Integrand size = 16, antiderivative size = 132 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5170, 98, 96, 95, 304, 209, 212} \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {1}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}+\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x} \]
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Rule 95
Rule 96
Rule 98
Rule 209
Rule 212
Rule 304
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt [4]{1-i a x}}{x^3 \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} (i a) \int \frac {\sqrt [4]{1-i a x}}{x^2 \sqrt [4]{1+i a x}} \, dx \\ & = \frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{8} a^2 \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = \frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{2} a^2 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ & = \frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}+\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ & = \frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {\sqrt [4]{1-i a x} \left (-2+i a x-3 a^2 x^2+2 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{4 x^2 \sqrt [4]{1+i a x}} \]
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\[\int \frac {1}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{3}}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-3 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \]
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\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {1}{x^{3} \sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}\, dx \]
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\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int { \frac {1}{x^{3} \sqrt {\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}}} \,d x } \]
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Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {1}{x^3\,\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}} \,d x \]
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