Integrand size = 21, antiderivative size = 76 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\frac {4 \sqrt {-x+x^2}}{x}+4 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x} \]
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Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2617, 2615, 6874, 654, 634, 212, 676, 678, 748, 857, 738} \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=4 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )+\frac {4 \sqrt {x^2-x}}{x}-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x}+4 \log (x)-4 \log (8 x+1) \]
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Rule 212
Rule 634
Rule 654
Rule 676
Rule 678
Rule 738
Rule 748
Rule 857
Rule 2615
Rule 2617
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x^2} \, dx \\ & = -\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}-8 \int \frac {1}{x \left (-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )\right )} \, dx \\ & = -\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}-8 \int \left (-\frac {1}{2 x}+\frac {4}{1+8 x}-\frac {x}{12 \sqrt {-x+x^2}}+\frac {\sqrt {-x+x^2}}{4 x^2}-\frac {5 \sqrt {-x+x^2}}{4 x}+\frac {32 \sqrt {-x+x^2}}{3 (1+8 x)}\right ) \, dx \\ & = 4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}+\frac {2}{3} \int \frac {x}{\sqrt {-x+x^2}} \, dx-2 \int \frac {\sqrt {-x+x^2}}{x^2} \, dx+10 \int \frac {\sqrt {-x+x^2}}{x} \, dx-\frac {256}{3} \int \frac {\sqrt {-x+x^2}}{1+8 x} \, dx \\ & = \frac {4 \sqrt {-x+x^2}}{x}+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}+\frac {1}{3} \int \frac {1}{\sqrt {-x+x^2}} \, dx-2 \int \frac {1}{\sqrt {-x+x^2}} \, dx-5 \int \frac {1}{\sqrt {-x+x^2}} \, dx+\frac {16}{3} \int \frac {-1+10 x}{(1+8 x) \sqrt {-x+x^2}} \, dx \\ & = \frac {4 \sqrt {-x+x^2}}{x}+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}+\frac {2}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-4 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )+\frac {20}{3} \int \frac {1}{\sqrt {-x+x^2}} \, dx-10 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-12 \int \frac {1}{(1+8 x) \sqrt {-x+x^2}} \, dx \\ & = \frac {4 \sqrt {-x+x^2}}{x}-\frac {40}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x}+\frac {40}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )+24 \text {Subst}\left (\int \frac {1}{36-x^2} \, dx,x,\frac {1-10 x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {4 \sqrt {-x+x^2}}{x}+4 \tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.87 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=2 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )+\frac {4 \sqrt {(-1+x) x}+\frac {4 \sqrt {-(-1+x)^2 x^2} \arcsin \left (\sqrt {1-x}\right )}{-1+x}+4 x \log (x)-2 x \log (1+8 x)-4 x \log \left (1-4 x-4 \sqrt {(-1+x) x}\right )+4 x \log \left (1-2 x-2 \sqrt {(-1+x) x}\right )-\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x} \]
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Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.74
method | result | size |
parts | \(-\frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{x}+\frac {4 \sqrt {x^{2}-x}}{x}+4 \,\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )+10 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )-4 \ln \left (1+8 x \right )+4 \ln \left (x \right )-16 \sqrt {x^{2}-x}+2 \sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}-10 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )\) | \(132\) |
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Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.51 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=-\frac {7 \, x \log \left (8 \, x + 1\right ) + 2 \, {\left (x + 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 8 \, x \log \left (x\right ) + x \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + 7 \, x \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - 7 \, x \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) - 8 \, x - 8 \, \sqrt {x^{2} - x}}{2 \, x} \]
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\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int \frac {\log {\left (4 x + 4 \sqrt {x^{2} - x} - 1 \right )}}{x^{2}}\, dx \]
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\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int { \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{x^{2}} \,d x } \]
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Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=-\frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{x} + \frac {4}{x - \sqrt {x^{2} - x}} - 4 \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + 4 \, \log \left ({\left | x \right |}\right ) - 4 \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + 4 \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
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Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int \frac {\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right )}{x^2} \,d x \]
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