\(\int e^{-\frac {5}{2} i \arctan (a x)} x \, dx\) [108]

Optimal result

Integrand size = 14, antiderivative size = 324 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}+\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2} \]

[Out]

-2*(1-I*a*x)^(9/4)/a^2/(1+I*a*x)^(1/4)-25/4*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a^2-5/2*(1-I*a*x)^(5/4)*(1+I*a*x)^
(3/4)/a^2-25/8*arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^2*2^(1/2)+25/8*arctan(1+(1-I*a*x)^(1/4)*2^(
1/2)/(1+I*a*x)^(1/4))/a^2*2^(1/2)-25/16*ln(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)
^(1/2))/a^2*2^(1/2)+25/16*ln(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)^(1/2))/a^2*2^
(1/2)

Rubi [A] (verified)

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, 246, 217, 1179, 642, 1176, 631, 210} \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}+\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {5 (1+i a x)^{3/4} (1-i a x)^{5/4}}{2 a^2}-\frac {25 (1+i a x)^{3/4} \sqrt [4]{1-i a x}}{4 a^2}-\frac {25 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt {2} a^2} \]

[In]

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

[Out]

(-2*(1 - I*a*x)^(9/4))/(a^2*(1 + I*a*x)^(1/4)) - (25*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(4*a^2) - (5*(1 - I*
a*x)^(5/4)*(1 + I*a*x)^(3/4))/(2*a^2) - (25*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt
[2]*a^2) + (25*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt[2]*a^2) - (25*Log[1 + Sqrt[1
- I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2) + (25*Log[1 + Sqrt
[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 (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 {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac {(25 i) \int \frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{4 a} \\ & = -\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac {(25 i) \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{8 a} \\ & = -\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac {25 \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{2 a^2} \\ & = -\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac {25 \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{2 a^2} \\ & = -\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 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 \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 {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 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 {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 {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{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}+\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 {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{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 \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.19 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 (1-i a x)^{9/4} \left (-\frac {9}{\sqrt [4]{1+i a x}}+5\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-i a x)\right )\right )}{9 a^2} \]

[In]

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

[Out]

(2*(1 - I*a*x)^(9/4)*(-9/(1 + I*a*x)^(1/4) + 5*2^(3/4)*Hypergeometric2F1[1/4, 9/4, 13/4, (1 - I*a*x)/2]))/(9*a
^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.89 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} i \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} i \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} i \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} i \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \sqrt {a^{2} x^{2} + 1} {\left (2 i \, a^{2} x^{2} - 9 \, a x + 43 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{4 \, {\left (a^{3} x - i \, a^{2}\right )}} \]

[In]

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

[Out]

-1/4*(2*(a^3*x - I*a^2)*sqrt(625/16*I/a^4)*log(4/25*I*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x +
I))) - 2*(a^3*x - I*a^2)*sqrt(625/16*I/a^4)*log(-4/25*I*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*
x + I))) - 2*(a^3*x - I*a^2)*sqrt(-625/16*I/a^4)*log(4/25*I*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)
/(a*x + I))) + 2*(a^3*x - I*a^2)*sqrt(-625/16*I/a^4)*log(-4/25*I*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2
+ 1)/(a*x + I))) - sqrt(a^2*x^2 + 1)*(2*I*a^2*x^2 - 9*a*x + 43*I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/(a^3*x
- I*a^2)

Sympy [F]

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

[In]

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

[Out]

Integral(x/(I*(a*x - I)/sqrt(a**2*x**2 + 1))**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate(x/((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 81, a substitution variable should perhaps be p
urged.Warni

Mupad [F(-1)]

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

[In]

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

[Out]

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