Integrand size = 14, antiderivative size = 113 \[ \int e^{i \arctan (a x)} x^4 \, dx=-\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \text {arcsinh}(a x)}{8 a^5} \]
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Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5168, 847, 794, 221} \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {3 \text {arcsinh}(a x)}{8 a^5}+\frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}+\frac {x^3 \sqrt {a^2 x^2+1}}{4 a^2}+\frac {(-45 a x+64 i) \sqrt {a^2 x^2+1}}{120 a^5}-\frac {4 i x^2 \sqrt {a^2 x^2+1}}{15 a^3} \]
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Rule 221
Rule 794
Rule 847
Rule 5168
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^4 (1+i a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = \frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x^3 \left (-4 i a+5 a^2 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{5 a^2} \\ & = \frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x^2 \left (-15 a^2-16 i a^3 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{20 a^4} \\ & = -\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x \left (32 i a^3-45 a^4 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{60 a^6} \\ & = -\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{8 a^4} \\ & = -\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \text {arcsinh}(a x)}{8 a^5} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {\sqrt {1+a^2 x^2} \left (64 i-45 a x-32 i a^2 x^2+30 a^3 x^3+24 i a^4 x^4\right )+45 \text {arcsinh}(a x)}{120 a^5} \]
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Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {i \left (24 a^{4} x^{4}-30 i a^{3} x^{3}-32 a^{2} x^{2}+45 i a x +64\right ) \sqrt {a^{2} x^{2}+1}}{120 a^{5}}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{4} \sqrt {a^{2}}}\) | \(84\) |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right ) \sqrt {a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{4 a^{5}}}{2 a^{4} \sqrt {\pi }\, \sqrt {a^{2}}}+\frac {i \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}-8 a^{2} x^{2}+16\right ) \sqrt {a^{2} x^{2}+1}}{15}\right )}{2 a^{5} \sqrt {\pi }}\) | \(117\) |
default | \(\frac {x^{3} \sqrt {a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 \left (\frac {x \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}+i a \left (\frac {x^{4} \sqrt {a^{2} x^{2}+1}}{5 a^{2}}-\frac {4 \left (\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}}\right )}{5 a^{2}}\right )\) | \(142\) |
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Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.59 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {{\left (24 i \, a^{4} x^{4} + 30 \, a^{3} x^{3} - 32 i \, a^{2} x^{2} - 45 \, a x + 64 i\right )} \sqrt {a^{2} x^{2} + 1} - 45 \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{120 \, a^{5}} \]
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Time = 0.56 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int e^{i \arctan (a x)} x^4 \, dx=\begin {cases} \sqrt {a^{2} x^{2} + 1} \left (\frac {i x^{4}}{5 a} + \frac {x^{3}}{4 a^{2}} - \frac {4 i x^{2}}{15 a^{3}} - \frac {3 x}{8 a^{4}} + \frac {8 i}{15 a^{5}}\right ) + \frac {3 \log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{8 a^{4} \sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{6}}{6} + \frac {x^{5}}{5} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{4}}{5 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a^{2}} - \frac {4 i \, \sqrt {a^{2} x^{2} + 1} x^{2}}{15 \, a^{3}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{5}} + \frac {8 i \, \sqrt {a^{2} x^{2} + 1}}{15 \, a^{5}} \]
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Exception generated. \[ \int e^{i \arctan (a x)} x^4 \, dx=\text {Exception raised: TypeError} \]
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Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {x^3\,{\left (a^2\right )}^{3/2}}{4\,a^4}-\frac {3\,x\,\sqrt {a^2}}{8\,a^4}+\frac {a\,8{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}-\frac {a^3\,x^2\,4{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}+\frac {a^5\,x^4\,1{}\mathrm {i}}{5\,{\left (a^2\right )}^{5/2}}\right )}{\sqrt {a^2}}+\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{8\,a^4\,\sqrt {a^2}} \]
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