Integrand size = 33, antiderivative size = 48 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {c \sqrt {1-d^2 x^2}}{d^2}+\frac {b \arcsin (d x)}{d}-a \text {arctanh}\left (\sqrt {1-d^2 x^2}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1623, 1823, 858, 222, 272, 65, 214} \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=-a \text {arctanh}\left (\sqrt {1-d^2 x^2}\right )+\frac {b \arcsin (d x)}{d}-\frac {c \sqrt {1-d^2 x^2}}{d^2} \]
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Rule 65
Rule 214
Rule 222
Rule 272
Rule 858
Rule 1623
Rule 1823
Rubi steps \begin{align*} \text {integral}& = \int \frac {a+b x+c x^2}{x \sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {c \sqrt {1-d^2 x^2}}{d^2}-\frac {\int \frac {-a d^2-b d^2 x}{x \sqrt {1-d^2 x^2}} \, dx}{d^2} \\ & = -\frac {c \sqrt {1-d^2 x^2}}{d^2}+a \int \frac {1}{x \sqrt {1-d^2 x^2}} \, dx+b \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {c \sqrt {1-d^2 x^2}}{d^2}+\frac {b \sin ^{-1}(d x)}{d}+\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \sqrt {1-d^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {c \sqrt {1-d^2 x^2}}{d^2}+\frac {b \sin ^{-1}(d x)}{d}-\frac {a \text {Subst}\left (\int \frac {1}{\frac {1}{d^2}-\frac {x^2}{d^2}} \, dx,x,\sqrt {1-d^2 x^2}\right )}{d^2} \\ & = -\frac {c \sqrt {1-d^2 x^2}}{d^2}+\frac {b \sin ^{-1}(d x)}{d}-a \tanh ^{-1}\left (\sqrt {1-d^2 x^2}\right ) \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {c \sqrt {1-d^2 x^2}}{d^2}+\frac {2 b \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{d}-a \log (x)+a \log \left (-1+\sqrt {1-d^2 x^2}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.61 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.00
method | result | size |
default | \(\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (-\operatorname {csgn}\left (d \right ) \operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) a \,d^{2}-\operatorname {csgn}\left (d \right ) \sqrt {-d^{2} x^{2}+1}\, c +\arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-\left (d x +1\right ) \left (d x -1\right )}}\right ) b d \right ) \operatorname {csgn}\left (d \right )}{d^{2} \sqrt {-d^{2} x^{2}+1}}\) | \(96\) |
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Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.69 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {a d^{2} \log \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{x}\right ) - 2 \, b d \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right ) - \sqrt {d x + 1} \sqrt {-d x + 1} c}{d^{2}} \]
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Result contains complex when optimal does not.
Time = 28.85 (sec) , antiderivative size = 245, normalized size of antiderivative = 5.10 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {i a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {i b {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=-a \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {b \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} c}{d^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (44) = 88\).
Time = 0.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 4.08 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {a d^{2} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} + 2 \right |}\right ) - a d^{2} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} - 2 \right |}\right ) - {\left (\pi + 2 \, \arctan \left (\frac {\sqrt {d x + 1} {\left (\frac {{\left (\sqrt {2} - \sqrt {-d x + 1}\right )}^{2}}{d x + 1} - 1\right )}}{2 \, {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}\right )\right )} b d + \sqrt {d x + 1} \sqrt {-d x + 1} c}{d^{2}} \]
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Time = 4.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.54 \[ \int \frac {a+b x+c x^2}{x \sqrt {1-d x} \sqrt {1+d x}} \, dx=a\,\left (\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )-\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\right )-\frac {\sqrt {1-d\,x}\,\left (\frac {c}{d^2}+\frac {c\,x}{d}\right )}{\sqrt {d\,x+1}}-\frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {1-d\,x}-1\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {d^2}}\right )}{\sqrt {d^2}} \]
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