\(\int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx\) [133]

Optimal result

Integrand size = 16, antiderivative size = 364 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=-\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt {2}}+\frac {a^2 \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt {2}} \]

[Out]

-1/8*I*a*(1-I*a*x)^(7/8)*(1+I*a*x)^(1/8)/x-1/2*(1-I*a*x)^(7/8)*(1+I*a*x)^(9/8)/x^2+1/16*a^2*arctan((1+I*a*x)^(
1/8)/(1-I*a*x)^(1/8))+1/16*a^2*arctanh((1+I*a*x)^(1/8)/(1-I*a*x)^(1/8))-1/32*a^2*arctan(1-(1+I*a*x)^(1/8)*2^(1
/2)/(1-I*a*x)^(1/8))*2^(1/2)+1/32*a^2*arctan(1+(1+I*a*x)^(1/8)*2^(1/2)/(1-I*a*x)^(1/8))*2^(1/2)-1/64*a^2*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)+1/64*a^2*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 = 0.11 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5170, 98, 96, 95, 220, 218, 212, 209, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt {2}}+\frac {a^2 \log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt {2}}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x} \]

[In]

Int[E^((I/4)*ArcTan[a*x])/x^3,x]

[Out]

((-1/8*I)*a*(1 - I*a*x)^(7/8)*(1 + I*a*x)^(1/8))/x - ((1 - I*a*x)^(7/8)*(1 + I*a*x)^(9/8))/(2*x^2) + (a^2*ArcT
an[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*ArcTan[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/
(16*Sqrt[2]) + (a^2*ArcTan[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/(16*Sqrt[2]) + (a^2*ArcTanh[(1
+ I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*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)])/(32*Sqrt[2]) + (a^2*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)])/(32*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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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^3 \sqrt [8]{1-i a x}} \, dx \\ & = -\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{8} (i a) \int \frac {\sqrt [8]{1+i a x}}{x^2 \sqrt [8]{1-i a x}} \, dx \\ & = -\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac {1}{32} a^2 \int \frac {1}{x \sqrt [8]{1-i a x} (1+i a x)^{7/8}} \, dx \\ & = -\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac {1}{4} a^2 \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 {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{8} a^2 \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}{8} a^2 \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 {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \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}{16} a^2 \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}{16} a^2 \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}{16} a^2 \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 {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{32} a^2 \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}{32} a^2 \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 {a^2 \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 )}{32 \sqrt {2}}-\frac {a^2 \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 )}{32 \sqrt {2}} \\ & = -\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \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 )}{32 \sqrt {2}}+\frac {a^2 \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 )}{32 \sqrt {2}}+\frac {a^2 \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 )}{16 \sqrt {2}}-\frac {a^2 \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 )}{16 \sqrt {2}} \\ & = -\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \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 )}{32 \sqrt {2}}+\frac {a^2 \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 )}{32 \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 = 84, normalized size of antiderivative = 0.23 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\frac {(1-i a x)^{7/8} \left (7 \left (-4-9 i a x+5 a^2 x^2\right )+2 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{8},1,\frac {15}{8},\frac {i+a x}{i-a x}\right )\right )}{56 x^2 (1+i a x)^{7/8}} \]

[In]

Integrate[E^((I/4)*ArcTan[a*x])/x^3,x]

[Out]

((1 - I*a*x)^(7/8)*(7*(-4 - (9*I)*a*x + 5*a^2*x^2) + 2*a^2*x^2*Hypergeometric2F1[7/8, 1, 15/8, (I + a*x)/(I -
a*x)]))/(56*x^2*(1 + I*a*x)^(7/8))

Maple [F]

\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {1}{4}}}{x^{3}}d x\]

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + 1\right ) + i \, a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + i\right ) - i \, a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - i\right ) - a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - 1\right ) + \sqrt {i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + \sqrt {i \, a^{4}}}{a^{2}}\right ) - \sqrt {i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - \sqrt {i \, a^{4}}}{a^{2}}\right ) + \sqrt {-i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + \sqrt {-i \, a^{4}}}{a^{2}}\right ) - \sqrt {-i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - \sqrt {-i \, a^{4}}}{a^{2}}\right ) - 4 \, {\left (5 \, a^{2} x^{2} + i \, a x + 4\right )} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}}{32 \, x^{2}} \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="fricas")

[Out]

1/32*(a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + 1) + I*a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(
1/4) + I) - I*a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) - I) - a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x +
I))^(1/4) - 1) + sqrt(I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + sqrt(I*a^4))/a^2) - sqrt(I*
a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) - sqrt(I*a^4))/a^2) + sqrt(-I*a^4)*x^2*log((a^2*(I*sqr
t(a^2*x^2 + 1)/(a*x + I))^(1/4) + sqrt(-I*a^4))/a^2) - sqrt(-I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I
))^(1/4) - sqrt(-I*a^4))/a^2) - 4*(5*a^2*x^2 + I*a*x + 4)*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4))/x^2

Sympy [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int \frac {\sqrt [4]{\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}{x^{3}}\, dx \]

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/4)/x**3,x)

[Out]

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(1/4)/x**3, x)

Maxima [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {1}{4}}}{x^{3}} \,d x } \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(1/4)/x^3, x)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,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. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{1/4}}{x^3} \,d x \]

[In]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^3,x)

[Out]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^3, x)