Integrand size = 16, antiderivative size = 11 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (2-x)+\log (5+x) \]
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Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (2-x)+\log (x+5) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{-2+x}+\frac {1}{5+x}\right ) \, dx \\ & = \log (2-x)+\log (5+x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (-2+x)+\log (5+x) \]
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Time = 0.80 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82
method | result | size |
default | \(\ln \left (\left (x -2\right ) \left (5+x \right )\right )\) | \(9\) |
norman | \(\ln \left (5+x \right )+\ln \left (x -2\right )\) | \(10\) |
risch | \(\ln \left (x^{2}+3 x -10\right )\) | \(10\) |
parallelrisch | \(\ln \left (5+x \right )+\ln \left (x -2\right )\) | \(10\) |
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none
Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left (x^{2} + 3 \, x - 10\right ) \]
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Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log {\left (x^{2} + 3 x - 10 \right )} \]
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none
Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left (x + 5\right ) + \log \left (x - 2\right ) \]
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none
Time = 0.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left ({\left | x + 5 \right |}\right ) + \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 9.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\ln \left (x^2+3\,x-10\right ) \]
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