Integrand size = 42, antiderivative size = 19 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=x+2 \left (-4-e \left (16+e^x\right )+x\right ) \log (3+x) \]
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Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(19)=38\).
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.47, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6873, 6874, 2326, 45, 2436, 2332} \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=x+2 (x+3) \log (x+3)-2 (7+16 e) \log (x+3)-\frac {2 e^{x+1} (x \log (x+3)+3 \log (x+3))}{x+3} \]
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Rule 45
Rule 2326
Rule 2332
Rule 2436
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 e^{1+x}-5 \left (1+\frac {32 e}{5}\right )+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx \\ & = \int \left (-\frac {2 e^{1+x} (1+3 \log (3+x)+x \log (3+x))}{3+x}+\frac {-5 \left (1+\frac {32 e}{5}\right )+3 x+6 \log (3+x)+2 x \log (3+x)}{3+x}\right ) \, dx \\ & = -\left (2 \int \frac {e^{1+x} (1+3 \log (3+x)+x \log (3+x))}{3+x} \, dx\right )+\int \frac {-5 \left (1+\frac {32 e}{5}\right )+3 x+6 \log (3+x)+2 x \log (3+x)}{3+x} \, dx \\ & = -\frac {2 e^{1+x} (3 \log (3+x)+x \log (3+x))}{3+x}+\int \left (\frac {-5-32 e+3 x}{3+x}+2 \log (3+x)\right ) \, dx \\ & = -\frac {2 e^{1+x} (3 \log (3+x)+x \log (3+x))}{3+x}+2 \int \log (3+x) \, dx+\int \frac {-5-32 e+3 x}{3+x} \, dx \\ & = -\frac {2 e^{1+x} (3 \log (3+x)+x \log (3+x))}{3+x}+2 \text {Subst}(\int \log (x) \, dx,x,3+x)+\int \left (3-\frac {2 (7+16 e)}{3+x}\right ) \, dx \\ & = x-2 (7+16 e) \log (3+x)+2 (3+x) \log (3+x)-\frac {2 e^{1+x} (3 \log (3+x)+x \log (3+x))}{3+x} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=x-2 \left (4+16 e+e^{1+x}-x\right ) \log (3+x) \]
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Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63
method | result | size |
norman | \(x +\left (-8-32 \,{\mathrm e}\right ) \ln \left (3+x \right )+2 x \ln \left (3+x \right )-2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (3+x \right )\) | \(31\) |
risch | \(\left (-2 \,{\mathrm e}^{1+x}+2 x \right ) \ln \left (3+x \right )-32 \,{\mathrm e} \ln \left (3+x \right )-8 \ln \left (3+x \right )+x\) | \(32\) |
default | \(-2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (3+x \right )+x -\left (14+32 \,{\mathrm e}\right ) \ln \left (3+x \right )+2 \left (3+x \right ) \ln \left (3+x \right )-6\) | \(35\) |
parallelrisch | \(-2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (3+x \right )-32 \,{\mathrm e} \ln \left (3+x \right )+2 x \ln \left (3+x \right )+x -8 \ln \left (3+x \right )-6\) | \(35\) |
parts | \(-2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \left (3+x \right )+x -\left (14+32 \,{\mathrm e}\right ) \ln \left (3+x \right )+2 \left (3+x \right ) \ln \left (3+x \right )-6\) | \(35\) |
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Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=2 \, {\left (x - 16 \, e - e^{\left (x + 1\right )} - 4\right )} \log \left (x + 3\right ) + x \]
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Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=2 x \log {\left (x + 3 \right )} + x - 2 e e^{x} \log {\left (x + 3 \right )} - 8 \cdot \left (1 + 4 e\right ) \log {\left (x + 3 \right )} \]
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\[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=\int { -\frac {2 \, {\left ({\left (x + 3\right )} e^{\left (x + 1\right )} - x - 3\right )} \log \left (x + 3\right ) - 3 \, x + 32 \, e + 2 \, e^{\left (x + 1\right )} + 5}{x + 3} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=2 \, x \log \left (x + 3\right ) - 32 \, e \log \left (x + 3\right ) - 2 \, e^{\left (x + 1\right )} \log \left (x + 3\right ) + x - 8 \, \log \left (x + 3\right ) \]
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Time = 11.90 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {-5-32 e-2 e^{1+x}+3 x+\left (6+e^{1+x} (-6-2 x)+2 x\right ) \log (3+x)}{3+x} \, dx=x-8\,\ln \left (x+3\right )-32\,\ln \left (x+3\right )\,\mathrm {e}+2\,x\,\ln \left (x+3\right )-2\,\ln \left (x+3\right )\,{\mathrm {e}}^{x+1} \]
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