Optimal. Leaf size=25 \[ -x+\log ^2\left (4-\frac {x}{2}-4 x \left (1-e^x+x\right )\right ) \]
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Rubi [F] time = 5.90, 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+9 x-8 e^x x+8 x^2+\left (-18-32 x+e^x (16+16 x)\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{8-9 x+8 e^x x-8 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-8-8 x+x^2+8 x^3\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x \left (8-9 x+8 e^x x-8 x^2\right )}+\frac {-x+2 \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )+2 x \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )}{x}\right ) \, dx\\ &=2 \int \frac {\left (-8-8 x+x^2+8 x^3\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x \left (8-9 x+8 e^x x-8 x^2\right )} \, dx+\int \frac {-x+2 \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )+2 x \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )}{x} \, dx\\ &=-\left (2 \int \frac {\left (9+16 x-8 e^x (1+x)\right ) \left (-8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx+\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{8+\left (-9+8 e^x\right ) x-8 x^2} \, dx\right )-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx+\int \left (-1+\frac {2 (1+x) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x}\right ) \, dx\\ &=-x+2 \int \frac {(1+x) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx-2 \int \left (\frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x}-\frac {\left (-8-8 x+x^2+8 x^3\right ) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )}\right ) \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 \int \left (\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )+\frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x}\right ) \, dx-2 \int \frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x} \, dx+2 \int \frac {\left (-8-8 x+x^2+8 x^3\right ) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 \int \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right ) \, dx+2 \int \frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx+2 \int \left (-\frac {8 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}-\frac {8 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )}+\frac {x \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}+\frac {8 x^2 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}\right ) \, dx-2 \int \left (\frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x}-\frac {8 (1+x) \int \frac {x^2}{-8+\left (9-8 e^x\right ) x+8 x^2} \, dx}{x}\right ) \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 x \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )-2 \int \frac {x \left (-9-16 x+8 e^x (1+x)\right )}{8-\left (9-8 e^x\right ) x-8 x^2} \, dx+2 \int \frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx-2 \int \frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x} \, dx+2 \int \frac {x \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2} \, dx-16 \int \frac {8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx}{-8+9 x-8 e^x x+8 x^2} \, dx-16 \int \frac {8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx+16 \int \frac {x^2 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2} \, dx+16 \int \frac {(1+x) \int \frac {x^2}{-8+\left (9-8 e^x\right ) x+8 x^2} \, dx}{x} \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} -x+\log ^2\left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 22, normalized size = 0.88 \begin {gather*} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right )^{2} - x \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 {8 \, x^{2} - 8 \, x e^{x} + 2 \, {\left (8 \, {\left (x + 1\right )} e^{x} - 16 \, x - 9\right )} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right ) + 9 \, x - 8}{8 \, x^{2} - 8 \, x e^{x} + 9 \, x - 8}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 23, normalized size = 0.92
method | result | size |
norman | \(\ln \left (4 \,{\mathrm e}^{x} x -4 x^{2}-\frac {9 x}{2}+4\right )^{2}-x\) | \(23\) |
risch | \(\ln \left (4 \,{\mathrm e}^{x} x -4 x^{2}-\frac {9 x}{2}+4\right )^{2}-x\) | \(23\) |
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*} -\int \frac {8 \, x^{2} - 8 \, x e^{x} + 2 \, {\left (8 \, {\left (x + 1\right )} e^{x} - 16 \, x - 9\right )} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right ) + 9 \, x - 8}{8 \, x^{2} - 8 \, x e^{x} + 9 \, x - 8}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.78, size = 22, normalized size = 0.88 \begin {gather*} {\ln \left (4\,x\,{\mathrm {e}}^x-\frac {9\,x}{2}-4\,x^2+4\right )}^2-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.44, size = 22, normalized size = 0.88 \begin {gather*} - x + \log {\left (- 4 x^{2} + 4 x e^{x} - \frac {9 x}{2} + 4 \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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