\(\int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx\) [149]

Optimal result

Integrand size = 16, antiderivative size = 36 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\frac {x^{1+m} \operatorname {AppellF1}\left (1+m,-\frac {5}{4},\frac {5}{4},2+m,i a x,-i a x\right )}{1+m} \]

[Out]

x^(1+m)*AppellF1(1+m,5/4,-5/4,2+m,-I*a*x,I*a*x)/(1+m)

Rubi [A] (verified)

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} \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\frac {x^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {5}{4},\frac {5}{4},m+2,i a x,-i a x\right )}{m+1} \]

[In]

Int[x^m/E^(((5*I)/2)*ArcTan[a*x]),x]

[Out]

(x^(1 + m)*AppellF1[1 + m, -5/4, 5/4, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 5170

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m (1-i a x)^{5/4}}{(1+i a x)^{5/4}} \, dx \\ & = \frac {x^{1+m} \operatorname {AppellF1}\left (1+m,-\frac {5}{4},\frac {5}{4},2+m,i a x,-i a x\right )}{1+m} \\ \end{align*}

Mathematica [F]

\[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx \]

[In]

Integrate[x^m/E^(((5*I)/2)*ArcTan[a*x]),x]

[Out]

Integrate[x^m/E^(((5*I)/2)*ArcTan[a*x]), x]

Maple [F]

\[\int \frac {x^{m}}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}d x\]

[In]

int(x^m/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2),x)

[Out]

int(x^m/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2),x)

Fricas [F]

\[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^m/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\text {Timed out} \]

[In]

integrate(x**m/((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^m/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2),x, algorithm="maxima")

[Out]

integrate(x^m/((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), x)

Giac [F(-2)]

Exception generated. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^m/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]Warning, replacing 0 by -8, a substitution variable should perhaps be p
urged.Warni

Mupad [F(-1)]

Timed out. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^m \, dx=\int \frac {x^m}{{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \]

[In]

int(x^m/((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2),x)

[Out]

int(x^m/((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2), x)