Optimal. Leaf size=30 \[ \frac {\left (6 \left (2+e^{x/3}+\frac {5 x}{6}\right )-\log (x)\right )^2}{4 x^2} \]
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Rubi [B] time = 0.65, antiderivative size = 117, normalized size of antiderivative = 3.90, number of steps used = 28, number of rules used = 13, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {12, 14, 2197, 6742, 37, 2334, 43, 2305, 2304, 2199, 2177, 2178, 2554} \begin {gather*} \frac {13 (5 x+12)^2}{48 x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {3}{x^2}+\frac {\log ^2(x)}{4 x^2}-\frac {(x+5)^2 \log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}+\frac {\log (x)}{4 x^2}+\frac {15 e^{x/3}}{x}-\frac {5}{2 x}+\frac {\log (x)}{4} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 37
Rule 43
Rule 2177
Rule 2178
Rule 2197
Rule 2199
Rule 2304
Rule 2305
Rule 2334
Rule 2554
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-156-65 x+e^{2 x/3} (-36+12 x)+e^{x/3} \left (-150-6 x+10 x^2\right )+\left (25+e^{x/3} (12-2 x)+5 x\right ) \log (x)-\log ^2(x)}{x^3} \, dx\\ &=\frac {1}{2} \int \left (\frac {12 e^{2 x/3} (-3+x)}{x^3}+\frac {(12+5 x-\log (x)) (-13+\log (x))}{x^3}+\frac {2 e^{x/3} \left (-75-3 x+5 x^2+6 \log (x)-x \log (x)\right )}{x^3}\right ) \, dx\\ &=\frac {1}{2} \int \frac {(12+5 x-\log (x)) (-13+\log (x))}{x^3} \, dx+6 \int \frac {e^{2 x/3} (-3+x)}{x^3} \, dx+\int \frac {e^{x/3} \left (-75-3 x+5 x^2+6 \log (x)-x \log (x)\right )}{x^3} \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}+\frac {1}{2} \int \left (-\frac {13 (12+5 x)}{x^3}+\frac {5 (5+x) \log (x)}{x^3}-\frac {\log ^2(x)}{x^3}\right ) \, dx+\int \left (\frac {e^{x/3} \left (-75-3 x+5 x^2\right )}{x^3}-\frac {e^{x/3} (-6+x) \log (x)}{x^3}\right ) \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}-\frac {1}{2} \int \frac {\log ^2(x)}{x^3} \, dx+\frac {5}{2} \int \frac {(5+x) \log (x)}{x^3} \, dx-\frac {13}{2} \int \frac {12+5 x}{x^3} \, dx+\int \frac {e^{x/3} \left (-75-3 x+5 x^2\right )}{x^3} \, dx-\int \frac {e^{x/3} (-6+x) \log (x)}{x^3} \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}+\frac {13 (12+5 x)^2}{48 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}-\frac {1}{2} \int \frac {\log (x)}{x^3} \, dx-\frac {5}{2} \int -\frac {(5+x)^2}{10 x^3} \, dx+\int \left (-\frac {75 e^{x/3}}{x^3}-\frac {3 e^{x/3}}{x^2}+\frac {5 e^{x/3}}{x}\right ) \, dx+\int \frac {3 e^{x/3}}{x^3} \, dx\\ &=\frac {1}{8 x^2}+\frac {9 e^{2 x/3}}{x^2}+\frac {13 (12+5 x)^2}{48 x^2}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{4} \int \frac {(5+x)^2}{x^3} \, dx+3 \int \frac {e^{x/3}}{x^3} \, dx-3 \int \frac {e^{x/3}}{x^2} \, dx+5 \int \frac {e^{x/3}}{x} \, dx-75 \int \frac {e^{x/3}}{x^3} \, dx\\ &=\frac {1}{8 x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}+\frac {3 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+5 \text {Ei}\left (\frac {x}{3}\right )+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{4} \int \left (\frac {25}{x^3}+\frac {10}{x^2}+\frac {1}{x}\right ) \, dx+\frac {1}{2} \int \frac {e^{x/3}}{x^2} \, dx-\frac {25}{2} \int \frac {e^{x/3}}{x^2} \, dx-\int \frac {e^{x/3}}{x} \, dx\\ &=-\frac {3}{x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {5}{2 x}+\frac {15 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+4 \text {Ei}\left (\frac {x}{3}\right )+\frac {\log (x)}{4}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{6} \int \frac {e^{x/3}}{x} \, dx-\frac {25}{6} \int \frac {e^{x/3}}{x} \, dx\\ &=-\frac {3}{x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {5}{2 x}+\frac {15 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+\frac {\log (x)}{4}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 41, normalized size = 1.37 \begin {gather*} \frac {\left (6 \left (2+e^{x/3}\right )-\log (x)\right ) \left (12+6 e^{x/3}+10 x-\log (x)\right )}{4 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 46, normalized size = 1.53 \begin {gather*} \frac {12 \, {\left (5 \, x + 12\right )} e^{\left (\frac {1}{3} \, x\right )} - 2 \, {\left (5 \, x + 6 \, e^{\left (\frac {1}{3} \, x\right )} + 12\right )} \log \relax (x) + \log \relax (x)^{2} + 120 \, x + 36 \, e^{\left (\frac {2}{3} \, x\right )} + 144}{4 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {12 \, {\left (x - 3\right )} e^{\left (\frac {2}{3} \, x\right )} + 2 \, {\left (5 \, x^{2} - 3 \, x - 75\right )} e^{\left (\frac {1}{3} \, x\right )} - {\left (2 \, {\left (x - 6\right )} e^{\left (\frac {1}{3} \, x\right )} - 5 \, x - 25\right )} \log \relax (x) - \log \relax (x)^{2} - 65 \, x - 156}{2 \, x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 58, normalized size = 1.93
method | result | size |
risch | \(\frac {\ln \relax (x )^{2}}{4 x^{2}}-\frac {\left (5 x +12+6 \,{\mathrm e}^{\frac {x}{3}}\right ) \ln \relax (x )}{2 x^{2}}+\frac {15 x \,{\mathrm e}^{\frac {x}{3}}+9 \,{\mathrm e}^{\frac {2 x}{3}}+30 x +36 \,{\mathrm e}^{\frac {x}{3}}+36}{x^{2}}\) | \(58\) |
default | \(\frac {30 x \,{\mathrm e}^{\frac {x}{3}}-6 \ln \relax (x ) {\mathrm e}^{\frac {x}{3}}+72 \,{\mathrm e}^{\frac {x}{3}}}{2 x^{2}}+\frac {\ln \relax (x )^{2}}{4 x^{2}}-\frac {6 \ln \relax (x )}{x^{2}}+\frac {36}{x^{2}}+\frac {30}{x}-\frac {5 \ln \relax (x )}{2 x}+\frac {9 \,{\mathrm e}^{\frac {2 x}{3}}}{x^{2}}\) | \(73\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {5 \, \log \relax (x)}{2 \, x} - \frac {24 \, e^{\left (\frac {1}{3} \, x\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) - 1}{8 \, x^{2}} + \frac {30}{x} - \frac {25 \, \log \relax (x)}{4 \, x^{2}} + \frac {287}{8 \, x^{2}} + 5 \, {\rm Ei}\left (\frac {1}{3} \, x\right ) - \Gamma \left (-1, -\frac {1}{3} \, x\right ) + 4 \, \Gamma \left (-1, -\frac {2}{3} \, x\right ) + \frac {25}{3} \, \Gamma \left (-2, -\frac {1}{3} \, x\right ) + 8 \, \Gamma \left (-2, -\frac {2}{3} \, x\right ) + 3 \, \int \frac {e^{\left (\frac {1}{3} \, x\right )}}{x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.69, size = 32, normalized size = 1.07 \begin {gather*} \frac {\left (6\,{\mathrm {e}}^{x/3}-\ln \relax (x)+12\right )\,\left (10\,x+6\,{\mathrm {e}}^{x/3}-\ln \relax (x)+12\right )}{4\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.42, size = 73, normalized size = 2.43 \begin {gather*} - \frac {- 30 x - 36}{x^{2}} + \frac {\left (- 5 x - 12\right ) \log {\relax (x )}}{2 x^{2}} + \frac {\log {\relax (x )}^{2}}{4 x^{2}} + \frac {9 x^{2} e^{\frac {2 x}{3}} + \left (15 x^{3} - 3 x^{2} \log {\relax (x )} + 36 x^{2}\right ) e^{\frac {x}{3}}}{x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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