Integrand size = 10, antiderivative size = 60 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {2 i (1+i a x)^2}{a \sqrt {1+a^2 x^2}}-\frac {3 i \sqrt {1+a^2 x^2}}{a}-\frac {3 \text {arcsinh}(a x)}{a} \]
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Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5167, 867, 683, 655, 221} \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {2 i (1+i a x)^2}{a \sqrt {a^2 x^2+1}}-\frac {3 i \sqrt {a^2 x^2+1}}{a}-\frac {3 \text {arcsinh}(a x)}{a} \]
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Rule 221
Rule 655
Rule 683
Rule 867
Rule 5167
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx \\ & = \int \frac {(1+i a x)^3}{\left (1+a^2 x^2\right )^{3/2}} \, dx \\ & = -\frac {2 i (1+i a x)^2}{a \sqrt {1+a^2 x^2}}-3 \int \frac {1+i a x}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {2 i (1+i a x)^2}{a \sqrt {1+a^2 x^2}}-\frac {3 i \sqrt {1+a^2 x^2}}{a}-3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {2 i (1+i a x)^2}{a \sqrt {1+a^2 x^2}}-\frac {3 i \sqrt {1+a^2 x^2}}{a}-\frac {3 \text {arcsinh}(a x)}{a} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int e^{3 i \arctan (a x)} \, dx=\frac {\sqrt {1+a^2 x^2} \left (-i+\frac {4}{i+a x}\right )}{a}-\frac {3 \text {arcsinh}(a x)}{a} \]
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Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.55
method | result | size |
risch | \(-\frac {i \sqrt {a^{2} x^{2}+1}}{a}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {4 \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{2} \left (x +\frac {i}{a}\right )}\) | \(93\) |
default | \(\frac {x}{\sqrt {a^{2} x^{2}+1}}-3 a^{2} \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )-i a^{3} \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )-\frac {3 i}{a \sqrt {a^{2} x^{2}+1}}\) | \(128\) |
meijerg | \(\frac {x}{\sqrt {a^{2} x^{2}+1}}+\frac {3 i \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {3 \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {3}{2}} \operatorname {arcsinh}\left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {a^{2}}}-\frac {i \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}\) | \(137\) |
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Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int e^{3 i \arctan (a x)} \, dx=\frac {4 \, a x + 3 \, {\left (a x + i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-i \, a x + 5\right )} + 4 i}{a^{2} x + i \, a} \]
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\[ \int e^{3 i \arctan (a x)} \, dx=- i \left (\int \frac {i}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
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Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {i \, a x^{2}}{\sqrt {a^{2} x^{2} + 1}} + \frac {4 \, x}{\sqrt {a^{2} x^{2} + 1}} - \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a} - \frac {5 i}{\sqrt {a^{2} x^{2} + 1} a} \]
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\[ \int e^{3 i \arctan (a x)} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
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Time = 0.47 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}+\frac {4\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]
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