Optimal. Leaf size=29 \[ \left (2+e^{3 x}\right ) \left (\frac {1}{5} \left (-4+\frac {e^{2 x}}{x}\right )^2+x+\log (4)\right ) \]
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Rubi [B] time = 0.19, antiderivative size = 79, normalized size of antiderivative = 2.72, number of steps used = 9, number of rules used = 5, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 2197, 2176, 2194} \begin {gather*} \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)) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2176
Rule 2194
Rule 2197
Rubi steps
\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}
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Mathematica [B] time = 0.11, size = 62, normalized size = 2.14 \begin {gather*} \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 ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 63, normalized size = 2.17 \begin {gather*} \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 \relax (2) + 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 64, normalized size = 2.21 \begin {gather*} \frac {5 \, x^{3} e^{\left (3 \, x\right )} + 10 \, x^{2} e^{\left (3 \, x\right )} \log \relax (2) + 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 61, normalized size = 2.10
method | result | size |
default | \(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}+2 \ln \relax (2) {\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{2 x}}{5 x}\) | \(61\) |
risch | \(2 \ln \relax (2) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}+\frac {16 \,{\mathrm e}^{3 x}}{5}+2 x +\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{4 x}}{5 x^{2}}-\frac {8 \left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{2 x} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )}{2}}}{5 x}\) | \(143\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.41, size = 83, normalized size = 2.86 \begin {gather*} \frac {1}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 2 \, e^{\left (3 \, x\right )} \log \relax (2) + 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 ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.23, size = 53, normalized size = 1.83 \begin {gather*} \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 \relax (2)+\frac {16}{5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.24, size = 71, normalized size = 2.45 \begin {gather*} 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 {\relax (2 )} + 10000 x^{6}\right ) e^{3 x}}{3125 x^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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