Integrand size = 70, antiderivative size = 23 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=x+\frac {-1+e^{-4+3 x-x^4}}{\log (5+x)} \]
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Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(23)=46\).
Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.26, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6874, 2326, 6820, 2437, 2339, 30} \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=\frac {e^{-x^4+3 x-4} \left (-4 x^4 \log (x+5)-20 x^3 \log (x+5)+3 x \log (x+5)+15 \log (x+5)\right )}{(x+5) \left (3-4 x^3\right ) \log ^2(x+5)}+x-\frac {1}{\log (x+5)} \]
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Rule 30
Rule 2326
Rule 2339
Rule 2437
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {e^{-4+3 x-x^4} \left (1-15 \log (5+x)-3 x \log (5+x)+20 x^3 \log (5+x)+4 x^4 \log (5+x)\right )}{(5+x) \log ^2(5+x)}+\frac {1+5 \log ^2(5+x)+x \log ^2(5+x)}{(5+x) \log ^2(5+x)}\right ) \, dx \\ & = -\int \frac {e^{-4+3 x-x^4} \left (1-15 \log (5+x)-3 x \log (5+x)+20 x^3 \log (5+x)+4 x^4 \log (5+x)\right )}{(5+x) \log ^2(5+x)} \, dx+\int \frac {1+5 \log ^2(5+x)+x \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx \\ & = \frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\int \left (1+\frac {1}{(5+x) \log ^2(5+x)}\right ) \, dx \\ & = x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\int \frac {1}{(5+x) \log ^2(5+x)} \, dx \\ & = x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5+x\right ) \\ & = x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (5+x)\right ) \\ & = x-\frac {1}{\log (5+x)}+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=\frac {-1+e^{-4+3 x-x^4}+x \log (5+x)}{\log (5+x)} \]
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Time = 0.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {{\mathrm e}^{-x^{4}+3 x -4}-1}{\ln \left (5+x \right )}+x\) | \(23\) |
parallelrisch | \(-\frac {-10 x \ln \left (5+x \right )+10+25 \ln \left (5+x \right )-10 \,{\mathrm e}^{-x^{4}+3 x -4}}{10 \ln \left (5+x \right )}\) | \(37\) |
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Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=\frac {x \log \left (x + 5\right ) + e^{\left (-x^{4} + 3 \, x - 4\right )} - 1}{\log \left (x + 5\right )} \]
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Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=x + \frac {e^{- x^{4} + 3 x - 4}}{\log {\left (x + 5 \right )}} - \frac {1}{\log {\left (x + 5 \right )}} \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=\frac {{\left (x e^{4} \log \left (x + 5\right ) + e^{\left (-x^{4} + 3 \, x\right )}\right )} e^{\left (-4\right )}}{\log \left (x + 5\right )} - \frac {1}{\log \left (x + 5\right )} \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=\frac {{\left (x e^{4} \log \left (x + 5\right ) - e^{4} + e^{\left (-x^{4} + 3 \, x\right )}\right )} e^{\left (-4\right )}}{\log \left (x + 5\right )} \]
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Time = 15.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} \left (15+3 x-20 x^3-4 x^4\right ) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx=x-\frac {1}{\ln \left (x+5\right )}+\frac {{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-x^4}}{\ln \left (x+5\right )} \]
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