Integrand size = 14, antiderivative size = 5 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=x^{x^2} \]
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\[ \int x^{1+x^2} (1+2 \log (x)) \, dx=\int x^{1+x^2} (1+2 \log (x)) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (x^{1+x^2}+2 x^{1+x^2} \log (x)\right ) \, dx \\ & = 2 \int x^{1+x^2} \log (x) \, dx+\int x^{1+x^2} \, dx \\ & = -\left (2 \int \frac {\int x^{1+x^2} \, dx}{x} \, dx\right )+(2 \log (x)) \int x^{1+x^2} \, dx+\int x^{1+x^2} \, dx \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=x^{x^2} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(11\) vs. \(2(5)=10\).
Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 2.40
method | result | size |
risch | \(\frac {x^{x^{2}+1}}{x}\) | \(12\) |
parallelrisch | \(\frac {x^{x^{2}+1}}{x}\) | \(12\) |
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Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (5) = 10\).
Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 2.20 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=\frac {x^{x^{2} + 1}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 7 vs. \(2 (3) = 6\).
Time = 0.07 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.40 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=\frac {x^{x^{2} + 1}}{x} \]
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none
Time = 0.24 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=x^{\left (x^{2}\right )} \]
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\[ \int x^{1+x^2} (1+2 \log (x)) \, dx=\int { x^{x^{2} + 1} {\left (2 \, \log \left (x\right ) + 1\right )} \,d x } \]
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Time = 15.46 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int x^{1+x^2} (1+2 \log (x)) \, dx=x^{x^2} \]
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