\(\int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x} \, dx\) [110]

Optimal result

Integrand size = 16, antiderivative size = 293 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x} \, dx=\frac {8 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {\log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}} \]

[Out]

8*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)+2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))-2*arctanh((1+I*a*x)^(1/4)/(1-I*a*x
)^(1/4))+1/2*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))*2^(1/2)-1/2*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))*2^(1/2)+arctan(1-(1-I*a*x)^(1/4)*2^(1/2
)/(1+I*a*x)^(1/4))*2^(1/2)-arctan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))*2^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5170, 100, 21, 132, 65, 246, 217, 1179, 642, 1176, 631, 210, 95, 304, 209, 212} \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x} \, dx=2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {8 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}}-\frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}} \]

[In]

Int[1/(E^(((5*I)/2)*ArcTan[a*x])*x),x]

[Out]

(8*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4) + 2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] + Sqrt[2]*ArcTan[1 - (
Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)] - Sqrt[2]*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/
4)] - 2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] + 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)]/Sqrt[2] - 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)]/Sqrt[2]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 100

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
-1]))

Rule 209

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

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 304

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

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

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x} \, dx=\frac {\sqrt [4]{1-i a x} \left (20-20 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )+2^{3/4} (1-i a x) \sqrt [4]{1+i a x} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2} (1-i a x)\right )\right )}{5 \sqrt [4]{1+i a x}} \]

[In]

Integrate[1/(E^(((5*I)/2)*ArcTan[a*x])*x),x]

[Out]

((1 - I*a*x)^(1/4)*(20 - 20*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x)] + 2^(3/4)*(1 - I*a*x)*(1 + I*a
*x)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, (1 - I*a*x)/2]))/(5*(1 + I*a*x)^(1/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x} \, dx=\frac {\sqrt {4 i} {\left (a x - i\right )} \log \left (\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \sqrt {4 i} {\left (a x - i\right )} \log \left (-\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \sqrt {-4 i} {\left (a x - i\right )} \log \left (\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + \sqrt {-4 i} {\left (a x - i\right )} \log \left (-\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, {\left (a x - i\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 2 \, {\left (-i \, a x - 1\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 2 \, {\left (i \, a x + 1\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 2 \, {\left (a x - i\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 16 i \, \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, {\left (a x - i\right )}} \]

[In]

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

[Out]

1/2*(sqrt(4*I)*(a*x - I)*log(1/2*I*sqrt(4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - sqrt(4*I)*(a*x - I)*log(
-1/2*I*sqrt(4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - sqrt(-4*I)*(a*x - I)*log(1/2*I*sqrt(-4*I) + sqrt(I*s
qrt(a^2*x^2 + 1)/(a*x + I))) + sqrt(-4*I)*(a*x - I)*log(-1/2*I*sqrt(-4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)
)) - 2*(a*x - I)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - 2*(-I*a*x - 1)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a
*x + I)) + I) - 2*(I*a*x + 1)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 2*(a*x - I)*log(sqrt(I*sqrt(a^2*x
^2 + 1)/(a*x + I)) - 1) - 16*I*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/(a*x - I)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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