Optimal. Leaf size=24 \[ e x-5 \left (-x+\frac {8 x^3}{1+e^{-1+x}}\right ) \]
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Rubi [A] time = 1.06, antiderivative size = 19, normalized size of antiderivative = 0.79, number of steps used = 30, number of rules used = 11, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {6741, 12, 6742, 2184, 2190, 2531, 2282, 6589, 6609, 2185, 2191} \begin {gather*} (5+e) x-\frac {40 e x^3}{e^x+e} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2282
Rule 2531
Rule 6589
Rule 6609
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (5 \left (1+\frac {e}{5}\right )+e^{-2+2 x} (5+e)-120 x^2+e^{-1+x} \left (10+2 e-120 x^2+40 x^3\right )\right )}{\left (e+e^x\right )^2} \, dx\\ &=e^2 \int \frac {5 \left (1+\frac {e}{5}\right )+e^{-2+2 x} (5+e)-120 x^2+e^{-1+x} \left (10+2 e-120 x^2+40 x^3\right )}{\left (e+e^x\right )^2} \, dx\\ &=e^2 \int \left (\frac {5+e}{e^2}+\frac {40 (-3+x) x^2}{e \left (e+e^x\right )}-\frac {40 x^3}{\left (e+e^x\right )^2}\right ) \, dx\\ &=(5+e) x+(40 e) \int \frac {(-3+x) x^2}{e+e^x} \, dx-\left (40 e^2\right ) \int \frac {x^3}{\left (e+e^x\right )^2} \, dx\\ &=(5+e) x+(40 e) \int \frac {e^x x^3}{\left (e+e^x\right )^2} \, dx-(40 e) \int \frac {x^3}{e+e^x} \, dx+(40 e) \int \left (-\frac {3 x^2}{e+e^x}+\frac {x^3}{e+e^x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-10 x^4+40 \int \frac {e^x x^3}{e+e^x} \, dx+(40 e) \int \frac {x^3}{e+e^x} \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+40 x^3 \log \left (1+e^{-1+x}\right )-40 \int \frac {e^x x^3}{e+e^x} \, dx-120 \int x^2 \log \left (1+e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+120 x^2 \text {Li}_2\left (-e^{-1+x}\right )+120 \int x^2 \log \left (1+e^{-1+x}\right ) \, dx-240 \int x \text {Li}_2\left (-e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-240 x \text {Li}_3\left (-e^{-1+x}\right )+240 \int x \text {Li}_2\left (-e^{-1+x}\right ) \, dx+240 \int \text {Li}_3\left (-e^{-1+x}\right ) \, dx\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}-240 \int \text {Li}_3\left (-e^{-1+x}\right ) \, dx+240 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-1+x}\right )\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}+240 \text {Li}_4\left (-e^{-1+x}\right )-240 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-1+x}\right )\\ &=(5+e) x-\frac {40 e x^3}{e+e^x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 19, normalized size = 0.79 \begin {gather*} (5+e) x-\frac {40 e x^3}{e+e^x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 38, normalized size = 1.58 \begin {gather*} -\frac {40 \, x^{3} - x e - {\left (x e + 5 \, x\right )} e^{\left (x - 1\right )} - 5 \, x}{e^{\left (x - 1\right )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.24, size = 137, normalized size = 5.71 \begin {gather*} -\frac {40 \, x^{3} e - x e^{2} - 5 \, x e - x e^{\left (x + 1\right )} - 5 \, x e^{x} + e^{2} \log \left (e + e^{x}\right ) + 5 \, e \log \left (e + e^{x}\right ) + e^{\left (x + 1\right )} \log \left (e + e^{x}\right ) + 5 \, e^{x} \log \left (e + e^{x}\right ) - e^{2} \log \left (-e - e^{x}\right ) - 5 \, e \log \left (-e - e^{x}\right ) - e^{\left (x + 1\right )} \log \left (-e - e^{x}\right ) - 5 \, e^{x} \log \left (-e - e^{x}\right )}{e + e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 22, normalized size = 0.92
method | result | size |
risch | \(x \,{\mathrm e}-\frac {40 x^{3}}{{\mathrm e}^{x -1}+1}+5 x\) | \(22\) |
norman | \(\frac {\left ({\mathrm e}+5\right ) x +\left ({\mathrm e}+5\right ) x \,{\mathrm e}^{x -1}-40 x^{3}}{{\mathrm e}^{x -1}+1}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 81, normalized size = 3.38 \begin {gather*} {\left (x + \frac {1}{e^{\left (x - 1\right )} + 1} - \log \left (e^{\left (x - 1\right )} + 1\right ) - 1\right )} e + {\left (e + 5\right )} \log \left (e + e^{x}\right ) + 5 \, x - \frac {40 \, x^{3} e + e^{2} + 5 \, e}{e + e^{x}} + \frac {5}{e^{\left (x - 1\right )} + 1} - 5 \, \log \left (e^{\left (x - 1\right )} + 1\right ) - 5 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.09, size = 20, normalized size = 0.83 \begin {gather*} x\,\left (\mathrm {e}+5\right )-\frac {40\,x^3}{{\mathrm {e}}^{x-1}+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.11, size = 17, normalized size = 0.71 \begin {gather*} - \frac {40 x^{3}}{e^{x - 1} + 1} + x \left (e + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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