left (e^{2}\right )}\right )} - e^{\left (x e^{\left (e^{2}\right )} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 25, normalized size = 1.00
method | result | size |
risch | \(\frac {5 x}{x -{\mathrm e}^{-\left (2 \,{\mathrm e}^{x}-x -100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\) | \(25\) |
norman | \(\frac {5 \,{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}{x -{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 41, normalized size = 1.64 \begin {gather*} \frac {5 \, e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}}{x e^{\left (2 \, e^{\left (x + e^{2}\right )}\right )} - e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.86, size = 29, normalized size = 1.16 \begin {gather*} \frac {5\,x}{x-{\mathrm {e}}^{x\,{\mathrm {e}}^{{\mathrm {e}}^2}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{100\,{\mathrm {e}}^{{\mathrm {e}}^2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 20, normalized size = 0.80 \begin {gather*} - \frac {5 x}{- x + e^{\left (x - 2 e^{x} + 100\right ) e^{e^{2}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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