Optimal. Leaf size=19 \[ x+\frac {1}{\log \left (\frac {1+x}{2+\frac {1}{x}-7 x}\right )} \]
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Rubi [F] time = 1.70, 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 {1+2 x+9 x^2+\left (-x-3 x^2+5 x^3+7 x^4\right ) \log ^2\left (\frac {-x-x^2}{-1-2 x+7 x^2}\right )}{\left (-x-3 x^2+5 x^3+7 x^4\right ) \log ^2\left (\frac {-x-x^2}{-1-2 x+7 x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1-2 x-9 x^2-\left (-x-3 x^2+5 x^3+7 x^4\right ) \log ^2\left (\frac {-x-x^2}{-1-2 x+7 x^2}\right )}{x \left (1+3 x-5 x^2-7 x^3\right ) \log ^2\left (\frac {(-1-x) x}{-1-2 x+7 x^2}\right )} \, dx\\ &=\int \left (1+\frac {1+2 x+9 x^2}{x (1+x) \left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}\right ) \, dx\\ &=x+\int \frac {1+2 x+9 x^2}{x (1+x) \left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ &=x+\int \left (-\frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}-\frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}+\frac {2 (-1+7 x)}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}\right ) \, dx\\ &=x+2 \int \frac {-1+7 x}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ &=x+2 \int \left (-\frac {1}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}+\frac {7 x}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}\right ) \, dx-\int \frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ &=x-2 \int \frac {1}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx+14 \int \frac {x}{\left (-1-2 x+7 x^2\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ &=x-2 \int \left (-\frac {7}{2 \sqrt {2} \left (2+4 \sqrt {2}-14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}-\frac {7}{2 \sqrt {2} \left (-2+4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}\right ) \, dx+14 \int \left (\frac {1+\frac {1}{2 \sqrt {2}}}{\left (-2-4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}+\frac {1-\frac {1}{2 \sqrt {2}}}{\left (-2+4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )}\right ) \, dx-\int \frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ &=x+\frac {7 \int \frac {1}{\left (2+4 \sqrt {2}-14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx}{\sqrt {2}}+\frac {7 \int \frac {1}{\left (-2+4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx}{\sqrt {2}}+\frac {1}{2} \left (7 \left (4-\sqrt {2}\right )\right ) \int \frac {1}{\left (-2+4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx+\frac {1}{2} \left (7 \left (4+\sqrt {2}\right )\right ) \int \frac {1}{\left (-2-4 \sqrt {2}+14 x\right ) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{x \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx-\int \frac {1}{(1+x) \log ^2\left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 22, normalized size = 1.16 \begin {gather*} x+\frac {1}{\log \left (\frac {x (1+x)}{1+2 x-7 x^2}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.12, size = 47, normalized size = 2.47 \begin {gather*} \frac {x \log \left (-\frac {x^{2} + x}{7 \, x^{2} - 2 \, x - 1}\right ) + 1}{\log \left (-\frac {x^{2} + x}{7 \, x^{2} - 2 \, x - 1}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.60, size = 24, normalized size = 1.26 \begin {gather*} x + \frac {1}{\log \left (-\frac {x^{2} + x}{7 \, x^{2} - 2 \, x - 1}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 25, normalized size = 1.32
method | result | size |
default | \(x +\frac {1}{\ln \left (\frac {\left (-x -1\right ) x}{7 x^{2}-2 x -1}\right )}\) | \(25\) |
risch | \(x +\frac {1}{\ln \left (\frac {-x^{2}-x}{7 x^{2}-2 x -1}\right )}\) | \(28\) |
norman | \(\frac {1+x \ln \left (\frac {-x^{2}-x}{7 x^{2}-2 x -1}\right )}{\ln \left (\frac {-x^{2}-x}{7 x^{2}-2 x -1}\right )}\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 52, normalized size = 2.74 \begin {gather*} \frac {x \log \left (-7 \, x^{2} + 2 \, x + 1\right ) - x \log \left (x + 1\right ) - x \log \relax (x) - 1}{\log \left (-7 \, x^{2} + 2 \, x + 1\right ) - \log \left (x + 1\right ) - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.84, size = 23, normalized size = 1.21 \begin {gather*} x+\frac {1}{\ln \left (\frac {x^2+x}{-7\,x^2+2\,x+1}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.20, size = 20, normalized size = 1.05 \begin {gather*} x + \frac {1}{\log {\left (\frac {- x^{2} - x}{7 x^{2} - 2 x - 1} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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