Integrand size = 14, antiderivative size = 324 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2} \]
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Time = 0.15 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5170, 79, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {25 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt {2} a^2} \]
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Rule 52
Rule 65
Rule 79
Rule 210
Rule 303
Rule 338
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1+i a x)^{5/4}}{(1-i a x)^{5/4}} \, dx \\ & = -\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {(5 i) \int \frac {(1+i a x)^{5/4}}{\sqrt [4]{1-i a x}} \, dx}{a} \\ & = -\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {(25 i) \int \frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx}{4 a} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {(25 i) \int \frac {1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{8 a} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac {25 \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{2 a^2} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac {25 \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 )}{2 a^2} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {25 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^2}-\frac {25 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^2} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac {25 \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^2}-\frac {25 \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^2}-\frac {25 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}-\frac {25 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}-\frac {25 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}+\frac {25 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2} \\ & = -\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.22 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 \left (-3 (1+i a x)^{9/4}+20 i \sqrt [4]{2} (i+a x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-i a x)\right )\right )}{3 a^2 \sqrt [4]{1-i a x}} \]
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\[\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x d x\]
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none
Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.73 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (2 \, a^{2} x^{2} - 9 i \, a x + 43\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{4 \, a^{2}} \]
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Timed out. \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\text {Timed out} \]
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\[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\int { x \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}} \,d x } \]
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Exception generated. \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\int x\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2} \,d x \]
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