Optimal. Leaf size=25 \[ \frac {8 \left (e^x+\frac {2}{3} \log ^2\left (\frac {5}{2 x}\right )\right )}{5 x} \]
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Rubi [B] time = 0.12, antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 14, 2197, 2304, 2366, 2303} \begin {gather*} \frac {8 e^x}{5 x}+\frac {16 \left (\log \left (\frac {5}{2 x}\right )+2\right ) \log \left (\frac {5}{2 x}\right )}{15 x}-\frac {32 \log \left (\frac {5}{2 x}\right )}{15 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2197
Rule 2303
Rule 2304
Rule 2366
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} \int \frac {e^x (-24+24 x)-32 \log \left (\frac {5}{2 x}\right )-16 \log ^2\left (\frac {5}{2 x}\right )}{x^2} \, dx\\ &=\frac {1}{15} \int \left (\frac {24 e^x (-1+x)}{x^2}-\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {16}{15} \int \frac {\log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{x^2} \, dx\right )+\frac {8}{5} \int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {8 e^x}{5 x}-\frac {16 \log \left (\frac {5}{2 x}\right )}{15 x}+\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{15 x}-\frac {16}{15} \int \frac {-1-\log \left (\frac {5}{2 x}\right )}{x^2} \, dx\\ &=\frac {8 e^x}{5 x}-\frac {32 \log \left (\frac {5}{2 x}\right )}{15 x}+\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{15 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 28, normalized size = 1.12 \begin {gather*} \frac {8}{15} \left (\frac {3 e^x}{x}+\frac {2 \log ^2\left (\frac {5}{2 x}\right )}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 \, {\left (2 \, \log \left (\frac {5}{2 \, x}\right )^{2} + 3 \, e^{x}\right )}}{15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 \, {\left (2 \, \log \left (\frac {5}{2 \, x}\right )^{2} + 3 \, e^{x}\right )}}{15 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 20, normalized size = 0.80
method | result | size |
norman | \(\frac {\frac {16 \ln \left (\frac {5}{2 x}\right )^{2}}{15}+\frac {8 \,{\mathrm e}^{x}}{5}}{x}\) | \(20\) |
default | \(\frac {16 \ln \left (\frac {5}{2 x}\right )^{2}}{15 x}+\frac {8 \,{\mathrm e}^{x}}{5 x}\) | \(22\) |
risch | \(\frac {16 \ln \relax (x )^{2}}{15 x}+\frac {16 \left (2 \ln \relax (2)-2 \ln \relax (5)\right ) \ln \relax (x )}{15 x}+\frac {\frac {16 \ln \relax (2)^{2}}{15}-\frac {32 \ln \relax (2) \ln \relax (5)}{15}+\frac {16 \ln \relax (5)^{2}}{15}+\frac {8 \,{\mathrm e}^{x}}{5}}{x}\) | \(55\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} \frac {16 \, \log \left (\frac {5}{2 \, x}\right )^{2}}{15 \, x} + \frac {8}{5} \, {\rm Ei}\relax (x) - \frac {8}{5} \, \Gamma \left (-1, -x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.92, size = 20, normalized size = 0.80 \begin {gather*} \frac {8\,\left (2\,{\ln \left (\frac {5}{2\,x}\right )}^2+3\,{\mathrm {e}}^x\right )}{15\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.30, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 e^{x}}{5 x} + \frac {16 \log {\left (\frac {5}{2 x} \right )}^{2}}{15 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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