Integrand size = 16, antiderivative size = 36 \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\frac {x^{1+m} \operatorname {AppellF1}\left (1+m,-\frac {3}{4},\frac {3}{4},2+m,i a x,-i a x\right )}{1+m} \]
[Out]
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5170, 138} \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\frac {x^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{4},\frac {3}{4},m+2,i a x,-i a x\right )}{m+1} \]
[In]
[Out]
Rule 138
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m (1-i a x)^{3/4}}{(1+i a x)^{3/4}} \, dx \\ & = \frac {x^{1+m} \operatorname {AppellF1}\left (1+m,-\frac {3}{4},\frac {3}{4},2+m,i a x,-i a x\right )}{1+m} \\ \end{align*}
\[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx \]
[In]
[Out]
\[\int \frac {x^{m}}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}}}d x\]
[In]
[Out]
\[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}} \,d x } \]
[In]
[Out]
\[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\int \frac {x^{m}}{\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}\, dx \]
[In]
[Out]
\[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}} \,d x } \]
[In]
[Out]
Exception generated. \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\text {Exception raised: TypeError} \]
[In]
[Out]
Timed out. \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^m \, dx=\int \frac {x^m}{{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}} \,d x \]
[In]
[Out]