Optimal. Leaf size=63 \[ -\frac {4 a \sqrt {a^2 x^2+1}}{a x+i}-\frac {\sqrt {a^2 x^2+1}}{x}-3 i a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]
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Rubi [A] time = 0.56, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5060, 6742, 264, 266, 63, 208, 651} \[ -\frac {4 a \sqrt {a^2 x^2+1}}{a x+i}-\frac {\sqrt {a^2 x^2+1}}{x}-3 i a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 264
Rule 266
Rule 651
Rule 5060
Rule 6742
Rubi steps
\begin {align*} \int \frac {e^{3 i \tan ^{-1}(a x)}}{x^2} \, dx &=\int \frac {(1+i a x)^2}{x^2 (1-i a x) \sqrt {1+a^2 x^2}} \, dx\\ &=\int \left (\frac {1}{x^2 \sqrt {1+a^2 x^2}}+\frac {3 i a}{x \sqrt {1+a^2 x^2}}-\frac {4 i a^2}{(i+a x) \sqrt {1+a^2 x^2}}\right ) \, dx\\ &=(3 i a) \int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx-\left (4 i a^2\right ) \int \frac {1}{(i+a x) \sqrt {1+a^2 x^2}} \, dx+\int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{x}-\frac {4 a \sqrt {1+a^2 x^2}}{i+a x}+\frac {1}{2} (3 i a) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1+a^2 x^2}}{x}-\frac {4 a \sqrt {1+a^2 x^2}}{i+a x}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2}} \, dx,x,\sqrt {1+a^2 x^2}\right )}{a}\\ &=-\frac {\sqrt {1+a^2 x^2}}{x}-\frac {4 a \sqrt {1+a^2 x^2}}{i+a x}-3 i a \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.05, size = 61, normalized size = 0.97 \[ \sqrt {a^2 x^2+1} \left (-\frac {1}{x}-\frac {4 a}{a x+i}\right )-3 i a \log \left (\sqrt {a^2 x^2+1}+1\right )+3 i a \log (x) \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.44, size = 109, normalized size = 1.73 \[ -\frac {5 \, a^{2} x^{2} + 5 i \, a x + 3 \, {\left (i \, a^{2} x^{2} - a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + 3 \, {\left (-i \, a^{2} x^{2} + a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (5 \, a x + i\right )}}{a x^{2} + i \, x} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 80, normalized size = 1.27 \[ \frac {i a}{\sqrt {a^{2} x^{2}+1}}-\frac {5 a^{2} x}{\sqrt {a^{2} x^{2}+1}}+3 i a \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )-\frac {1}{x \sqrt {a^{2} x^{2}+1}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 60, normalized size = 0.95 \[ -\frac {5 \, a^{2} x}{\sqrt {a^{2} x^{2} + 1}} - 3 i \, a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {4 i \, a}{\sqrt {a^{2} x^{2} + 1}} - \frac {1}{\sqrt {a^{2} x^{2} + 1} x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.06, size = 75, normalized size = 1.19 \[ -a\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )\,3{}\mathrm {i}-\frac {\sqrt {a^2\,x^2+1}}{x}-\frac {4\,a^2\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - i \left (\int \frac {i}{a^{2} x^{4} \sqrt {a^{2} x^{2} + 1} + x^{2} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{4} \sqrt {a^{2} x^{2} + 1} + x^{2} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{4} \sqrt {a^{2} x^{2} + 1} + x^{2} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{4} \sqrt {a^{2} x^{2} + 1} + x^{2} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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