Integrand size = 10, antiderivative size = 132 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{1+a x}}}+\frac {1}{8} a^3 \sqrt {\frac {1}{1+a x}} \sqrt {1+a x} \text {arctanh}\left (\sqrt {1-a x} \sqrt {1+a x}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6470, 30, 105, 12, 94, 214} \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {1}{8} a^3 \sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \text {arctanh}\left (\sqrt {1-a x} \sqrt {a x+1}\right )+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{a x+1}}}+\frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{a x+1}}} \]
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Rule 12
Rule 30
Rule 94
Rule 105
Rule 214
Rule 6470
Rubi steps \begin{align*} \text {integral}& = -\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}-\frac {\int \frac {1}{x^5} \, dx}{3 a}-\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{x^5 \sqrt {1-a x} \sqrt {1+a x}} \, dx}{3 a} \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int -\frac {3 a^2}{x^3 \sqrt {1-a x} \sqrt {1+a x}} \, dx}{12 a} \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}-\frac {1}{4} \left (a \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{x^3 \sqrt {1-a x} \sqrt {1+a x}} \, dx \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{1+a x}}}-\frac {1}{8} \left (a \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {a^2}{x \sqrt {1-a x} \sqrt {1+a x}} \, dx \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{1+a x}}}-\frac {1}{8} \left (a^3 \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{x \sqrt {1-a x} \sqrt {1+a x}} \, dx \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{1+a x}}}+\frac {1}{8} \left (a^4 \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \text {Subst}\left (\int \frac {1}{a-a x^2} \, dx,x,\sqrt {1-a x} \sqrt {1+a x}\right ) \\ & = \frac {1}{12 a x^4}-\frac {e^{\text {sech}^{-1}(a x)}}{3 x^3}+\frac {\sqrt {1-a x}}{12 a x^4 \sqrt {\frac {1}{1+a x}}}+\frac {a \sqrt {1-a x}}{8 x^2 \sqrt {\frac {1}{1+a x}}}+\frac {1}{8} a^3 \sqrt {\frac {1}{1+a x}} \sqrt {1+a x} \text {arctanh}\left (\sqrt {1-a x} \sqrt {1+a x}\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {-2+\sqrt {\frac {1-a x}{1+a x}} \left (-2-2 a x+a^2 x^2+a^3 x^3\right )-a^4 x^4 \log (x)+a^4 x^4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a x^4} \]
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Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) a^{4} x^{4}+a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}-2 \sqrt {-a^{2} x^{2}+1}\right )}{8 x^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {1}{4 a \,x^{4}}\) | \(110\) |
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Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) + 2 \, {\left (a^{3} x^{3} - 2 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 4}{16 \, a x^{4}} \]
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\[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {\int \frac {1}{x^{5}}\, dx + \int \frac {a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{4}}\, dx}{a} \]
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\[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}{x^{4}} \,d x } \]
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\[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}{x^{4}} \,d x } \]
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Time = 19.50 (sec) , antiderivative size = 602, normalized size of antiderivative = 4.56 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {a^3\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )}{2}-\frac {\frac {35\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^3}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^3}+\frac {273\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^5}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^5}+\frac {715\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^7}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^7}+\frac {715\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^9}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^9}+\frac {273\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{11}}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{11}}+\frac {35\,a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{13}}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{13}}+\frac {a^3\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{15}}{2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{15}}+\frac {a^3\,\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}{2\,\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}}{1+\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {56\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {70\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}-\frac {56\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}+\frac {28\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{12}}-\frac {8\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{14}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{14}}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{16}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{16}}-\frac {8\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}}-\frac {1}{4\,a\,x^4} \]
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