Optimal. Leaf size=13 \[ e^x (4+x) \log \left (5+e^x\right ) \]
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Rubi [A]
time = 0.27, antiderivative size = 25, normalized size of antiderivative = 1.92, number of steps
used = 20, number of rules used = 10, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6820, 6874,
2286, 2207, 2225, 2215, 2221, 2317, 2438, 2634} \begin {gather*} e^x (x+5) \log \left (e^x+5\right )-e^x \log \left (e^x+5\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2207
Rule 2215
Rule 2221
Rule 2225
Rule 2286
Rule 2317
Rule 2438
Rule 2634
Rule 6820
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int e^x \left (\frac {e^x (4+x)}{5+e^x}+(5+x) \log \left (5+e^x\right )\right ) \, dx\\ &=\int \left (\frac {e^{2 x} (4+x)}{5+e^x}+e^x (5+x) \log \left (5+e^x\right )\right ) \, dx\\ &=\int \frac {e^{2 x} (4+x)}{5+e^x} \, dx+\int e^x (5+x) \log \left (5+e^x\right ) \, dx\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-\int \frac {e^{2 x} (4+x)}{5+e^x} \, dx+\int \left (-5 (4+x)+e^x (4+x)+\frac {25 (4+x)}{5+e^x}\right ) \, dx\\ &=-\frac {5}{2} (4+x)^2-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )+25 \int \frac {4+x}{5+e^x} \, dx+\int e^x (4+x) \, dx-\int \left (-5 (4+x)+e^x (4+x)+\frac {25 (4+x)}{5+e^x}\right ) \, dx\\ &=e^x (4+x)+\frac {5}{2} (4+x)^2-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \int \frac {e^x (4+x)}{5+e^x} \, dx-25 \int \frac {4+x}{5+e^x} \, dx-\int e^x \, dx-\int e^x (4+x) \, dx\\ &=-e^x-5 (4+x) \log \left (1+\frac {e^x}{5}\right )-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )+5 \int \frac {e^x (4+x)}{5+e^x} \, dx+5 \int \log \left (1+\frac {e^x}{5}\right ) \, dx+\int e^x \, dx\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \int \log \left (1+\frac {e^x}{5}\right ) \, dx+5 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )-5 \text {Li}_2\left (-\frac {e^x}{5}\right )-5 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-e^x \log \left (5+e^x\right )+e^x (5+x) \log \left (5+e^x\right )\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.06, size = 13, normalized size = 1.00 \begin {gather*} e^x (4+x) \log \left (5+e^x\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.15, size = 12, normalized size = 0.92
method | result | size |
risch | \(\ln \left ({\mathrm e}^{x}+5\right ) \left (4+x \right ) {\mathrm e}^{x}\) | \(12\) |
norman | \({\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+5\right ) x +4 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+5\right )\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.31, size = 22, normalized size = 1.69 \begin {gather*} {\left ({\left (x + 4\right )} e^{x} + 20\right )} \log \left (e^{x} + 5\right ) - 20 \, \log \left (e^{x} + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 11, normalized size = 0.85 \begin {gather*} {\left (x + 4\right )} e^{x} \log \left (e^{x} + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 19, normalized size = 1.46 \begin {gather*} x e^{x} \log \left (e^{x} + 5\right ) + 4 \, e^{x} \log \left (e^{x} + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.25, size = 78, normalized size = 6.00 \begin {gather*} \frac {40\,{\mathrm {e}}^{2\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+4\,{\mathrm {e}}^{3\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+100\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+5\right )+25\,x\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+5\right )+10\,x\,{\mathrm {e}}^{2\,x}\,\ln \left ({\mathrm {e}}^x+5\right )+x\,{\mathrm {e}}^{3\,x}\,\ln \left ({\mathrm {e}}^x+5\right )}{{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x+25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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