\(\int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx\) [132]

Optimal result

Integrand size = 16, antiderivative size = 328 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-\frac {1}{2} i a \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {i a \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {1}{2} i a \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {2}}-\frac {i a \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {2}} \]

[Out]

-(1-I*a*x)^(7/8)*(1+I*a*x)^(1/8)/x-1/2*I*a*arctan((1+I*a*x)^(1/8)/(1-I*a*x)^(1/8))-1/2*I*a*arctanh((1+I*a*x)^(
1/8)/(1-I*a*x)^(1/8))+1/4*I*a*arctan(1-(1+I*a*x)^(1/8)*2^(1/2)/(1-I*a*x)^(1/8))*2^(1/2)-1/4*I*a*arctan(1+(1+I*
a*x)^(1/8)*2^(1/2)/(1-I*a*x)^(1/8))*2^(1/2)+1/8*I*a*ln(1+(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4)-(1+I*a*x)^(1/8)*2^(1/
2)/(1-I*a*x)^(1/8))*2^(1/2)-1/8*I*a*ln(1+(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4)+(1+I*a*x)^(1/8)*2^(1/2)/(1-I*a*x)^(1/
8))*2^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {5170, 96, 95, 220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=-\frac {1}{2} i a \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {i a \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {1}{2} i a \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}+\frac {i a \log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{4 \sqrt {2}}-\frac {i a \log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{4 \sqrt {2}} \]

[In]

Int[E^((I/4)*ArcTan[a*x])/x^2,x]

[Out]

-(((1 - I*a*x)^(7/8)*(1 + I*a*x)^(1/8))/x) - (I/2)*a*ArcTan[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)] + ((I/2)*a*Ar
cTan[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/Sqrt[2] - ((I/2)*a*ArcTan[1 + (Sqrt[2]*(1 + I*a*x)^(1
/8))/(1 - I*a*x)^(1/8)])/Sqrt[2] - (I/2)*a*ArcTanh[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)] + ((I/4)*a*Log[1 - (Sq
rt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/Sqrt[2] - ((I/4)*a*Log[1 +
(Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/Sqrt[2]

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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -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 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 220

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a, b},
x] && IGtQ[n/4, 1] &&  !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 {\sqrt [8]{1+i a x}}{x^2 \sqrt [8]{1-i a x}} \, dx \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}+\frac {1}{4} (i a) \int \frac {1}{x \sqrt [8]{1-i a x} (1+i a x)^{7/8}} \, dx \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}+(2 i a) \text {Subst}\left (\int \frac {1}{-1+x^8} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right ) \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-(i a) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-(i a) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right ) \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-\frac {1}{2} (i a) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} (i a) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} (i a) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} (i a) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right ) \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-\frac {1}{2} i a \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} i a \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{4} (i a) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{4} (i a) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {(i a) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{4 \sqrt {2}}+\frac {(i a) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{4 \sqrt {2}} \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-\frac {1}{2} i a \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} i a \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {2}}-\frac {i a \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {2}}-\frac {(i a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}+\frac {(i a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}} \\ & = -\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}-\frac {1}{2} i a \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {i a \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{2 \sqrt {2}}-\frac {1}{2} i a \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {i a \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {2}}-\frac {i a \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{4 \sqrt {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 = 71, normalized size of antiderivative = 0.22 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=-\frac {i (1-i a x)^{7/8} \left (-7 i+7 a x+2 a x \operatorname {Hypergeometric2F1}\left (\frac {7}{8},1,\frac {15}{8},\frac {i+a x}{i-a x}\right )\right )}{7 x (1+i a x)^{7/8}} \]

[In]

Integrate[E^((I/4)*ArcTan[a*x])/x^2,x]

[Out]

((-1/7*I)*(1 - I*a*x)^(7/8)*(-7*I + 7*a*x + 2*a*x*Hypergeometric2F1[7/8, 1, 15/8, (I + a*x)/(I - a*x)]))/(x*(1
+ I*a*x)^(7/8))

Maple [F]

\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {1}{4}}}{x^{2}}d x\]

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^2,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=\frac {-i \, a x \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + 1\right ) + a x \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + i\right ) - a x \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - i\right ) + i \, a x \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - 1\right ) + \sqrt {i \, a^{2}} x \log \left (\frac {a \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + i \, \sqrt {i \, a^{2}}}{a}\right ) - \sqrt {i \, a^{2}} x \log \left (\frac {a \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - i \, \sqrt {i \, a^{2}}}{a}\right ) + \sqrt {-i \, a^{2}} x \log \left (\frac {a \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + i \, \sqrt {-i \, a^{2}}}{a}\right ) - \sqrt {-i \, a^{2}} x \log \left (\frac {a \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - i \, \sqrt {-i \, a^{2}}}{a}\right ) - 4 \, {\left (-i \, a x + 1\right )} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}}{4 \, x} \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/4)/x**2,x)

[Out]

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(1/4)/x**2, x)

Maxima [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {1}{4}}}{x^{2}} \,d x } \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(1/4)/x^2, x)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^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 -28, a substitution variable should perhaps be
purged.Warn

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{1/4}}{x^2} \,d x \]

[In]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^2,x)

[Out]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^2, x)