Integrand size = 59, antiderivative size = 26 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=-x+\frac {4+e^x+\left (-1+e^5\right )^{e^{8 x}}}{4+x} \]
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Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(26)=52\).
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {27, 6874, 45, 2228, 2326} \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=-x+\frac {e^{-8 x} \left (e^5-1\right )^{e^{8 x}} \left (e^{8 x} x+4 e^{8 x}\right )}{(x+4)^2}+\frac {e^x}{x+4}+\frac {4}{x+4} \]
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Rule 27
Rule 45
Rule 2228
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{(4+x)^2} \, dx \\ & = \int \left (-\frac {20}{(4+x)^2}-\frac {8 x}{(4+x)^2}-\frac {x^2}{(4+x)^2}+\frac {e^x (3+x)}{(4+x)^2}+\frac {\left (-1+e^5\right )^{e^{8 x}} \left (-1+32 e^{8 x} \log \left (-1+e^5\right )+8 e^{8 x} x \log \left (-1+e^5\right )\right )}{(4+x)^2}\right ) \, dx \\ & = \frac {20}{4+x}-8 \int \frac {x}{(4+x)^2} \, dx-\int \frac {x^2}{(4+x)^2} \, dx+\int \frac {e^x (3+x)}{(4+x)^2} \, dx+\int \frac {\left (-1+e^5\right )^{e^{8 x}} \left (-1+32 e^{8 x} \log \left (-1+e^5\right )+8 e^{8 x} x \log \left (-1+e^5\right )\right )}{(4+x)^2} \, dx \\ & = \frac {20}{4+x}+\frac {e^x}{4+x}+\frac {e^{-8 x} \left (-1+e^5\right )^{e^{8 x}} \left (4 e^{8 x}+e^{8 x} x\right )}{(4+x)^2}-8 \int \left (-\frac {4}{(4+x)^2}+\frac {1}{4+x}\right ) \, dx-\int \left (1+\frac {16}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx \\ & = -x+\frac {4}{4+x}+\frac {e^x}{4+x}+\frac {e^{-8 x} \left (-1+e^5\right )^{e^{8 x}} \left (4 e^{8 x}+e^{8 x} x\right )}{(4+x)^2} \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=\frac {4+e^x+\left (-1+e^5\right )^{e^{8 x}}-4 x-x^2}{4+x} \]
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Time = 0.96 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(-\frac {x^{2}-20-{\mathrm e}^{x}-{\mathrm e}^{{\mathrm e}^{8 x} \ln \left ({\mathrm e}^{5}-1\right )}}{4+x}\) | \(32\) |
risch | \(\frac {4}{4+x}-x +\frac {{\mathrm e}^{x}}{4+x}+\frac {\left ({\mathrm e}^{5}-1\right )^{{\mathrm e}^{8 x}}}{4+x}\) | \(35\) |
parts | \(\frac {x \,{\mathrm e}^{{\mathrm e}^{8 x} \ln \left ({\mathrm e}^{5}-1\right )}+4 \,{\mathrm e}^{{\mathrm e}^{8 x} \ln \left ({\mathrm e}^{5}-1\right )}}{\left (4+x \right )^{2}}+\frac {{\mathrm e}^{x}}{4+x}+\frac {4}{4+x}-x\) | \(57\) |
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=-\frac {x^{2} + 4 \, x - {\left (e^{5} - 1\right )}^{e^{\left (8 \, x\right )}} - e^{x} - 4}{x + 4} \]
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Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=- x + \frac {e^{x}}{x + 4} + \frac {e^{e^{8 x} \log {\left (-1 + e^{5} \right )}}}{x + 4} + \frac {4}{x + 4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (23) = 46\).
Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=-\frac {x^{2} + 4 \, x - e^{\left (e^{\left (8 \, x\right )} \log \left (e^{4} + e^{3} + e^{2} + e + 1\right ) + e^{\left (8 \, x\right )} \log \left (e - 1\right )\right )} - e^{x} + 16}{x + 4} + \frac {20}{x + 4} \]
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\[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=\int { -\frac {x^{2} - {\left (8 \, {\left (x + 4\right )} e^{\left (8 \, x\right )} \log \left (e^{5} - 1\right ) - 1\right )} {\left (e^{5} - 1\right )}^{e^{\left (8 \, x\right )}} - {\left (x + 3\right )} e^{x} + 8 \, x + 20}{x^{2} + 8 \, x + 16} \,d x } \]
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Time = 8.62 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{16+8 x+x^2} \, dx=\frac {4}{x+4}-x+\frac {{\mathrm {e}}^x}{x+4}+\frac {{\left ({\mathrm {e}}^5-1\right )}^{{\mathrm {e}}^{8\,x}}}{x+4} \]
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