Integrand size = 17, antiderivative size = 95 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}-\frac {1}{16} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {7}{8} \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {2617, 2613, 6874, 654, 634, 212, 748, 857, 738} \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {1}{16} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )-\frac {7}{8} \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )-\frac {\sqrt {x^2-x}}{2}+x \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {x}{2}+\frac {1}{16} \log (8 x+1) \]
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Rule 212
Rule 634
Rule 654
Rule 738
Rule 748
Rule 857
Rule 2613
Rule 2617
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \, dx \\ & = x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+8 \int \frac {x}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx \\ & = x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+8 \int \left (-\frac {1}{16}+\frac {1}{16 (1+8 x)}-\frac {x}{12 \sqrt {-x+x^2}}+\frac {\sqrt {-x+x^2}}{6 (1+8 x)}\right ) \, dx \\ & = -\frac {x}{2}+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {2}{3} \int \frac {x}{\sqrt {-x+x^2}} \, dx+\frac {4}{3} \int \frac {\sqrt {-x+x^2}}{1+8 x} \, dx \\ & = -\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {1}{12} \int \frac {-1+10 x}{(1+8 x) \sqrt {-x+x^2}} \, dx-\frac {1}{3} \int \frac {1}{\sqrt {-x+x^2}} \, dx \\ & = -\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {5}{48} \int \frac {1}{\sqrt {-x+x^2}} \, dx+\frac {3}{16} \int \frac {1}{(1+8 x) \sqrt {-x+x^2}} \, dx-\frac {2}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right ) \\ & = -\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}-\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {5}{24} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-\frac {3}{8} \text {Subst}\left (\int \frac {1}{36-x^2} \, dx,x,\frac {1-10 x}{\sqrt {-x+x^2}}\right ) \\ & = -\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}-\frac {1}{16} \tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {7}{8} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{16} \left (-8 x-8 \sqrt {(-1+x) x}+2 \log (1+8 x)-7 \log \left (1-2 x-2 \sqrt {(-1+x) x}\right )+16 x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )-\log \left (1-10 x+6 \sqrt {(-1+x) x}\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84
method | result | size |
default | \(x \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )-\frac {7 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{16}-\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{16}-\frac {\sqrt {x^{2}-x}}{2}-\frac {x}{2}+\frac {\ln \left (1+8 x \right )}{16}\) | \(80\) |
parts | \(x \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )-\frac {19 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{32}-\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{16}+\frac {\sqrt {x^{2}-x}}{4}-x \sqrt {x^{2}-x}-\frac {x}{2}+\frac {\ln \left (1+8 x \right )}{16}+\frac {\left (2 x -1\right ) \sqrt {x^{2}-x}}{2}-\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{32}+\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{32}\) | \(142\) |
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Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx={\left (x + 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - x} - \frac {7}{16} \, \log \left (8 \, x + 1\right ) + \frac {15}{16} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) - \frac {7}{16} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) + \frac {7}{16} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]
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\[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int \log {\left (4 x + 4 \sqrt {x \left (x - 1\right )} - 1 \right )}\, dx \]
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\[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \]
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Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=x \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - x} + \frac {1}{16} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {7}{16} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) - \frac {1}{16} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
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Timed out. \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int \ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]
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