Integrand size = 81, antiderivative size = 29 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\left (2+e^{3 x}\right ) \left (\frac {1}{5} \left (-4+\frac {e^{2 x}}{x}\right )^2+x+\log (4)\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(29)=58\).
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.72, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 2228, 2207, 2225} \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}+2 x-\frac {e^{3 x}}{3}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+\frac {1}{15} e^{3 x} (15 x+53+15 \log (4)) \]
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Rule 12
Rule 14
Rule 2207
Rule 2225
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{x} \, dx \\ & = \frac {1}{5} \int \left (10+\frac {4 e^{4 x} (-1+2 x)}{x^3}-\frac {16 e^{2 x} (-1+2 x)}{x^2}-\frac {8 e^{5 x} (-1+5 x)}{x^2}+\frac {e^{7 x} (-2+7 x)}{x^3}+e^{3 x} (53+15 x+15 \log (4))\right ) \, dx \\ & = 2 x+\frac {1}{5} \int \frac {e^{7 x} (-2+7 x)}{x^3} \, dx+\frac {1}{5} \int e^{3 x} (53+15 x+15 \log (4)) \, dx+\frac {4}{5} \int \frac {e^{4 x} (-1+2 x)}{x^3} \, dx-\frac {8}{5} \int \frac {e^{5 x} (-1+5 x)}{x^2} \, dx-\frac {16}{5} \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx \\ & = \frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+2 x+\frac {1}{15} e^{3 x} (53+15 x+15 \log (4))-\int e^{3 x} \, dx \\ & = -\frac {e^{3 x}}{3}+\frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+2 x+\frac {1}{15} e^{3 x} (53+15 x+15 \log (4)) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(29)=58\).
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {1}{5} \left (\frac {2 e^{4 x}}{x^2}+\frac {e^{7 x}}{x^2}-\frac {16 e^{2 x}}{x}-\frac {8 e^{5 x}}{x}+10 x+e^{3 x} (16+5 x+5 \log (4))\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(72\) vs. \(2(31)=62\).
Time = 2.56 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52
method | result | size |
default | \(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(73\) |
parts | \(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(73\) |
parallelrisch | \(\frac {{\mathrm e}^{3 x} {\mathrm e}^{4 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{3 x} {\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}+2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(93\) |
risch | \(2 \ln \left (2\right ) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}+\frac {16 \,{\mathrm e}^{3 x}}{5}+2 x +\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{4 x}}{5 x^{2}}-\frac {8 \left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{2 x} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}}}{5 x}\) | \(143\) |
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Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {{\left (10 \, x^{3} e^{\left (-7 \, x\right )} - 8 \, x e^{\left (-2 \, x\right )} + {\left (5 \, x^{3} + 10 \, x^{2} \log \left (2\right ) + 16 \, x^{2}\right )} e^{\left (-4 \, x\right )} - 16 \, x e^{\left (-5 \, x\right )} + 2 \, e^{\left (-3 \, x\right )} + 1\right )} e^{\left (7 \, x\right )}}{5 \, x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=2 x + \frac {- 5000 x^{5} e^{5 x} - 10000 x^{5} e^{2 x} + 625 x^{4} e^{7 x} + 1250 x^{4} e^{4 x} + \left (3125 x^{7} + 6250 x^{6} \log {\left (2 \right )} + 10000 x^{6}\right ) e^{3 x}}{3125 x^{6}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.86 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {1}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 2 \, e^{\left (3 \, x\right )} \log \left (2\right ) + 2 \, x - 8 \, {\rm Ei}\left (5 \, x\right ) - \frac {32}{5} \, {\rm Ei}\left (2 \, x\right ) + \frac {53}{15} \, e^{\left (3 \, x\right )} + \frac {32}{5} \, \Gamma \left (-1, -2 \, x\right ) + \frac {32}{5} \, \Gamma \left (-1, -4 \, x\right ) + 8 \, \Gamma \left (-1, -5 \, x\right ) + \frac {49}{5} \, \Gamma \left (-1, -7 \, x\right ) + \frac {64}{5} \, \Gamma \left (-2, -4 \, x\right ) + \frac {98}{5} \, \Gamma \left (-2, -7 \, x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {5 \, x^{3} e^{\left (3 \, x\right )} + 10 \, x^{2} e^{\left (3 \, x\right )} \log \left (2\right ) + 10 \, x^{3} + 16 \, x^{2} e^{\left (3 \, x\right )} - 8 \, x e^{\left (5 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + e^{\left (7 \, x\right )} + 2 \, e^{\left (4 \, x\right )}}{5 \, x^{2}} \]
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Time = 14.98 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {\frac {2\,{\mathrm {e}}^{4\,x}}{5}+\frac {{\mathrm {e}}^{7\,x}}{5}-x\,\left (\frac {16\,{\mathrm {e}}^{2\,x}}{5}+\frac {8\,{\mathrm {e}}^{5\,x}}{5}\right )}{x^2}+x\,\left ({\mathrm {e}}^{3\,x}+2\right )+{\mathrm {e}}^{3\,x}\,\left (2\,\ln \left (2\right )+\frac {16}{5}\right ) \]
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