Integrand size = 30, antiderivative size = 63 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}+\frac {\left (c+2 a d^2\right ) \arcsin (d x)}{2 d^3} \]
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Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {913, 1829, 655, 222} \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {\left (2 a d^2+c\right ) \arcsin (d x)}{2 d^3}-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2} \]
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Rule 222
Rule 655
Rule 913
Rule 1829
Rubi steps \begin{align*} \text {integral}& = \int \frac {a+b x+c x^2}{\sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}-\frac {\int \frac {-c-2 a d^2-2 b d^2 x}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2} \\ & = -\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}-\frac {\left (-c-2 a d^2\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2} \\ & = -\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}+\frac {\left (c+2 a d^2\right ) \sin ^{-1}(d x)}{2 d^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {(-2 b-c x) \sqrt {1-d^2 x^2}}{2 d^2}+\frac {\left (c+2 a d^2\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{d^3} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.46 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.86
method | result | size |
default | \(-\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (\sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d c x -2 \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) a \,d^{2}+2 \,\operatorname {csgn}\left (d \right ) d \sqrt {-d^{2} x^{2}+1}\, b -\arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) c \right ) \operatorname {csgn}\left (d \right )}{2 d^{3} \sqrt {-d^{2} x^{2}+1}}\) | \(117\) |
risch | \(\frac {\left (c x +2 b \right ) \left (d x -1\right ) \sqrt {d x +1}\, \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{2 d^{2} \sqrt {-\left (d x -1\right ) \left (d x +1\right )}\, \sqrt {-d x +1}}+\frac {\left (2 a \,d^{2}+c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{2 d^{2} \sqrt {d^{2}}\, \sqrt {-d x +1}\, \sqrt {d x +1}}\) | \(129\) |
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Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (c d x + 2 \, b d\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 2 \, {\left (2 \, a d^{2} + c\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{2 \, d^{3}} \]
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Timed out. \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\text {Timed out} \]
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Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {a \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} c x}{2 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} b}{d^{2}} + \frac {c \arcsin \left (d x\right )}{2 \, d^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left ({\left (d x + 1\right )} c + 2 \, b d - c\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 2 \, {\left (2 \, a d^{2} + c\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{2 \, d^{3}} \]
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Time = 16.60 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.68 \[ \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\sqrt {1-d\,x}\,\left (\frac {b}{d^2}+\frac {b\,x}{d}\right )}{\sqrt {d\,x+1}}-\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {1-d\,x}-1\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {d^2}}\right )}{\sqrt {d^2}}-\frac {2\,c\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {\frac {14\,c\,{\left (\sqrt {1-d\,x}-1\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {14\,c\,{\left (\sqrt {1-d\,x}-1\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,c\,{\left (\sqrt {1-d\,x}-1\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {2\,c\,\left (\sqrt {1-d\,x}-1\right )}{\sqrt {d\,x+1}-1}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^4} \]
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