Integrand size = 12, antiderivative size = 116 \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {14 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {11 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]
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Time = 0.77 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6304, 6874, 277, 270, 272, 44, 65, 214, 665} \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=-\frac {3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}+\frac {14 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^2}+\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {1}{3} x^3 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {11 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]
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Rule 44
Rule 65
Rule 214
Rule 270
Rule 272
Rule 277
Rule 665
Rule 6304
Rule 6874
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2}{x^4 \left (1+\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\text {Subst}\left (\int \left (\frac {1}{x^4 \sqrt {1-\frac {x^2}{a^2}}}-\frac {3}{a x^3 \sqrt {1-\frac {x^2}{a^2}}}+\frac {4}{a^2 x^2 \sqrt {1-\frac {x^2}{a^2}}}-\frac {4}{a^3 x \sqrt {1-\frac {x^2}{a^2}}}+\frac {4}{a^3 (a+x) \sqrt {1-\frac {x^2}{a^2}}}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = \frac {4 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^3}-\frac {4 \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^3}-\frac {4 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^2}+\frac {3 \text {Subst}\left (\int \frac {1}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a}-\text {Subst}\left (\int \frac {1}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {4 \sqrt {1-\frac {1}{a^2 x^2}} x}{a^2}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {2 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{a^3}-\frac {2 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{3 a^2}+\frac {3 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{2 a} \\ & = \frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {14 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {3 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a^3}-\frac {4 \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \\ & = \frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {14 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a^3}-\frac {3 \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \\ & = \frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a+\frac {1}{x}\right )}+\frac {14 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}-\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {11 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (52+19 a x-7 a^2 x^2+2 a^3 x^3\right )}{1+a x}-33 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{6 a^3} \]
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Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.25
method | result | size |
risch | \(\frac {\left (2 a^{2} x^{2}-9 a x +28\right ) \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{6 a^{3}}+\frac {\left (-\frac {11 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{2 a^{2} \sqrt {a^{2}}}+\frac {4 \sqrt {a^{2} \left (x +\frac {1}{a}\right )^{2}-2 a \left (x +\frac {1}{a}\right )}}{a^{4} \left (x +\frac {1}{a}\right )}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{a x -1}\) | \(145\) |
default | \(-\frac {\left (9 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{3} x^{3}-2 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} a^{2} x^{2}+18 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}-9 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-4 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} a x -42 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+42 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+9 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x -18 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x +10 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-84 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x +84 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -9 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -42 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+42 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{6 a^{3} \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )}\) | \(471\) |
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Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\frac {{\left (2 \, a^{3} x^{3} - 7 \, a^{2} x^{2} + 19 \, a x + 52\right )} \sqrt {\frac {a x - 1}{a x + 1}} - 33 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 33 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{6 \, a^{3}} \]
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\[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\int x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.60 \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{6} \, a {\left (\frac {2 \, {\left (39 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 52 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 21 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {33 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{4}} - \frac {33 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{4}} - \frac {24 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{4}}\right )} \]
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\[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\int { x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \]
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Time = 0.06 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.34 \[ \int e^{-3 \coth ^{-1}(a x)} x^2 \, dx=\frac {7\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {52\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+13\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {4\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a^3}-\frac {11\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^3} \]
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