Integrand size = 14, antiderivative size = 92 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+\frac {9}{2} a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
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Time = 0.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5168, 6874, 272, 44, 65, 214, 270, 665} \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\frac {9}{2} a^2 \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {4 i a^2 \sqrt {a^2 x^2+1}}{a x+i}-\frac {3 i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2} \]
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Rule 44
Rule 65
Rule 214
Rule 270
Rule 272
Rule 665
Rule 5168
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{x^3 (1-i a x) \sqrt {1+a^2 x^2}} \, dx \\ & = \int \left (\frac {1}{x^3 \sqrt {1+a^2 x^2}}+\frac {3 i a}{x^2 \sqrt {1+a^2 x^2}}-\frac {4 a^2}{x \sqrt {1+a^2 x^2}}+\frac {4 a^3}{(i+a x) \sqrt {1+a^2 x^2}}\right ) \, dx \\ & = (3 i a) \int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx-\left (4 a^2\right ) \int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx+\left (4 a^3\right ) \int \frac {1}{(i+a x) \sqrt {1+a^2 x^2}} \, dx+\int \frac {1}{x^3 \sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+a^2 x}} \, dx,x,x^2\right )-\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}-4 \text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2}} \, dx,x,\sqrt {1+a^2 x^2}\right )-\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+4 a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2}} \, dx,x,\sqrt {1+a^2 x^2}\right ) \\ & = -\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+\frac {9}{2} a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.86 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\sqrt {1+a^2 x^2} \left (-\frac {1}{2 x^2}-\frac {3 i a}{x}-\frac {4 i a^2}{i+a x}\right )-\frac {9}{2} a^2 \log (x)+\frac {9}{2} a^2 \log \left (1+\sqrt {1+a^2 x^2}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14
method | result | size |
default | \(-\frac {1}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {9 a^{2} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}+3 i a \left (-\frac {1}{x \sqrt {a^{2} x^{2}+1}}-\frac {2 a^{2} x}{\sqrt {a^{2} x^{2}+1}}\right )\) | \(105\) |
risch | \(-\frac {i \left (6 a^{3} x^{3}-i a^{2} x^{2}+6 a x -i\right )}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {a^{2} \left (-9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {8 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a \left (x +\frac {i}{a}\right )}\right )}{2}\) | \(108\) |
meijerg | \(\frac {a^{2} \left (\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \left (24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{\sqrt {\pi }}-\frac {3 i a \left (2 a^{2} x^{2}+1\right )}{x \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{2} \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }}-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}\) | \(229\) |
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Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\frac {-14 i \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-14 i \, a^{2} x^{2} + 5 \, a x - i\right )}}{2 \, {\left (a x^{3} + i \, x^{2}\right )}} \]
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\[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=- i \left (\int \frac {i}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
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Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {7 i \, a^{3} x}{\sqrt {a^{2} x^{2} + 1}} + \frac {9}{2} \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {9 \, a^{2}}{2 \, \sqrt {a^{2} x^{2} + 1}} - \frac {3 i \, a}{\sqrt {a^{2} x^{2} + 1} x} - \frac {1}{2 \, \sqrt {a^{2} x^{2} + 1} x^{2}} \]
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\[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
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Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {a^2\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}-\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{x}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]
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