Integrand size = 33, antiderivative size = 71 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x}-\frac {1}{2} \left (2 c+a d^2\right ) \text {arctanh}\left (\sqrt {1-d^2 x^2}\right ) \]
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Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1623, 1821, 821, 272, 65, 214} \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {1}{2} \left (a d^2+2 c\right ) \text {arctanh}\left (\sqrt {1-d^2 x^2}\right )-\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x} \]
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Rule 65
Rule 214
Rule 272
Rule 821
Rule 1623
Rule 1821
Rubi steps \begin{align*} \text {integral}& = \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {1}{2} \int \frac {-2 b-\left (2 c+a d^2\right ) x}{x^2 \sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x}-\frac {1}{2} \left (-2 c-a d^2\right ) \int \frac {1}{x \sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x}-\frac {1}{4} \left (-2 c-a d^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-d^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x}-\frac {1}{2} \left (a+\frac {2 c}{d^2}\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{d^2}-\frac {x^2}{d^2}} \, dx,x,\sqrt {1-d^2 x^2}\right ) \\ & = -\frac {a \sqrt {1-d^2 x^2}}{2 x^2}-\frac {b \sqrt {1-d^2 x^2}}{x}-\frac {1}{2} \left (2 c+a d^2\right ) \tanh ^{-1}\left (\sqrt {1-d^2 x^2}\right ) \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {1}{2} \left (-\frac {(a+2 b x) \sqrt {1-d^2 x^2}}{x^2}-\left (2 c+a d^2\right ) \log (x)+\left (2 c+a d^2\right ) \log \left (-1+\sqrt {1-d^2 x^2}\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.61 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.52
method | result | size |
default | \(-\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )^{2} \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) a \,d^{2} x^{2}+2 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) c \,x^{2}+2 \sqrt {-d^{2} x^{2}+1}\, b x +\sqrt {-d^{2} x^{2}+1}\, a \right )}{2 \sqrt {-d^{2} x^{2}+1}\, x^{2}}\) | \(108\) |
risch | \(\frac {\sqrt {d x +1}\, \left (d x -1\right ) \left (2 b x +a \right ) \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{2 x^{2} \sqrt {-\left (d x +1\right ) \left (d x -1\right )}\, \sqrt {-d x +1}}-\frac {\left (c +\frac {a \,d^{2}}{2}\right ) \operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{\sqrt {-d x +1}\, \sqrt {d x +1}}\) | \(113\) |
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Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {{\left (a d^{2} + 2 \, c\right )} x^{2} \log \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{x}\right ) - {\left (2 \, b x + a\right )} \sqrt {d x + 1} \sqrt {-d x + 1}}{2 \, x^{2}} \]
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Timed out. \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\text {Timed out} \]
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Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.38 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {1}{2} \, a d^{2} \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - c \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\sqrt {-d^{2} x^{2} + 1} b}{x} - \frac {\sqrt {-d^{2} x^{2} + 1} a}{2 \, x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (61) = 122\).
Time = 0.42 (sec) , antiderivative size = 407, normalized size of antiderivative = 5.73 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (a d^{3} + 2 \, c d\right )} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} + 2 \right |}\right ) - {\left (a d^{3} + 2 \, c d\right )} \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} + \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}} - 2 \right |}\right ) - \frac {4 \, {\left (a d^{3} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{3} - 2 \, b d^{2} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{3} + 4 \, a d^{3} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )} + 8 \, b d^{2} {\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}\right )}}{{\left ({\left (\frac {\sqrt {2} - \sqrt {-d x + 1}}{\sqrt {d x + 1}} - \frac {\sqrt {d x + 1}}{\sqrt {2} - \sqrt {-d x + 1}}\right )}^{2} - 4\right )}^{2}}}{2 \, d} \]
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Time = 6.15 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.39 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {1-d x} \sqrt {1+d x}} \, dx=c\,\left (\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )-\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\right )-\frac {\frac {a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {a\,d^2}{2}+\frac {15\,a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^4}{2\,{\left (\sqrt {d\,x+1}-1\right )}^4}}{\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {32\,{\left (\sqrt {1-d\,x}-1\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {16\,{\left (\sqrt {1-d\,x}-1\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}}+\frac {a\,d^2\,\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )}{2}-\frac {a\,d^2\,\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2}-\frac {b\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{x}+\frac {a\,d^2\,{\left (\sqrt {1-d\,x}-1\right )}^2}{32\,{\left (\sqrt {d\,x+1}-1\right )}^2} \]
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