Integrand size = 16, antiderivative size = 202 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}+\frac {123}{64} a^4 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {123}{64} a^4 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5170, 101, 156, 12, 95, 304, 209, 212} \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=\frac {123}{64} a^4 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {123}{64} a^4 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3} \]
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Rule 12
Rule 95
Rule 101
Rule 156
Rule 209
Rule 212
Rule 304
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^{3/4}}{x^5 (1-i a x)^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}+\frac {1}{4} \int \frac {\frac {9 i a}{2}-3 a^2 x}{x^4 (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}-\frac {1}{12} \int \frac {\frac {45 a^2}{4}+9 i a^3 x}{x^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {1}{24} \int \frac {-\frac {189 i a^3}{8}+\frac {45 a^4 x}{4}}{x^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}-\frac {1}{24} \int -\frac {369 a^4}{16 x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}+\frac {1}{128} \left (123 a^4\right ) \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}+\frac {1}{32} \left (123 a^4\right ) \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 {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}-\frac {1}{64} \left (123 a^4\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}{64} \left (123 a^4\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 {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x^4}-\frac {3 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 x^3}+\frac {15 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{32 x^2}+\frac {63 i a^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 x}+\frac {123}{64} a^4 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {123}{64} a^4 \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 = 0.49 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {\sqrt [4]{1-i a x} \left (16+40 i a x-54 a^2 x^2-93 i a^3 x^3+63 a^4 x^4+246 a^4 x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{64 x^4 \sqrt [4]{1+i a x}} \]
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\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}}}{x^{5}}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {123 \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 123 i \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 123 i \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 123 \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) + 2 \, {\left (-63 i \, a^{3} x^{3} - 30 \, a^{2} x^{2} + 24 i \, a x + 16\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{128 \, x^{4}} \]
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\[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=\int \frac {\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]
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\[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{5}} \,d x } \]
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Exception generated. \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^5} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}}{x^5} \,d x \]
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