Integrand size = 29, antiderivative size = 52 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {(2 b+c x) \sqrt {-1+d x} \sqrt {1+d x}}{2 d^2}+\frac {\left (c+2 a d^2\right ) \text {arccosh}(d x)}{2 d^3} \]
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Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(52)=104\).
Time = 0.05 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.60, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {915, 1829, 655, 223, 212} \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {d^2 x^2-1} \left (2 a d^2+c\right ) \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{2 d^3 \sqrt {d x-1} \sqrt {d x+1}}-\frac {b \left (1-d^2 x^2\right )}{d^2 \sqrt {d x-1} \sqrt {d x+1}}-\frac {c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt {d x-1} \sqrt {d x+1}} \]
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Rule 212
Rule 223
Rule 655
Rule 915
Rule 1829
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-1+d^2 x^2} \int \frac {a+b x+c x^2}{\sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\sqrt {-1+d^2 x^2} \int \frac {c+2 a d^2+2 b d^2 x}{\sqrt {-1+d^2 x^2}} \, dx}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {b \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (\left (c+2 a d^2\right ) \sqrt {-1+d^2 x^2}\right ) \int \frac {1}{\sqrt {-1+d^2 x^2}} \, dx}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {b \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (\left (c+2 a d^2\right ) \sqrt {-1+d^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{1-d^2 x^2} \, dx,x,\frac {x}{\sqrt {-1+d^2 x^2}}\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {b \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {c x \left (1-d^2 x^2\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (c+2 a d^2\right ) \sqrt {-1+d^2 x^2} \tanh ^{-1}\left (\frac {d x}{\sqrt {-1+d^2 x^2}}\right )}{2 d^3 \sqrt {-1+d x} \sqrt {1+d x}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.21 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {d (2 b+c x) \sqrt {-1+d x} \sqrt {1+d x}+2 \left (c+2 a d^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{2 d^3} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(95\) vs. \(2(44)=88\).
Time = 5.70 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.85
method | result | size |
risch | \(\frac {\left (c x +2 b \right ) \sqrt {d x -1}\, \sqrt {d x +1}}{2 d^{2}}+\frac {\left (2 a \,d^{2}+c \right ) \ln \left (\frac {x \,d^{2}}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-1}\right ) \sqrt {\left (d x +1\right ) \left (d x -1\right )}}{2 d^{2} \sqrt {d^{2}}\, \sqrt {d x -1}\, \sqrt {d x +1}}\) | \(96\) |
default | \(\frac {\sqrt {d x -1}\, \sqrt {d x +1}\, \left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right ) d c x +2 \ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) a \,d^{2}+2 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-1}\, b +\ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) c \right ) \operatorname {csgn}\left (d \right )}{2 d^{3} \sqrt {d^{2} x^{2}-1}}\) | \(120\) |
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Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {{\left (c d x + 2 \, b d\right )} \sqrt {d x + 1} \sqrt {d x - 1} - {\left (2 \, a d^{2} + c\right )} \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right )}{2 \, d^{3}} \]
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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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Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (44) = 88\).
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.73 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} c x}{2 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} b}{d^{2}} + \frac {c \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{2 \, d^{3}} \]
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Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.54 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {d x + 1} \sqrt {d x - 1} {\left (\frac {{\left (d x + 1\right )} c}{d^{2}} + \frac {2 \, b d^{5} - c d^{4}}{d^{6}}\right )} - \frac {2 \, {\left (2 \, a d^{2} + c\right )} \log \left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}{d^{2}}}{2 \, d} \]
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Time = 20.38 (sec) , antiderivative size = 312, normalized size of antiderivative = 6.00 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {b\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}+\frac {2\,c\,\mathrm {atanh}\left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-\frac {\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {2\,c\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\sqrt {d\,x+1}-1}}{d^3-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}} \]
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