Optimal. Leaf size=63 \[ -\frac {i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a^2 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]
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Rubi [A] time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5060, 835, 807, 266, 63, 208} \[ -\frac {i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a^2 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 266
Rule 807
Rule 835
Rule 5060
Rubi steps
\begin {align*} \int \frac {e^{i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac {1+i a x}{x^3 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {1}{2} \int \frac {-2 i a+a^2 x}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {i a \sqrt {1+a^2 x^2}}{x}-\frac {1}{2} a^2 \int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {i a \sqrt {1+a^2 x^2}}{x}-\frac {1}{4} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {i a \sqrt {1+a^2 x^2}}{x}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2}} \, dx,x,\sqrt {1+a^2 x^2}\right )\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {i a \sqrt {1+a^2 x^2}}{x}+\frac {1}{2} a^2 \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.04, size = 57, normalized size = 0.90 \[ \frac {1}{2} \left (\frac {(-1-2 i a x) \sqrt {a^2 x^2+1}}{x^2}+a^2 \log \left (\sqrt {a^2 x^2+1}+1\right )+a^2 (-\log (x))\right ) \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 1.20, size = 83, normalized size = 1.32 \[ \frac {a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - 2 i \, a^{2} x^{2} + \sqrt {a^{2} x^{2} + 1} {\left (-2 i \, a x - 1\right )}}{2 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.15, size = 155, normalized size = 2.46 \[ \frac {1}{2} \, a^{2} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) - \frac {1}{2} \, a^{2} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) + \frac {{\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{3} a^{2} + 2 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} a i {\left | a \right |} + {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )} a^{2} - 2 \, a i {\left | a \right |}}{{\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 53, normalized size = 0.84 \[ -\frac {i a \sqrt {a^{2} x^{2}+1}}{x}-\frac {\sqrt {a^{2} x^{2}+1}}{2 x^{2}}+\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 48, normalized size = 0.76 \[ \frac {1}{2} \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {i \, \sqrt {a^{2} x^{2} + 1} a}{x} - \frac {\sqrt {a^{2} x^{2} + 1}}{2 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.04, size = 52, normalized size = 0.83 \[ \frac {a^2\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}-\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{x} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.91, size = 48, normalized size = 0.76 \[ - i a^{2} \sqrt {1 + \frac {1}{a^{2} x^{2}}} + \frac {a^{2} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{2 x} \]
Verification of antiderivative is not currently implemented for this CAS.
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