\(\int e^{i \arctan (a x)} x^m \, dx\) [141]

Optimal result

Integrand size = 14, antiderivative size = 79 \[ \int e^{i \arctan (a x)} x^m \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}+\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m} \]

[Out]

x^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],-a^2*x^2)/(1+m)+I*a*x^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],
-a^2*x^2)/(2+m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5168, 822, 371} \[ \int e^{i \arctan (a x)} x^m \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}+\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2} \]

[In]

Int[E^(I*ArcTan[a*x])*x^m,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) + (I*a*x^(2 + m)*Hypergeometric2F
1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 5168

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m (1+i a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = (i a) \int \frac {x^{1+m}}{\sqrt {1+a^2 x^2}} \, dx+\int \frac {x^m}{\sqrt {1+a^2 x^2}} \, dx \\ & = \frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}+\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08 \[ \int e^{i \arctan (a x)} x^m \, dx=\frac {i x^{1+m} \sqrt {1-i a x} \sqrt {-i+a x} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {1}{2},2+m,-i a x,i a x\right )}{(1+m) \sqrt {1+i a x} \sqrt {i+a x}} \]

[In]

Integrate[E^(I*ArcTan[a*x])*x^m,x]

[Out]

(I*x^(1 + m)*Sqrt[1 - I*a*x]*Sqrt[-I + a*x]*AppellF1[1 + m, -1/2, 1/2, 2 + m, (-I)*a*x, I*a*x])/((1 + m)*Sqrt[
1 + I*a*x]*Sqrt[I + a*x])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90

method	result	size
meijerg	\(\frac {x^{1+m} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {i a \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {1}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m}\)	\(71\)

[In]

int((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^m,x,method=_RETURNVERBOSE)

[Out]

x^(1+m)*hypergeom([1/2,1/2+1/2*m],[3/2+1/2*m],-a^2*x^2)/(1+m)+I*a*x^(2+m)*hypergeom([1/2,1+1/2*m],[2+1/2*m],-a
^2*x^2)/(2+m)

Fricas [F]

\[ \int e^{i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )} x^{m}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^m,x, algorithm="fricas")

[Out]

integral(I*sqrt(a^2*x^2 + 1)*x^m/(a*x + I), x)

Sympy [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int e^{i \arctan (a x)} x^m \, dx=\frac {i a x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {a^{2} x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {a^{2} x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]

[In]

integrate((1+I*a*x)/(a**2*x**2+1)**(1/2)*x**m,x)

[Out]

I*a*x**(m + 2)*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), a**2*x**2*exp_polar(I*pi))/(2*gamma(m/2 + 2))
+ x**(m + 1)*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,), a**2*x**2*exp_polar(I*pi))/(2*gamma(m/2 +
3/2))

Maxima [F]

\[ \int e^{i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )} x^{m}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^m,x, algorithm="maxima")

[Out]

integrate((I*a*x + 1)*x^m/sqrt(a^2*x^2 + 1), x)

Giac [F]

\[ \int e^{i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )} x^{m}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^m,x, algorithm="giac")

[Out]

integrate((I*a*x + 1)*x^m/sqrt(a^2*x^2 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int e^{i \arctan (a x)} x^m \, dx=\int \frac {x^m\,\left (1+a\,x\,1{}\mathrm {i}\right )}{\sqrt {a^2\,x^2+1}} \,d x \]

[In]

int((x^m*(a*x*1i + 1))/(a^2*x^2 + 1)^(1/2),x)

[Out]

int((x^m*(a*x*1i + 1))/(a^2*x^2 + 1)^(1/2), x)