Integrand size = 12, antiderivative size = 142 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2-\frac {\arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \]
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Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6306, 98, 96, 95, 304, 209, 212} \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {1}{2} x^2 \left (1-\frac {1}{a x}\right )^{5/4} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{4 a} \]
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Rule 95
Rule 96
Rule 98
Rule 209
Rule 212
Rule 304
Rule 6306
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{x^3 \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2+\frac {\text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{x^2 \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{4 a} \\ & = -\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2-\frac {\text {Subst}\left (\int \frac {1}{x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{8 a^2} \\ & = -\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2-\frac {\text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{2 a^2} \\ & = -\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \\ & = -\frac {\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{4 a}+\frac {1}{2} \left (1-\frac {1}{a x}\right )^{5/4} \left (1+\frac {1}{a x}\right )^{3/4} x^2-\frac {\arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {-\frac {2 e^{\frac {3}{2} \coth ^{-1}(a x)} \left (-5+e^{2 \coth ^{-1}(a x)}\right )}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}-\arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+\text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \]
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\[\int x \left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}}d x\]
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Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {2 \, {\left (2 \, a^{2} x^{2} - a x - 3\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{8 \, a^{2}} \]
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\[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\int x \sqrt [4]{\frac {a x - 1}{a x + 1}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {4 \, {\left (5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {2 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {\log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {\log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (\frac {5 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \]
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Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x \, dx=\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{2}-\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}+\frac {\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2} \]
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