3.121 \(\int \frac{e^{\frac{1}{3} i \tan ^{-1}(x)}}{x^4} \, dx\)

Optimal. Leaf size=319 \[ -\frac{7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}+\frac{11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac{19}{324} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac{19}{324} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{19 i \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{54 \sqrt{3}}+\frac{19 i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{54 \sqrt{3}}+\frac{19}{81} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

[Out]

-((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/(3*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)]

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Rubi [A]  time = 0.200022, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5062, 99, 151, 12, 93, 210, 634, 618, 204, 628, 206} \[ -\frac{7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}+\frac{11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac{19}{324} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac{19}{324} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{19 i \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{54 \sqrt{3}}+\frac{19 i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{54 \sqrt{3}}+\frac{19}{81} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/(3*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 5062

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]

Rule 99

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 151

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] && LtQ[m, -1] && IntegerQ[m]

Rule 12

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

Rule 93

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 210

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*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
 Dist[(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 634

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 618

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 204

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

Rule 628

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 206

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

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{3} i \tan ^{-1}(x)}}{x^4} \, dx &=\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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{19}{324} i \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \tanh ^{-1}\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 \operatorname{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 \operatorname{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 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{54 \sqrt{3}}+\frac{19 i \tan ^{-1}\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 \tanh ^{-1}\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]  time = 0.022007, size = 81, normalized size = 0.25 \[ \frac{(1-i x)^{5/6} \left (38 i x^3 \text{Hypergeometric2F1}\left (\frac{5}{6},1,\frac{11}{6},\frac{x+i}{-x+i}\right )+5 \left (22 i x^3+43 x^2-39 i x-18\right )\right )}{270 (1+i x)^{5/6} x^3} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\sqrt [3]{{(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.73565, size = 733, normalized size = 2.3 \begin{align*} \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 ) -{\left (19 \, \sqrt{3} x^{3} - 19 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 ) -{\left (19 \, \sqrt{3} x^{3} + 19 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 ) +{\left (19 \, \sqrt{3} x^{3} + 19 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 ) +{\left (19 \, \sqrt{3} x^{3} - 19 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 ) +{\left (-132 i \, x^{3} + 6 \, x^{2} - 18 i \, x - 108\right )} \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}}}{324 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 19*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 + 19
*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 + 19*I*x^3)*log(-1/2*I*sq
rt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + (19*sqrt(3)*x^3 - 19*I*x^3)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2
+ 1)/(x + I))^(1/3) - 1/2) + (-132*I*x^3 + 6*x^2 - 18*I*x - 108)*(I*sqrt(x^2 + 1)/(x + I))^(1/3))/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)