Integrand size = 10, antiderivative size = 130 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]
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Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6305, 96, 95, 304, 209, 212} \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}+\frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}+\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{\frac {1}{a x}+1}} \]
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Rule 95
Rule 96
Rule 209
Rule 212
Rule 304
Rule 6305
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{5/4}}{x^2 \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{x \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{2 a} \\ & = \frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \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 )}{2 a} \\ & = \frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {10 \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 )}{a} \\ & = \frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {5 \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 )}{a}+\frac {5 \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 )}{a} \\ & = \frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.24 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\frac {8 e^{-\frac {1}{2} \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},e^{2 \coth ^{-1}(a x)}\right )}{a} \]
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\[\int \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}d x\]
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Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\frac {2 \, {\left (a x + 9\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{2 \, a} \]
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\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\int \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=-\frac {1}{2} \, a {\left (\frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{2}} - \frac {16 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2}}\right )} \]
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Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=-\frac {1}{2} \, a {\left (\frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {5 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{2}} - \frac {16 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2}} + \frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2} {\left (\frac {a x - 1}{a x + 1} - 1\right )}}\right )} \]
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Time = 4.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx=\frac {2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}+\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a}-\frac {5\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a}+\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{a} \]
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