Integrand size = 134, antiderivative size = 26 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {(-5-x)^2 x^6}{\log \left (-e^3+e^x-x\right )} \]
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\[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x^5 (5+x) \left (\left (-1+e^x\right ) x (5+x)-2 \left (-e^3+e^x-x\right ) (15+4 x) \log \left (-e^3+e^x-x\right )\right )}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = \int \left (-\frac {x^6 (5+x)^2 \left (-1+e^3+x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )}-\frac {x^5 (5+x) \left (5 x+x^2-30 \log \left (-e^3+e^x-x\right )-8 x \log \left (-e^3+e^x-x\right )\right )}{\log ^2\left (-e^3+e^x-x\right )}\right ) \, dx \\ & = -\int \frac {x^6 (5+x)^2 \left (-1+e^3+x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \frac {x^5 (5+x) \left (5 x+x^2-30 \log \left (-e^3+e^x-x\right )-8 x \log \left (-e^3+e^x-x\right )\right )}{\log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = -\int \left (\frac {x^9}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )}-\frac {25 \left (-1+e^3\right ) x^6}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )}-\frac {5 \left (3+2 e^3\right ) x^7}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )}-\frac {\left (9+e^3\right ) x^8}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )}\right ) \, dx-\int \frac {x^5 (5+x) \left (x (5+x)-2 (15+4 x) \log \left (-e^3+e^x-x\right )\right )}{\log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = -\left (\left (-9-e^3\right ) \int \frac {x^8}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx\right )-\left (25 \left (1-e^3\right )\right ) \int \frac {x^6}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx+\left (5 \left (3+2 e^3\right )\right ) \int \frac {x^7}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \left (\frac {x^6 (5+x)^2}{\log ^2\left (-e^3+e^x-x\right )}-\frac {2 x^5 (5+x) (15+4 x)}{\log \left (-e^3+e^x-x\right )}\right ) \, dx-\int \frac {x^9}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = 2 \int \frac {x^5 (5+x) (15+4 x)}{\log \left (-e^3+e^x-x\right )} \, dx-\left (-9-e^3\right ) \int \frac {x^8}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\left (25 \left (1-e^3\right )\right ) \int \frac {x^6}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx+\left (5 \left (3+2 e^3\right )\right ) \int \frac {x^7}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \frac {x^9}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \frac {x^6 (5+x)^2}{\log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = 2 \int \left (\frac {75 x^5}{\log \left (-e^3+e^x-x\right )}+\frac {35 x^6}{\log \left (-e^3+e^x-x\right )}+\frac {4 x^7}{\log \left (-e^3+e^x-x\right )}\right ) \, dx-\left (-9-e^3\right ) \int \frac {x^8}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\left (25 \left (1-e^3\right )\right ) \int \frac {x^6}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx+\left (5 \left (3+2 e^3\right )\right ) \int \frac {x^7}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \left (\frac {25 x^6}{\log ^2\left (-e^3+e^x-x\right )}+\frac {10 x^7}{\log ^2\left (-e^3+e^x-x\right )}+\frac {x^8}{\log ^2\left (-e^3+e^x-x\right )}\right ) \, dx-\int \frac {x^9}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx \\ & = 8 \int \frac {x^7}{\log \left (-e^3+e^x-x\right )} \, dx-10 \int \frac {x^7}{\log ^2\left (-e^3+e^x-x\right )} \, dx-25 \int \frac {x^6}{\log ^2\left (-e^3+e^x-x\right )} \, dx+70 \int \frac {x^6}{\log \left (-e^3+e^x-x\right )} \, dx+150 \int \frac {x^5}{\log \left (-e^3+e^x-x\right )} \, dx-\left (-9-e^3\right ) \int \frac {x^8}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\left (25 \left (1-e^3\right )\right ) \int \frac {x^6}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx+\left (5 \left (3+2 e^3\right )\right ) \int \frac {x^7}{\left (e^3-e^x+x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx-\int \frac {x^8}{\log ^2\left (-e^3+e^x-x\right )} \, dx-\int \frac {x^9}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx \\ \end{align*}
Time = 0.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^6 (5+x)^2}{\log \left (-e^3+e^x-x\right )} \]
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Time = 1.75 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x^{6} \left (x^{2}+10 x +25\right )}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{3}-x \right )}\) | \(26\) |
parallelrisch | \(\frac {x^{8}+10 x^{7}+25 x^{6}}{\ln \left ({\mathrm e}^{x}-{\mathrm e}^{3}-x \right )}\) | \(29\) |
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Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^{8} + 10 \, x^{7} + 25 \, x^{6}}{\log \left (-x - e^{3} + e^{x}\right )} \]
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Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^{8} + 10 x^{7} + 25 x^{6}}{\log {\left (- x + e^{x} - e^{3} \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^{8} + 10 \, x^{7} + 25 \, x^{6}}{\log \left (-x - e^{3} + e^{x}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^{8} + 10 \, x^{7} + 25 \, x^{6}}{\log \left (-x - e^{3} + e^{x}\right )} \]
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Time = 0.44 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.00 \[ \int \frac {25 x^6+10 x^7+x^8+e^x \left (-25 x^6-10 x^7-x^8\right )+\left (-150 x^6-70 x^7-8 x^8+e^3 \left (-150 x^5-70 x^6-8 x^7\right )+e^x \left (150 x^5+70 x^6+8 x^7\right )\right ) \log \left (-e^3+e^x-x\right )}{\left (-e^3+e^x-x\right ) \log ^2\left (-e^3+e^x-x\right )} \, dx=\frac {x^6\,{\left (x+5\right )}^2+\frac {2\,x^5\,\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^3-x\right )\,\left (4\,x^2+35\,x+75\right )\,\left (x+{\mathrm {e}}^3-{\mathrm {e}}^x\right )}{{\mathrm {e}}^x-1}}{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^3-x\right )}-\frac {150\,x^5\,{\mathrm {e}}^3+70\,x^6\,{\mathrm {e}}^3+8\,x^7\,{\mathrm {e}}^3-150\,x^5+80\,x^6+62\,x^7+8\,x^8}{{\mathrm {e}}^x-1}+150\,x^5+70\,x^6+8\,x^7 \]
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