Integrand size = 12, antiderivative size = 86 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=-\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-\log (1+a x)-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right ) \]
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Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6472, 1643, 266} \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {2}{1-\sqrt {\frac {1-a x}{a x+1}}}-\frac {2}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}-\log (a x+1)-2 \log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \]
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Rule 266
Rule 1643
Rule 6472
Rubi steps \begin{align*} \text {integral}& = \int \frac {\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 )^2}{x} \, dx \\ & = 4 \text {Subst}\left (\int \frac {x (1+x)}{(-1+x)^3 \left (1+x^2\right )} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = 4 \text {Subst}\left (\int \left (\frac {1}{(-1+x)^3}+\frac {1}{2 (-1+x)^2}-\frac {1}{2 (-1+x)}+\frac {x}{2 \left (1+x^2\right )}\right ) \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )+2 \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\frac {2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}-\log (1+a x)-2 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=-\frac {1}{a^2 x^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a^2 x^2}-2 \log (x)+\log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.28
method | result | size |
default | \(\frac {-a^{2} \ln \left (x \right )-\frac {1}{2 x^{2}}}{a^{2}}+\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\sqrt {-a^{2} x^{2}+1}\right )}{a x \sqrt {-a^{2} x^{2}+1}}-\frac {1}{2 a^{2} x^{2}}\) | \(110\) |
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Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.60 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 2 \, a^{2} x^{2} \log \left (x\right ) - 2 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 2}{2 \, a^{2} x^{2}} \]
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Time = 2.89 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\frac {- 2 a^{2} \cdot \left (2 \sqrt {-1 + \frac {1}{a x}} \left (\frac {\left (1 + \frac {1}{a x}\right )^{\frac {3}{2}}}{4} - \frac {\sqrt {1 + \frac {1}{a x}}}{4}\right ) - \log {\left (2 \sqrt {-1 + \frac {1}{a x}} + 2 \sqrt {1 + \frac {1}{a x}} \right )}\right ) - a^{2} \log {\left (x \right )} - \frac {1}{x^{2}}}{a^{2}} \]
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\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x} \,d x } \]
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\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x} \,d x } \]
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Time = 13.88 (sec) , antiderivative size = 323, normalized size of antiderivative = 3.76 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x} \, dx=\ln \left (\frac {1}{x}\right )-4\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )+2\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )+\frac {\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^3}+\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^5}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^7}+\frac {4\,\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}{\sqrt {\frac {1}{a\,x}+1}-1}}{1+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}}-\frac {1}{a^2\,x^2} \]
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