\(\int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx\) [229]

Optimal result

Integrand size = 18, antiderivative size = 395 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 \sqrt [4]{i+a} \arctan \left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\frac {2 \sqrt [4]{i+a} \text {arctanh}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {5202, 492, 217, 1179, 642, 1176, 631, 210, 218, 214, 211} \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 \sqrt [4]{a+i} \arctan \left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{-a+i}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\frac {2 \sqrt [4]{a+i} \text {arctanh}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{-a+i}}-\frac {\log \left (\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}+1\right )}{\sqrt {2}}+\frac {\log \left (\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}+1\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

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 211

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

Rule 214

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

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 218

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

Rule 492

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(-a)*(e^n/(b*c -
a*d)), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[c*(e^n/(b*c - a*d)), Int[(e*x)^(m - n)/(c + d*x^n), x], x
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

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 5202

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Dist[4/(I^m*n*b^(m + 1)*c^(m + 1)), Sub
st[Int[x^(2/(I*n))*((1 - I*a*c - (1 + I*a*c)*x^(2/(I*n)))^m/(1 + x^(2/(I*n)))^(m + 2)), x], x, (1 - I*c*(a + b
*x))^(I*(n/2))/(1 + I*c*(a + b*x))^(I*(n/2))], x] /; FreeQ[{a, b, c}, x] && ILtQ[m, 0] && LtQ[-1, I*n, 1]

Rubi steps \begin{align*} \text {integral}& = -\left (8 \text {Subst}\left (\int \frac {x^4}{\left (1+x^4\right ) \left (1-i a-(1+i a) x^4\right )} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-(4 (1-i a)) \text {Subst}\left (\int \frac {1}{1-i a+(-1-i a) x^4} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right ) \\ & = 2 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )+2 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\left (2 \sqrt {i+a}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {i+a}-\sqrt {i-a} x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\left (2 \sqrt {i+a}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {i+a}+\sqrt {i-a} x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i+a} \arctan \left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {2 \sqrt [4]{i+a} \text {arctanh}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b 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+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{\sqrt {2}}+\text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )+\text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i+a} \arctan \left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {2 \sqrt [4]{i+a} \text {arctanh}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b 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+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i+a} \arctan \left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )-\frac {2 \sqrt [4]{i+a} \text {arctanh}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{\sqrt [4]{i-a}}-\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-i (a+b x)}}{\sqrt {1+i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b 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 = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.32 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\frac {2 \sqrt [4]{-i (i+a+b x)} \left (2^{3/4} \sqrt [4]{1+i a+i b x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},-\frac {1}{2} i (i+a+b x)\right )-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{\sqrt [4]{1+i a+i b x}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {1}{2} \, \sqrt {4 i} \log \left (\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \frac {1}{2} \, \sqrt {4 i} \log \left (-\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \frac {1}{2} \, \sqrt {-4 i} \log \left (\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac {1}{2} \, \sqrt {-4 i} \log \left (-\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \left (-\frac {a + i}{a - i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a - i\right )} \left (-\frac {a + i}{a - i}\right )^{\frac {3}{4}} + {\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{a + i}\right ) - \left (-\frac {a + i}{a - i}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (a - i\right )} \left (-\frac {a + i}{a - i}\right )^{\frac {3}{4}} - {\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{a + i}\right ) - i \, \left (-\frac {a + i}{a - i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (i \, a + 1\right )} \left (-\frac {a + i}{a - i}\right )^{\frac {3}{4}} + {\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{a + i}\right ) + i \, \left (-\frac {a + i}{a - i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (-i \, a - 1\right )} \left (-\frac {a + i}{a - i}\right )^{\frac {3}{4}} + {\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{a + i}\right ) \]

[In]

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

[Out]

-1/2*sqrt(4*I)*log(1/2*I*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 1/2*sqrt(4*I)*
log(-1/2*I*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 1/2*sqrt(-4*I)*log(1/2*I*sqr
t(-4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - 1/2*sqrt(-4*I)*log(-1/2*I*sqrt(-4*I) + sq
rt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + (-(a + I)/(a - I))^(1/4)*log(((a - I)*(-(a + I)/(a -
I))^(3/4) + (a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(a + I)) - (-(a + I)/(a - I))^(1/
4)*log(-((a - I)*(-(a + I)/(a - I))^(3/4) - (a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(
a + I)) - I*(-(a + I)/(a - I))^(1/4)*log(((I*a + 1)*(-(a + I)/(a - I))^(3/4) + (a + I)*sqrt(I*sqrt(b^2*x^2 + 2
*a*b*x + a^2 + 1)/(b*x + a + I)))/(a + I)) + I*(-(a + I)/(a - I))^(1/4)*log(((-I*a - 1)*(-(a + I)/(a - I))^(3/
4) + (a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(a + I))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/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=[0,0,0]Warning, replacing 0 by 14, a substitution variable should perhaps be
purged.War

Mupad [F(-1)]

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

[In]

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

[Out]

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