Optimal. Leaf size=14 \[ \frac {35}{8}+x+\frac {x}{5+e^x} \]
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Rubi [B] time = 0.30, antiderivative size = 77, normalized size of antiderivative = 5.50, number of steps used = 17, number of rules used = 12, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {6688, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} \frac {x^2}{10}-\frac {1}{10} (1-x)^2+\frac {x}{e^x+5}+\frac {4 x}{5}-\frac {1}{5} (1-x) \log \left (\frac {e^x}{5}+1\right )-\frac {1}{5} x \log \left (\frac {e^x}{5}+1\right )+\frac {1}{5} \log \left (e^x+5\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2279
Rule 2282
Rule 2391
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30+e^{2 x}-e^x (-11+x)}{\left (5+e^x\right )^2} \, dx\\ &=\int \left (1-\frac {-1+x}{5+e^x}+\frac {5 x}{\left (5+e^x\right )^2}\right ) \, dx\\ &=x+5 \int \frac {x}{\left (5+e^x\right )^2} \, dx-\int \frac {-1+x}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {1}{5} \int \frac {e^x (-1+x)}{5+e^x} \, dx-\int \frac {e^x x}{\left (5+e^x\right )^2} \, dx+\int \frac {x}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} \int \frac {e^x x}{5+e^x} \, dx-\frac {1}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx-\int \frac {1}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx-\frac {1}{5} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )-\operatorname {Subst}\left (\int \frac {1}{x (5+x)} \, dx,x,e^x\right )\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \text {Li}_2\left (-\frac {e^x}{5}\right )-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{5+x} \, dx,x,e^x\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-\frac {1}{10} (1-x)^2+\frac {4 x}{5}+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \log \left (5+e^x\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 11, normalized size = 0.79 \begin {gather*} x+\frac {x}{5+e^x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 15, normalized size = 1.07 \begin {gather*} \frac {x e^{x} + 6 \, x}{e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 15, normalized size = 1.07 \begin {gather*} \frac {x e^{x} + 6 \, x}{e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 11, normalized size = 0.79
method | result | size |
risch | \(x +\frac {x}{{\mathrm e}^{x}+5}\) | \(11\) |
norman | \(\frac {{\mathrm e}^{x} x +6 x}{{\mathrm e}^{x}+5}\) | \(16\) |
default | \(\frac {6 \ln \left ({\mathrm e}^{x}\right )}{5}-\frac {x \,{\mathrm e}^{x}}{5 \left ({\mathrm e}^{x}+5\right )}\) | \(18\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.39, size = 15, normalized size = 1.07 \begin {gather*} \frac {6}{5} \, x - \frac {x e^{x}}{5 \, {\left (e^{x} + 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.07, size = 10, normalized size = 0.71 \begin {gather*} x+\frac {x}{{\mathrm {e}}^x+5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.10, size = 7, normalized size = 0.50 \begin {gather*} x + \frac {x}{e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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