Optimal. Leaf size=118 \[ -\frac {2 \sqrt {x^2-x}}{\sqrt {x}}+2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {\sqrt {x^2-x} \tan ^{-1}\left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{\sqrt {2} \sqrt {x-1} \sqrt {x}}-2 \sqrt {x}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}} \]
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Rubi [A] time = 0.39, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2537, 2535, 6733, 6742, 203, 1588, 1146, 444, 50, 63} \[ -\frac {2 \sqrt {x^2-x}}{\sqrt {x}}+2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {\sqrt {x^2-x} \tan ^{-1}\left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{\sqrt {2} \sqrt {x-1} \sqrt {x}}-2 \sqrt {x}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 203
Rule 444
Rule 1146
Rule 1588
Rule 2535
Rule 2537
Rule 6733
Rule 6742
Rubi steps
\begin {align*} \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx &=\int \frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{\sqrt {x}} \, dx\\ &=2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+16 \int \frac {\sqrt {x}}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx\\ &=2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+32 \operatorname {Subst}\left (\int \frac {x^2}{-4 \left (1+2 x^2\right ) \sqrt {-x^2+x^4}+8 \left (-x^2+x^4\right )} \, dx,x,\sqrt {x}\right )\\ &=2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+32 \operatorname {Subst}\left (\int \left (-\frac {1}{16}+\frac {1}{16 \left (1+8 x^2\right )}-\frac {x^2}{12 \sqrt {-x^2+x^4}}+\frac {\sqrt {-x^2+x^4}}{6 \left (1+8 x^2\right )}\right ) \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {x}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+2 \operatorname {Subst}\left (\int \frac {1}{1+8 x^2} \, dx,x,\sqrt {x}\right )-\frac {8}{3} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {-x^2+x^4}} \, dx,x,\sqrt {x}\right )+\frac {16}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-x^2+x^4}}{1+8 x^2} \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {x}-\frac {8 \sqrt {-x+x^2}}{3 \sqrt {x}}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\left (16 \sqrt {-x+x^2}\right ) \operatorname {Subst}\left (\int \frac {x \sqrt {-1+x^2}}{1+8 x^2} \, dx,x,\sqrt {x}\right )}{3 \sqrt {-1+x} \sqrt {x}}\\ &=-2 \sqrt {x}-\frac {8 \sqrt {-x+x^2}}{3 \sqrt {x}}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\left (8 \sqrt {-x+x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{1+8 x} \, dx,x,x\right )}{3 \sqrt {-1+x} \sqrt {x}}\\ &=-2 \sqrt {x}-\frac {2 \sqrt {-x+x^2}}{\sqrt {x}}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {\left (3 \sqrt {-x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} (1+8 x)} \, dx,x,x\right )}{\sqrt {-1+x} \sqrt {x}}\\ &=-2 \sqrt {x}-\frac {2 \sqrt {-x+x^2}}{\sqrt {x}}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {\left (6 \sqrt {-x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{9+8 x^2} \, dx,x,\sqrt {-1+x}\right )}{\sqrt {-1+x} \sqrt {x}}\\ &=-2 \sqrt {x}-\frac {2 \sqrt {-x+x^2}}{\sqrt {x}}-\frac {\sqrt {-x+x^2} \tan ^{-1}\left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{\sqrt {2} \sqrt {-1+x} \sqrt {x}}+\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )\\ \end {align*}
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Mathematica [C] time = 0.57, size = 186, normalized size = 1.58 \[ \frac {1}{8} \left (-16 \sqrt {x}-\frac {16 \sqrt {(x-1) x}}{\sqrt {x}}-2 i \sqrt {2} \log \left (4 (8 x+1)^2\right )+i \sqrt {2} \log \left ((8 x+1) \left (-10 x-6 \sqrt {(x-1) x}+1\right )\right )+16 \sqrt {x} \log \left (4 x+4 \sqrt {(x-1) x}-1\right )+i \sqrt {2} \log \left ((8 x+1) \left (-10 x+6 \sqrt {(x-1) x}+1\right )\right )+4 \sqrt {2} \tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )-4 \sqrt {2} \tan ^{-1}\left (\frac {2 \sqrt {2} \sqrt {(x-1) x}}{3 \sqrt {x}}\right )\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.98, size = 84, normalized size = 0.71 \[ \frac {\sqrt {2} x \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) + \sqrt {2} x \arctan \left (\frac {3 \, \sqrt {2} \sqrt {x}}{4 \, \sqrt {x^{2} - x}}\right ) + 4 \, x^{\frac {3}{2}} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 4 \, x^{\frac {3}{2}} - 4 \, \sqrt {x^{2} - x} \sqrt {x}}{2 \, x} \]
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 \[ \int \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{\sqrt {x}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (4 x -1+4 \sqrt {\left (x -1\right ) x}\right )}{\sqrt {x}}\, dx \]
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 \[ 2 \, \sqrt {x} \log \left (4 \, \sqrt {x - 1} \sqrt {x} + 4 \, x - 1\right ) - 4 \, \sqrt {x} + \int \frac {2 \, x^{2} + x}{4 \, x^{\frac {7}{2}} - 5 \, x^{\frac {5}{2}} + 4 \, {\left (x^{3} - x^{2}\right )} \sqrt {x - 1} + x^{\frac {3}{2}}}\,{d x} + \log \left (\sqrt {x} + 1\right ) - \log \left (\sqrt {x} - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right )}{\sqrt {x}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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