Integrand size = 16, antiderivative size = 163 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5170, 98, 96, 95, 218, 212, 209} \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}} \]
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Rule 95
Rule 96
Rule 98
Rule 209
Rule 212
Rule 218
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^{5/4}}{x^3 (1-i a x)^{5/4}} \, dx \\ & = -\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {1}{4} (5 i a) \int \frac {(1+i a x)^{5/4}}{x^2 (1-i a x)^{5/4}} \, dx \\ & = -\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {\sqrt [4]{1+i a x}}{x (1-i a x)^{5/4}} \, dx \\ & = -\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx \\ & = -\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{2} \left (25 a^2\right ) \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 {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {1}{4} \left (25 a^2\right ) \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} \left (25 a^2\right ) \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 {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{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 = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.61 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {-6-33 i a x-102 a^2 x^2-129 i a^3 x^3+50 a^2 x^2 (1-i a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )}{12 x^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \]
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\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}{x^{3}}d x\]
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none
Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (43 \, a^{2} x^{2} + 9 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \]
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Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}{x^{3}} \,d x } \]
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Exception generated. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}}{x^3} \,d x \]
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