Integrand size = 148, antiderivative size = 31 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\frac {1}{x}+\log (x)+\log \left (x^2 \left (-e^x+x\right )^2-\log \left (x+5 x^2\right )\right ) \]
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\[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-x-10 x^2-x^4+25 x^6-e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )-e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )-\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{x^2 (1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx \\ & = \int \left (\frac {-1+3 x+2 x^2}{x^2}-\frac {1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))}{x (1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}\right ) \, dx \\ & = \int \frac {-1+3 x+2 x^2}{x^2} \, dx-\int \frac {1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))}{x (1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx \\ & = \int \left (2-\frac {1}{x^2}+\frac {3}{x}\right ) \, dx-\int \left (\frac {1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {5 \left (1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))\right )}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}\right ) \, dx \\ & = \frac {1}{x}+2 x+3 \log (x)+5 \int \frac {1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-\int \frac {1+10 x+2 e^x x^3-2 x^4+8 e^x x^4-8 x^5-10 e^x x^5+10 x^6-2 \log (x (1+5 x))-12 x \log (x (1+5 x))-10 x^2 \log (x (1+5 x))}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx \\ & = \frac {1}{x}+2 x+3 \log (x)+5 \int \left (\frac {1}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}+\frac {10 x}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}+\frac {2 e^x x^3}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {2 x^4}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}+\frac {8 e^x x^4}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {8 x^5}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {10 e^x x^5}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}+\frac {10 x^6}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {2 \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {12 x \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {10 x^2 \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}\right ) \, dx-\int \left (\frac {10}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}+\frac {1}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}+\frac {2 e^x x^2}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}-\frac {2 x^3}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}+\frac {8 e^x x^3}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}-\frac {8 x^4}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}-\frac {10 e^x x^4}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}+\frac {10 x^5}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}-\frac {12 \log (x (1+5 x))}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}-\frac {2 \log (x (1+5 x))}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )}-\frac {10 x \log (x (1+5 x))}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))}\right ) \, dx \\ & = \frac {1}{x}+2 x+3 \log (x)-2 \int \frac {e^x x^2}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+2 \int \frac {x^3}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+2 \int \frac {\log (x (1+5 x))}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx+5 \int \frac {1}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-8 \int \frac {e^x x^3}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+8 \int \frac {x^4}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx-10 \int \frac {1}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+10 \int \frac {e^x x^4}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx-10 \int \frac {x^5}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+10 \int \frac {e^x x^3}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-10 \int \frac {x^4}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx+10 \int \frac {x \log (x (1+5 x))}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx-10 \int \frac {\log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx+12 \int \frac {\log (x (1+5 x))}{e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))} \, dx+40 \int \frac {e^x x^4}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-40 \int \frac {x^5}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx+50 \int \frac {x}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-50 \int \frac {e^x x^5}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx+50 \int \frac {x^6}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-50 \int \frac {x^2 \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-60 \int \frac {x \log (x (1+5 x))}{(1+5 x) \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx-\int \frac {1}{x \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right )} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\frac {1}{x}+\log (x)+\log \left (e^{2 x} x^2-2 e^x x^3+x^4-\log (x (1+5 x))\right ) \]
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Time = 2.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(\frac {25+25 x \ln \left (x \right )+25 \ln \left ({\mathrm e}^{2 x} x^{2}-2 \,{\mathrm e}^{x} x^{3}+x^{4}-\ln \left (5 x^{2}+x \right )\right ) x}{25 x}\) | \(46\) |
risch | \(\frac {x \ln \left (x \right )+1}{x}+\ln \left (\ln \left (\frac {1}{5}+x \right )-\frac {i \left (-2 i x^{4}+4 i x^{3} {\mathrm e}^{x}-2 i x^{2} {\mathrm e}^{2 x}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{3}+2 i \ln \left (x \right )\right )}{2}\right )\) | \(123\) |
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Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\frac {x \log \left (-x^{4} + 2 \, x^{3} e^{x} - x^{2} e^{\left (2 \, x\right )} + \log \left (5 \, x^{2} + x\right )\right ) + x \log \left (x\right ) + 1}{x} \]
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Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=3 \log {\left (x \right )} + \log {\left (- 2 x e^{x} + e^{2 x} + \frac {x^{4} - \log {\left (5 x^{2} + x \right )}}{x^{2}} \right )} + \frac {1}{x} \]
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Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\frac {1}{x} + \log \left (-x^{4} + 2 \, x^{3} e^{x} - x^{2} e^{\left (2 \, x\right )} + \log \left (5 \, x + 1\right ) + \log \left (x\right )\right ) + \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\frac {x \log \left (-x^{4} + 2 \, x^{3} e^{x} - x^{2} e^{\left (2 \, x\right )} + \log \left (5 \, x^{2} + x\right )\right ) + x \log \left (x\right ) + 1}{x} \]
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Timed out. \[ \int \frac {x+10 x^2+x^4-25 x^6+e^{2 x} \left (x^2+2 x^3-17 x^4-10 x^5\right )+e^x \left (-2 x^3-2 x^4+42 x^5+10 x^6\right )+\left (-1-4 x+5 x^2\right ) \log \left (x+5 x^2\right )}{-x^6-5 x^7+e^{2 x} \left (-x^4-5 x^5\right )+e^x \left (2 x^5+10 x^6\right )+\left (x^2+5 x^3\right ) \log \left (x+5 x^2\right )} \, dx=\int -\frac {x-\ln \left (5\,x^2+x\right )\,\left (-5\,x^2+4\,x+1\right )+{\mathrm {e}}^{2\,x}\,\left (-10\,x^5-17\,x^4+2\,x^3+x^2\right )-{\mathrm {e}}^x\,\left (-10\,x^6-42\,x^5+2\,x^4+2\,x^3\right )+10\,x^2+x^4-25\,x^6}{{\mathrm {e}}^{2\,x}\,\left (5\,x^5+x^4\right )-{\mathrm {e}}^x\,\left (10\,x^6+2\,x^5\right )-\ln \left (5\,x^2+x\right )\,\left (5\,x^3+x^2\right )+x^6+5\,x^7} \,d x \]
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