Integrand size = 103, antiderivative size = 28 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=x^4 \left (-\frac {1}{5 x}-\log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right ) \]
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Time = 1.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6873, 12, 6820, 6874, 14, 2631, 45} \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=x^4 \left (-\log \left (x (x+5)^2-2 e^x x\right )\right )-\frac {x^3}{5} \]
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Rule 12
Rule 14
Rule 45
Rule 2631
Rule 6820
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-75 x^2-155 x^3-103 x^4-15 x^5-e^x \left (-6 x^2-10 x^3-10 x^4\right )-\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{5 \left (25-2 e^x+10 x+x^2\right )} \, dx \\ & = \frac {1}{5} \int \frac {-75 x^2-155 x^3-103 x^4-15 x^5-e^x \left (-6 x^2-10 x^3-10 x^4\right )-\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{25-2 e^x+10 x+x^2} \, dx \\ & = \frac {1}{5} \int \frac {x^2 \left (75+155 x+103 x^2+15 x^3-2 e^x \left (3+5 x+5 x^2\right )+20 x \left (-2 e^x+(5+x)^2\right ) \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right )}{2 e^x-(5+x)^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {5 x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2}-x^2 \left (3+5 x+5 x^2+20 x \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right )\right ) \, dx \\ & = -\left (\frac {1}{5} \int x^2 \left (3+5 x+5 x^2+20 x \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right ) \, dx\right )+\int \frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2} \, dx \\ & = -\left (\frac {1}{5} \int \left (x^2 \left (3+5 x+5 x^2\right )+20 x^3 \log \left (-2 e^x x+x (5+x)^2\right )\right ) \, dx\right )+\int \left (\frac {15 x^4}{25-2 e^x+10 x+x^2}+\frac {8 x^5}{25-2 e^x+10 x+x^2}+\frac {x^6}{25-2 e^x+10 x+x^2}\right ) \, dx \\ & = -\left (\frac {1}{5} \int x^2 \left (3+5 x+5 x^2\right ) \, dx\right )-4 \int x^3 \log \left (-2 e^x x+x (5+x)^2\right ) \, dx+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx \\ & = -x^4 \log \left (-2 e^x x+x (5+x)^2\right )-\frac {1}{5} \int \left (3 x^2+5 x^3+5 x^4\right ) \, dx+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^3 \left (-25-20 x-3 x^2+2 e^x (1+x)\right )}{2 e^x-(5+x)^2} \, dx \\ & = -\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \left (x^3 (1+x)-\frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2}\right ) \, dx \\ & = -\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int x^3 (1+x) \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx-\int \frac {x^4 \left (15+8 x+x^2\right )}{25-2 e^x+10 x+x^2} \, dx \\ & = -\frac {x^3}{5}-\frac {x^4}{4}-\frac {x^5}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right )+8 \int \frac {x^5}{25-2 e^x+10 x+x^2} \, dx+15 \int \frac {x^4}{25-2 e^x+10 x+x^2} \, dx+\int \frac {x^6}{25-2 e^x+10 x+x^2} \, dx+\int \left (x^3+x^4\right ) \, dx-\int \left (\frac {15 x^4}{25-2 e^x+10 x+x^2}+\frac {8 x^5}{25-2 e^x+10 x+x^2}+\frac {x^6}{25-2 e^x+10 x+x^2}\right ) \, dx \\ & = -\frac {x^3}{5}-x^4 \log \left (-2 e^x x+x (5+x)^2\right ) \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=\frac {1}{5} \left (-x^3-5 x^4 \log \left (x \left (-2 e^x+(5+x)^2\right )\right )\right ) \]
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Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\ln \left (-x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right ) x^{4}-\frac {x^{3}}{5}\) | \(30\) |
risch | \(-x^{4} \ln \left (x^{2}-2 \,{\mathrm e}^{x}+10 x +25\right )-x^{4} \ln \left (x \right )+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right ) \operatorname {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )}{2}-\frac {i \pi \,x^{4} \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )}^{2}}{2}+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right ) {\operatorname {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )}^{2}}{2}-\frac {i \pi \,x^{4} {\operatorname {csgn}\left (i x \left (-x^{2}+2 \,{\mathrm e}^{x}-10 x -25\right )\right )}^{3}}{2}-\frac {x^{3}}{5}\) | \(188\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=-x^{4} \log \left (x^{3} + 10 \, x^{2} - 2 \, x e^{x} + 25 \, x\right ) - \frac {1}{5} \, x^{3} \]
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Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=- x^{4} \log {\left (x^{3} + 10 x^{2} - 2 x e^{x} + 25 x \right )} - \frac {x^{3}}{5} \]
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Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=-x^{4} \log \left (x^{2} + 10 \, x - 2 \, e^{x} + 25\right ) - x^{4} \log \left (x\right ) - \frac {1}{5} \, x^{3} \]
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Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=-x^{4} \log \left (x^{3} + 10 \, x^{2} - 2 \, x e^{x} + 25 \, x\right ) - \frac {1}{5} \, x^{3} \]
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Time = 10.96 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {75 x^2+155 x^3+103 x^4+15 x^5+e^x \left (-6 x^2-10 x^3-10 x^4\right )+\left (500 x^3-40 e^x x^3+200 x^4+20 x^5\right ) \log \left (25 x-2 e^x x+10 x^2+x^3\right )}{-125+10 e^x-50 x-5 x^2} \, dx=-x^4\,\ln \left (25\,x-2\,x\,{\mathrm {e}}^x+10\,x^2+x^3\right )-\frac {x^3}{5} \]
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