Integrand size = 71, antiderivative size = 29 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=5 e^{2+3 \left (x-\log \left (4 \left (-1+\frac {x}{-1-\log \left (x^2\right )}\right )\right )\right )} \]
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\[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=\int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {15 e^{2+3 x} \left (1+\log \left (x^2\right )\right )^2 \left (-2-x-(1+x) \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{64 \left (1+x+\log \left (x^2\right )\right )^4} \, dx \\ & = \frac {15}{64} \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^2 \left (-2-x-(1+x) \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx \\ & = \frac {15}{64} \int \left (-e^{2+3 x}-\frac {e^{2+3 x} x^2 (2+x)}{\left (1+x+\log \left (x^2\right )\right )^4}+\frac {e^{2+3 x} x \left (4+3 x+x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {e^{2+3 x} \left (-2-3 x-3 x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^2}+\frac {e^{2+3 x} (1+3 x)}{1+x+\log \left (x^2\right )}\right ) \, dx \\ & = -\left (\frac {15}{64} \int e^{2+3 x} \, dx\right )-\frac {15}{64} \int \frac {e^{2+3 x} x^2 (2+x)}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} x \left (4+3 x+x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} \left (-2-3 x-3 x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} (1+3 x)}{1+x+\log \left (x^2\right )} \, dx \\ & = -\frac {5}{64} e^{2+3 x}-\frac {15}{64} \int \left (\frac {2 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^4}+\frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^4}\right ) \, dx+\frac {15}{64} \int \left (\frac {4 e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {3 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^3}\right ) \, dx+\frac {15}{64} \int \left (-\frac {2 e^{2+3 x}}{\left (1+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^2}\right ) \, dx+\frac {15}{64} \int \left (\frac {e^{2+3 x}}{1+x+\log \left (x^2\right )}+\frac {3 e^{2+3 x} x}{1+x+\log \left (x^2\right )}\right ) \, dx \\ & = -\frac {5}{64} e^{2+3 x}-\frac {15}{64} \int \frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx+\frac {15}{64} \int \frac {e^{2+3 x}}{1+x+\log \left (x^2\right )} \, dx-\frac {15}{32} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx-\frac {15}{32} \int \frac {e^{2+3 x}}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {45}{64} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx-\frac {45}{64} \int \frac {e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx-\frac {45}{64} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {45}{64} \int \frac {e^{2+3 x} x}{1+x+\log \left (x^2\right )} \, dx+\frac {15}{16} \int \frac {e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=-\frac {5 e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3}{64 \left (1+x+\log \left (x^2\right )\right )^3} \]
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Time = 1.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(5 \,{\mathrm e}^{-3 \ln \left (-\frac {4 \left (\ln \left (x^{2}\right )+x +1\right )}{\ln \left (x^{2}\right )+1}\right )+3 x +2}\) | \(29\) |
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Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=5 \, e^{\left (3 \, x - 3 \, \log \left (-\frac {4 \, {\left (x + \log \left (x^{2}\right ) + 1\right )}}{\log \left (x^{2}\right ) + 1}\right ) + 2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.45 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=\frac {\left (- 5 \log {\left (x^{2} \right )}^{3} - 15 \log {\left (x^{2} \right )}^{2} - 15 \log {\left (x^{2} \right )} - 5\right ) e^{3 x + 2}}{64 x^{3} + 192 x^{2} \log {\left (x^{2} \right )} + 192 x^{2} + 192 x \log {\left (x^{2} \right )}^{2} + 384 x \log {\left (x^{2} \right )} + 192 x + 64 \log {\left (x^{2} \right )}^{3} + 192 \log {\left (x^{2} \right )}^{2} + 192 \log {\left (x^{2} \right )} + 64} \]
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Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (24) = 48\).
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=-\frac {5 \, {\left (8 \, e^{2} \log \left (x\right )^{3} + 12 \, e^{2} \log \left (x\right )^{2} + 6 \, e^{2} \log \left (x\right ) + e^{2}\right )} e^{\left (3 \, x\right )}}{64 \, {\left (x^{3} + 12 \, {\left (x + 1\right )} \log \left (x\right )^{2} + 8 \, \log \left (x\right )^{3} + 3 \, x^{2} + 6 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right ) + 3 \, x + 1\right )}} \]
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Time = 0.54 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=5 \, e^{\left (3 \, x - 3 \, \log \left (-\frac {4 \, x}{\log \left (x^{2}\right ) + 1} - \frac {4 \, \log \left (x^{2}\right )}{\log \left (x^{2}\right ) + 1} - \frac {4}{\log \left (x^{2}\right ) + 1}\right ) + 2\right )} \]
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Timed out. \[ \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx=\int \frac {{\mathrm {e}}^{3\,x-3\,\ln \left (-\frac {4\,x+4\,\ln \left (x^2\right )+4}{\ln \left (x^2\right )+1}\right )+2}\,\left (15\,{\ln \left (x^2\right )}^2+\left (15\,x+15\right )\,\ln \left (x^2\right )+15\,x+30\right )}{{\ln \left (x^2\right )}^2+\left (x+2\right )\,\ln \left (x^2\right )+x+1} \,d x \]
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