Integrand size = 39, antiderivative size = 23 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=4-\frac {4}{5} \left (1+e^x\right )-x-\log (\log (4+2 x)) \]
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Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6820, 2225, 2437, 12, 2339, 29} \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x-\frac {4 e^x}{5}-\log (\log (2 (x+2))) \]
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Rule 12
Rule 29
Rule 2225
Rule 2339
Rule 2437
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (-1-\frac {4 e^x}{5}-\frac {1}{(2+x) \log (4+2 x)}\right ) \, dx \\ & = -x-\frac {4 \int e^x \, dx}{5}-\int \frac {1}{(2+x) \log (4+2 x)} \, dx \\ & = -\frac {4 e^x}{5}-x-\frac {1}{2} \text {Subst}\left (\int \frac {2}{x \log (x)} \, dx,x,4+2 x\right ) \\ & = -\frac {4 e^x}{5}-x-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4+2 x\right ) \\ & = -\frac {4 e^x}{5}-x-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (2 (2+x))\right ) \\ & = -\frac {4 e^x}{5}-x-\log (\log (2 (2+x))) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-\frac {4 e^x}{5}-x-\log (\log (2 (2+x))) \]
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Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
default | \(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\) | \(18\) |
norman | \(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\) | \(18\) |
risch | \(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\) | \(18\) |
parts | \(-x -\ln \left (\ln \left (4+2 x \right )\right )-\frac {4 \,{\mathrm e}^{x}}{5}\) | \(18\) |
parallelrisch | \(4-\ln \left (\ln \left (4+2 x \right )\right )-x -\frac {4 \,{\mathrm e}^{x}}{5}\) | \(19\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x - \frac {4}{5} \, e^{x} - \log \left (\log \left (2 \, x + 4\right )\right ) \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=- x - \frac {4 e^{x}}{5} - \log {\left (\log {\left (2 x + 4 \right )} \right )} \]
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\[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=\int { -\frac {{\left (4 \, {\left (x + 2\right )} e^{x} + 5 \, x + 10\right )} \log \left (2 \, x + 4\right ) + 5}{5 \, {\left (x + 2\right )} \log \left (2 \, x + 4\right )} \,d x } \]
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Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x - \frac {4}{5} \, e^{x} - \log \left (\log \left (2 \, x + 4\right )\right ) \]
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Time = 13.73 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-5+\left (-10+e^x (-8-4 x)-5 x\right ) \log (4+2 x)}{(10+5 x) \log (4+2 x)} \, dx=-x-\ln \left (\ln \left (2\,x+4\right )\right )-\frac {4\,{\mathrm {e}}^x}{5} \]
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