Integrand size = 16, antiderivative size = 121 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5170, 96, 95, 218, 212, 209} \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \]
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Rule 95
Rule 96
Rule 209
Rule 212
Rule 218
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^{5/4}}{x^2 (1-i a x)^{5/4}} \, dx \\ & = -\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac {1}{2} (5 i a) \int \frac {\sqrt [4]{1+i a x}}{x (1-i a x)^{5/4}} \, dx \\ & = \frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac {1}{2} (5 i a) \int \frac {1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx \\ & = \frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+(10 i a) \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ & = \frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-(5 i a) \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 )-(5 i a) \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 {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \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 = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {-3 \left (1-8 i a x+9 a^2 x^2\right )-10 a x (i+a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )}{3 x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \]
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\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}{x^{2}}d x\]
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none
Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {-5 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 5 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 5 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 5 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (-9 i \, a x + 1\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, x} \]
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Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}{x^{2}} \,d x } \]
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Exception generated. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}}{x^2} \,d x \]
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