Integrand size = 12, antiderivative size = 233 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {a^4}{6 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {2 a^4}{5 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {4 a^4}{3 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \]
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Time = 0.36 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6472, 1626} \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {3 a^4}{8 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {3 a^4}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}+\frac {a^4}{4 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {a^4}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}-\frac {a^4}{6 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3}-\frac {4 a^4}{3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}+\frac {a^4}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}-\frac {2 a^4}{5 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5} \]
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Rule 1626
Rule 6472
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^5 \left (\frac {1}{a x}+\sqrt {\frac {1-a x}{1+a x}}+\frac {\sqrt {\frac {1-a x}{1+a x}}}{a x}\right )} \, dx \\ & = -\left ((4 a) \text {Subst}\left (\int \frac {x \left (a+a x^2\right )^3}{(-1+x)^4 (1+x)^6} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )\right ) \\ & = -\left ((4 a) \text {Subst}\left (\int \left (\frac {a^3}{8 (-1+x)^4}+\frac {a^3}{8 (-1+x)^3}+\frac {3 a^3}{32 (-1+x)^2}-\frac {a^3}{2 (1+x)^6}+\frac {a^3}{(1+x)^5}-\frac {a^3}{(1+x)^4}+\frac {a^3}{2 (1+x)^3}-\frac {3 a^3}{32 (1+x)^2}\right ) \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )\right ) \\ & = -\frac {a^4}{6 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {2 a^4}{5 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {4 a^4}{3 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {3+\sqrt {\frac {1-a x}{1+a x}} (1+a x)^2 \left (-3+3 a x-2 a^2 x^2+2 a^3 x^3\right )}{15 a x^5} \]
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Time = 0.94 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.29
method | result | size |
default | \(a \left (-\frac {1}{5 a^{2} x^{5}}-\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right ) \left (2 a^{2} x^{2}+3\right )}{15 a \,x^{4}}\right )\) | \(68\) |
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Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {{\left (2 \, a^{5} x^{5} + a^{3} x^{3} - 3 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 3}{15 \, a x^{5}} \]
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\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=a \int \frac {1}{a x^{5} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x^{4}}\, dx \]
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\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=\int { \frac {1}{x^{5} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \]
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\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=\int { \frac {1}{x^{5} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \]
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Time = 5.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.32 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {1}{5\,a\,x^5}-\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {a\,x^2}{15}-\frac {x}{5}-\frac {1}{5\,a}+\frac {a^2\,x^3}{15}+\frac {2\,a^3\,x^4}{15}+\frac {2\,a^4\,x^5}{15}\right )}{x^5\,\sqrt {\frac {1}{a\,x}+1}} \]
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