Integrand size = 24, antiderivative size = 39 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+2 a \arctan \left (\sqrt {-1+x} \sqrt {1+x}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {156, 12, 94, 209} \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=2 a \arctan \left (\sqrt {x-1} \sqrt {x+1}\right )-\frac {\sqrt {x-1} \sqrt {x+1}}{x} \]
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Rule 12
Rule 94
Rule 156
Rule 209
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+\int \frac {2 a}{\sqrt {-1+x} x \sqrt {1+x}} \, dx \\ & = -\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+(2 a) \int \frac {1}{\sqrt {-1+x} x \sqrt {1+x}} \, dx \\ & = -\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+(2 a) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x} \sqrt {1+x}\right ) \\ & = -\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+2 a \tan ^{-1}\left (\sqrt {-1+x} \sqrt {1+x}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+4 a \arctan \left (\sqrt {\frac {-1+x}{1+x}}\right ) \]
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Time = 1.58 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {\left (-2 a x \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )-\sqrt {x^{2}-1}\right ) \sqrt {-1+x}\, \sqrt {1+x}}{x \sqrt {x^{2}-1}}\) | \(44\) |
risch | \(-\frac {\sqrt {-1+x}\, \sqrt {1+x}}{x}-\frac {2 a \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right ) \sqrt {\left (-1+x \right ) \left (1+x \right )}}{\sqrt {-1+x}\, \sqrt {1+x}}\) | \(47\) |
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none
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=\frac {4 \, a x \arctan \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) - \sqrt {x + 1} \sqrt {x - 1} - x}{x} \]
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Result contains complex when optimal does not.
Time = 22.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.00 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=- \frac {a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} + \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {3}{2}, \frac {3}{2}, 2 \\1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2 & 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 1 & \\\frac {3}{4}, \frac {5}{4} & \frac {1}{2}, 1, 1, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-2 \, a \arcsin \left (\frac {1}{{\left | x \right |}}\right ) - \frac {\sqrt {x^{2} - 1}}{x} \]
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none
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-4 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2}\right ) - \frac {8}{{\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{4} + 4} \]
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Time = 4.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.67 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {x-1}\,\sqrt {x+1}}{x}-a\,\left (\ln \left (\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )\right )\,2{}\mathrm {i} \]
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