Integrand size = 14, antiderivative size = 117 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt {1+a^2 x^2}}{3 x^3}-\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}+\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
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Time = 0.44 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5168, 6874, 277, 270, 272, 44, 65, 214, 665} \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {14 a^2 \sqrt {a^2 x^2+1}}{3 x}-\frac {3 i a \sqrt {a^2 x^2+1}}{2 x^2}-\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )+\frac {4 a^3 \sqrt {a^2 x^2+1}}{a x+i} \]
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Rule 44
Rule 65
Rule 214
Rule 270
Rule 272
Rule 277
Rule 665
Rule 5168
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{x^4 (1-i a x) \sqrt {1+a^2 x^2}} \, dx \\ & = \int \left (\frac {1}{x^4 \sqrt {1+a^2 x^2}}+\frac {3 i a}{x^3 \sqrt {1+a^2 x^2}}-\frac {4 a^2}{x^2 \sqrt {1+a^2 x^2}}-\frac {4 i a^3}{x \sqrt {1+a^2 x^2}}+\frac {4 i a^4}{(i+a x) \sqrt {1+a^2 x^2}}\right ) \, dx \\ & = (3 i a) \int \frac {1}{x^3 \sqrt {1+a^2 x^2}} \, dx-\left (4 a^2\right ) \int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx-\left (4 i a^3\right ) \int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx+\left (4 i a^4\right ) \int \frac {1}{(i+a x) \sqrt {1+a^2 x^2}} \, dx+\int \frac {1}{x^4 \sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {\sqrt {1+a^2 x^2}}{3 x^3}+\frac {4 a^2 \sqrt {1+a^2 x^2}}{x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}+\frac {1}{2} (3 i a) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+a^2 x}} \, dx,x,x^2\right )-\frac {1}{3} \left (2 a^2\right ) \int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx-\left (2 i a^3\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}}{3 x^3}-\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}-(4 i a) \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} \left (3 i a^3\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}}{3 x^3}-\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}+4 i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right )-\frac {1}{2} (3 i a) \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}}{3 x^3}-\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}+\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {1+a^2 x^2} \left (-2 i+7 a x+19 i a^2 x^2+52 a^3 x^3\right )}{x^3 (i+a x)}-33 i a^3 \log (x)+33 i a^3 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \]
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Time = 0.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99
method | result | size |
risch | \(\frac {28 a^{4} x^{4}-9 i a^{3} x^{3}+26 a^{2} x^{2}-9 i a x -2}{6 x^{3} \sqrt {a^{2} x^{2}+1}}-\frac {i a^{3} \left (-11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {8 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a \left (x +\frac {i}{a}\right )}\right )}{2}\) | \(116\) |
default | \(-\frac {1}{3 x^{3} \sqrt {a^{2} x^{2}+1}}-\frac {13 a^{2} \left (-\frac {1}{x \sqrt {a^{2} x^{2}+1}}-\frac {2 a^{2} x}{\sqrt {a^{2} x^{2}+1}}\right )}{3}-i a^{3} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )+3 i a \left (-\frac {1}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{2} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}\right )\) | \(141\) |
meijerg | \(-\frac {-8 a^{4} x^{4}-4 a^{2} x^{2}+1}{3 x^{3} \sqrt {a^{2} x^{2}+1}}+\frac {3 i a^{3} \left (\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \left (24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{\sqrt {\pi }}+\frac {3 a^{2} \left (2 a^{2} x^{2}+1\right )}{x \sqrt {a^{2} x^{2}+1}}-\frac {i a^{3} \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }}\) | \(249\) |
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Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {52 \, a^{4} x^{4} + 52 i \, a^{3} x^{3} - 33 \, {\left (-i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 33 \, {\left (i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + {\left (52 \, a^{3} x^{3} + 19 i \, a^{2} x^{2} + 7 \, a x - 2 i\right )} \sqrt {a^{2} x^{2} + 1}}{6 \, {\left (a x^{4} + i \, x^{3}\right )}} \]
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\[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=- i \left (\int \frac {i}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
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Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {26 \, a^{4} x}{3 \, \sqrt {a^{2} x^{2} + 1}} + \frac {11}{2} i \, a^{3} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {11 i \, a^{3}}{2 \, \sqrt {a^{2} x^{2} + 1}} + \frac {13 \, a^{2}}{3 \, \sqrt {a^{2} x^{2} + 1} x} - \frac {3 i \, a}{2 \, \sqrt {a^{2} x^{2} + 1} x^{2}} - \frac {1}{3 \, \sqrt {a^{2} x^{2} + 1} x^{3}} \]
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\[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
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Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {11\,a^3\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{3\,x^3}-\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{2\,x^2}+\frac {14\,a^2\,\sqrt {a^2\,x^2+1}}{3\,x}+\frac {4\,a^4\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]
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