Optimal. Leaf size=27 \[ 2 x+\log (5)-\log \left (x+9 e^{\frac {9}{16} (2-x)^2} x\right ) \]
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Rubi [F]
time = 2.25, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{8 x+72 e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} x} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{8 \left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {1}{8} \int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {1}{8} \int \left (\frac {8 (-1+2 x)}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}-\frac {9 e^{\frac {9}{16} (-2+x)^2} \left (8-34 x+9 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}\right ) \, dx\\ &=-\left (\frac {9}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} \left (8-34 x+9 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\right )+\int \frac {-1+2 x}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=-\left (\frac {9}{8} \int \left (-\frac {34 e^{\frac {9}{16} (-2+x)^2}}{1+9 e^{\frac {9}{16} (-2+x)^2}}+\frac {8 e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}+\frac {9 e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}}\right ) \, dx\right )+\int \left (\frac {2}{1+9 e^{\frac {9}{16} (-2+x)^2}}-\frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}\right ) \, dx\\ &=2 \int \frac {1}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \int \frac {e^{\frac {9}{16} (-2+x)^2}}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=2 \text {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \text {Subst}\left (\int \frac {e^{\frac {9 x^2}{16}}}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=2 \text {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \text {Subst}\left (\int \left (\frac {1}{9}-\frac {1}{9 \left (1+9 e^{\frac {9 x^2}{16}}\right )}\right ) \, dx,x,-2+x\right )-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {17 x}{4}+2 \text {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-\frac {17}{4} \text {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 2.22, size = 51, normalized size = 1.89 \begin {gather*} \frac {1}{8} \left (34 x-\frac {9 x^2}{2}+8 \log \left (e^{\frac {9}{16} (-2+x)^2}\right )-8 \log \left (1+9 e^{\frac {9}{16} (-2+x)^2}\right )-8 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.53, size = 23, normalized size = 0.85
method | result | size |
risch | \(2 x -\ln \left (x \right )+\frac {9}{4}-\ln \left ({\mathrm e}^{\frac {9 \left (x -2\right )^{2}}{16}}+\frac {1}{9}\right )\) | \(23\) |
norman | \(2 x -\ln \left (x \right )-\ln \left (9 \,{\mathrm e}^{\frac {9}{16} x^{2}-\frac {9}{4} x +\frac {9}{4}}+1\right )\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.31, size = 30, normalized size = 1.11 \begin {gather*} \frac {17}{4} \, x - \log \left (\frac {1}{9} \, {\left (9 \, e^{\left (\frac {9}{16} \, x^{2} + \frac {9}{4}\right )} + e^{\left (\frac {9}{4} \, x\right )}\right )} e^{\left (-\frac {9}{4}\right )}\right ) - \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 26, normalized size = 0.96 \begin {gather*} 2 \, x - \log \left (x\right ) - \log \left (9 \, e^{\left (\frac {9}{16} \, x^{2} - \frac {9}{4} \, x + \frac {9}{4}\right )} + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.18, size = 27, normalized size = 1.00 \begin {gather*} 2 x - \log {\left (x \right )} - \log {\left (e^{\frac {9 x^{2}}{16} - \frac {9 x}{4} + \frac {9}{4}} + \frac {1}{9} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 26, normalized size = 0.96 \begin {gather*} 2 \, x - \log \left (x\right ) - \log \left (9 \, e^{\left (\frac {9}{16} \, x^{2} - \frac {9}{4} \, x + \frac {9}{4}\right )} + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.15, size = 25, normalized size = 0.93 \begin {gather*} \frac {17\,x}{4}-\ln \left (x\,{\left ({\mathrm {e}}^x\right )}^{9/4}+9\,x\,{\left ({\mathrm {e}}^{x^2}\right )}^{9/16}\,{\mathrm {e}}^{9/4}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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