Integrand size = 33, antiderivative size = 25 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {5 e^{3+x} (-4+2 x)}{x^2 \left (3+9 \log ^2(4)\right )} \]
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Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6, 12, 2230, 2208, 2209} \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {10 e^{x+3}}{3 x \left (1+3 \log ^2(4)\right )}-\frac {20 e^{x+3}}{3 x^2 \left (1+3 \log ^2(4)\right )} \]
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Rule 6
Rule 12
Rule 2208
Rule 2209
Rule 2230
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{x^3 \left (3+9 \log ^2(4)\right )} \, dx \\ & = \frac {\int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{x^3} \, dx}{3 \left (1+3 \log ^2(4)\right )} \\ & = \frac {\int \left (\frac {40 e^{3+x}}{x^3}-\frac {30 e^{3+x}}{x^2}+\frac {10 e^{3+x}}{x}\right ) \, dx}{3 \left (1+3 \log ^2(4)\right )} \\ & = \frac {10 \int \frac {e^{3+x}}{x} \, dx}{3 \left (1+3 \log ^2(4)\right )}-\frac {10 \int \frac {e^{3+x}}{x^2} \, dx}{1+3 \log ^2(4)}+\frac {40 \int \frac {e^{3+x}}{x^3} \, dx}{3 \left (1+3 \log ^2(4)\right )} \\ & = -\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{x \left (1+3 \log ^2(4)\right )}+\frac {10 e^3 \text {Ei}(x)}{3 \left (1+3 \log ^2(4)\right )}+\frac {20 \int \frac {e^{3+x}}{x^2} \, dx}{3 \left (1+3 \log ^2(4)\right )}-\frac {10 \int \frac {e^{3+x}}{x} \, dx}{1+3 \log ^2(4)} \\ & = -\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{3 x \left (1+3 \log ^2(4)\right )}-\frac {20 e^3 \text {Ei}(x)}{3 \left (1+3 \log ^2(4)\right )}+\frac {20 \int \frac {e^{3+x}}{x} \, dx}{3 \left (1+3 \log ^2(4)\right )} \\ & = -\frac {20 e^{3+x}}{3 x^2 \left (1+3 \log ^2(4)\right )}+\frac {10 e^{3+x}}{3 x \left (1+3 \log ^2(4)\right )} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {10 e^{3+x} (-2+x)}{3 x^2 \left (1+3 \log ^2(4)\right )} \]
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Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
gosper | \(\frac {10 \left (-2+x \right ) {\mathrm e}^{3+x}}{3 x^{2} \left (12 \ln \left (2\right )^{2}+1\right )}\) | \(23\) |
risch | \(\frac {10 \left (-2+x \right ) {\mathrm e}^{3+x}}{3 x^{2} \left (12 \ln \left (2\right )^{2}+1\right )}\) | \(23\) |
parallelrisch | \(\frac {{\mathrm e}^{3} \left (10 \,{\mathrm e}^{x} x -20 \,{\mathrm e}^{x}\right )}{3 \left (12 \ln \left (2\right )^{2}+1\right ) x^{2}}\) | \(28\) |
norman | \(\frac {-\frac {20 \,{\mathrm e}^{3} {\mathrm e}^{x}}{3 \left (12 \ln \left (2\right )^{2}+1\right )}+\frac {10 \,{\mathrm e}^{3} x \,{\mathrm e}^{x}}{3 \left (12 \ln \left (2\right )^{2}+1\right )}}{x^{2}}\) | \(39\) |
default | \(\frac {10 \,{\mathrm e}^{3} \left (-\frac {\operatorname {Ei}_{1}\left (-x \right )}{12 \ln \left (2\right )^{2}+1}+\frac {-\frac {2 \,{\mathrm e}^{x}}{x^{2}}-\frac {2 \,{\mathrm e}^{x}}{x}-2 \,\operatorname {Ei}_{1}\left (-x \right )}{12 \ln \left (2\right )^{2}+1}-\frac {3 \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{12 \ln \left (2\right )^{2}+1}\right )}{3}\) | \(84\) |
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {10 \, {\left (x - 2\right )} e^{\left (x + 3\right )}}{3 \, {\left (12 \, x^{2} \log \left (2\right )^{2} + x^{2}\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {\left (10 x e^{3} - 20 e^{3}\right ) e^{x}}{3 x^{2} + 36 x^{2} \log {\left (2 \right )}^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {10 \, {\rm Ei}\left (x\right ) e^{3}}{3 \, {\left (12 \, \log \left (2\right )^{2} + 1\right )}} - \frac {10 \, e^{3} \Gamma \left (-1, -x\right )}{12 \, \log \left (2\right )^{2} + 1} - \frac {40 \, e^{3} \Gamma \left (-2, -x\right )}{3 \, {\left (12 \, \log \left (2\right )^{2} + 1\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=\frac {10 \, {\left (x e^{\left (x + 3\right )} - 2 \, e^{\left (x + 3\right )}\right )}}{3 \, {\left (12 \, x^{2} \log \left (2\right )^{2} + x^{2}\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{3+x} \left (40-30 x+10 x^2\right )}{3 x^3+9 x^3 \log ^2(4)} \, dx=-\frac {20\,{\mathrm {e}}^{x+3}-10\,x\,{\mathrm {e}}^{x+3}}{3\,x^2\,\left (12\,{\ln \left (2\right )}^2+1\right )} \]
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