\(\int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx\) [121]

Optimal result

Integrand size = 14, antiderivative size = 319 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19 i \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19 i \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{324} i \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\frac {19}{324} i \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \]

[Out]

-1/3*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x^3-7/18*I*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x^2+11/27*(1-I*x)^(5/6)*(1+I*x)^(1/6)/
x+19/81*I*arctanh((1+I*x)^(1/6)/(1-I*x)^(1/6))-19/324*I*ln(1-(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)
^(1/3))+19/324*I*ln(1+(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^(1/3))-19/162*I*arctan(1/3*(1-2*(1+I*x
)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)+19/162*I*arctan(1/3*(1+2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5170, 101, 156, 12, 95, 216, 648, 632, 210, 642, 212} \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=-\frac {19 i \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19 i \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19}{324} i \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac {19}{324} i \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right ) \]

[In]

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

[Out]

-1/3*((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x^3 - (((7*I)/18)*(1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x^2 + (11*(1 - I*x)^
(5/6)*(1 + I*x)^(1/6))/(27*x) - (((19*I)/54)*ArcTan[(1 - (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3
] + (((19*I)/54)*ArcTan[(1 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3] + ((19*I)/81)*ArcTanh[(1 +
I*x)^(1/6)/(1 - I*x)^(1/6)] - ((19*I)/324)*Log[1 - (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x
)^(1/3)] + ((19*I)/324)*Log[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 101

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[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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 216

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a/b, n]], s = Denominator[Rt[-a/b, n
]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 [6]{1+i x}}{\sqrt [6]{1-i x} x^4} \, dx \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}+\frac {1}{3} \int \frac {\frac {7 i}{3}-2 x}{\sqrt [6]{1-i x} (1+i x)^{5/6} x^3} \, dx \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}-\frac {1}{6} \int \frac {\frac {22}{9}+\frac {7 i x}{3}}{\sqrt [6]{1-i x} (1+i x)^{5/6} x^2} \, dx \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}+\frac {1}{6} \int -\frac {19 i}{27 \sqrt [6]{1-i x} (1+i x)^{5/6} x} \, dx \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19}{162} i \int \frac {1}{\sqrt [6]{1-i x} (1+i x)^{5/6} x} \, dx \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19}{27} i \text {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}+\frac {19}{81} i \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {19}{81} i \text {Subst}\left (\int \frac {1-\frac {x}{2}}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {19}{81} i \text {Subst}\left (\int \frac {1+\frac {x}{2}}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{324} i \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {19}{324} i \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {19}{108} i \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {19}{108} i \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{324} i \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\frac {19}{324} i \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {19}{54} i \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{54} i \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \\ & = -\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19 i \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19 i \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{324} i \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\frac {19}{324} i \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \\ \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 = 81, normalized size of antiderivative = 0.25 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\frac {(1-i x)^{5/6} \left (5 \left (-18-39 i x+43 x^2+22 i x^3\right )+38 i x^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{270 (1+i x)^{5/6} x^3} \]

[In]

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

[Out]

((1 - I*x)^(5/6)*(5*(-18 - (39*I)*x + 43*x^2 + (22*I)*x^3) + (38*I)*x^3*Hypergeometric2F1[5/6, 1, 11/6, (I + x
)/(I - x)]))/(270*(1 + I*x)^(5/6)*x^3)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\frac {38 i \, x^{3} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) - 38 i \, x^{3} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) - 19 \, {\left (\sqrt {3} x^{3} - i \, x^{3}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) - 19 \, {\left (\sqrt {3} x^{3} + i \, x^{3}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) + 19 \, {\left (\sqrt {3} x^{3} + i \, x^{3}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + 19 \, {\left (\sqrt {3} x^{3} - i \, x^{3}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - 6 \, {\left (22 i \, x^{3} - x^{2} + 3 i \, x + 18\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}}}{324 \, x^{3}} \]

[In]

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

[Out]

1/324*(38*I*x^3*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1) - 38*I*x^3*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1) -
19*(sqrt(3)*x^3 - I*x^3)*log(1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) - 19*(sqrt(3)*x^3 + I*x^3
)*log(1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) + 19*(sqrt(3)*x^3 + I*x^3)*log(-1/2*I*sqrt(3) + (
I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + 19*(sqrt(3)*x^3 - I*x^3)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I)
)^(1/3) - 1/2) - 6*(22*I*x^3 - x^2 + 3*I*x + 18)*(I*sqrt(x^2 + 1)/(x + I))^(1/3))/x^3

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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