\(\int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx\) [120]

Optimal result

Integrand size = 14, antiderivative size = 280 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{36} \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\frac {1}{36} \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \]

[Out]

-1/2*(1-I*x)^(5/6)*(1+I*x)^(7/6)/x^2-1/6*I*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x+1/9*arctanh((1+I*x)^(1/6)/(1-I*x)^(1/
6))-1/36*ln(1-(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^(1/3))+1/36*ln(1+(1+I*x)^(1/6)/(1-I*x)^(1/6)+(
1+I*x)^(1/3)/(1-I*x)^(1/3))-1/18*arctan(1/3*(1-2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)+1/18*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 = 280, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5170, 98, 96, 95, 216, 648, 632, 210, 642, 212} \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {1}{36} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac {1}{36} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right ) \]

[In]

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

[Out]

-1/2*((1 - I*x)^(5/6)*(1 + I*x)^(7/6))/x^2 - ((I/6)*(1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x - ArcTan[(1 - (2*(1 + I
*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]]/(6*Sqrt[3]) + ArcTan[(1 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]]/(
6*Sqrt[3]) + ArcTanh[(1 + I*x)^(1/6)/(1 - I*x)^(1/6)]/9 - Log[1 - (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^
(1/3)/(1 - I*x)^(1/3)]/36 + Log[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]/36

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 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^3} \, dx \\ & = -\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}+\frac {1}{6} i \int \frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x^2} \, dx \\ & = -\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {1}{18} \int \frac {1}{\sqrt [6]{1-i x} (1+i x)^{5/6} x} \, dx \\ & = -\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {1}{3} \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} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}+\frac {1}{9} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {1}{9} \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}{9} \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} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{36} \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 {1}{36} \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 {1}{12} \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}{12} \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} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{36} \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 {1}{36} \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 {1}{6} \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}{6} \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} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{36} \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 {1}{36} \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 = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.26 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=\frac {(1-i x)^{5/6} \left (5 \left (-3-7 i x+4 x^2\right )+2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{30 (1+i x)^{5/6} x^2} \]

[In]

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

[Out]

((1 - I*x)^(5/6)*(5*(-3 - (7*I)*x + 4*x^2) + 2*x^2*Hypergeometric2F1[5/6, 1, 11/6, (I + x)/(I - x)]))/(30*(1 +
I*x)^(5/6)*x^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=\frac {2 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) - 2 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) + {\left (i \, \sqrt {3} x^{2} + x^{2}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + {\left (i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) + {\left (-i \, \sqrt {3} x^{2} + x^{2}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + {\left (-i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - 6 \, {\left (4 \, x^{2} + i \, x + 3\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}}}{36 \, x^{2}} \]

[In]

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

[Out]

1/36*(2*x^2*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1) - 2*x^2*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1) + (I*sqr
t(3)*x^2 + x^2)*log(1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + (I*sqrt(3)*x^2 - x^2)*log(1/2*I*s
qrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) + (-I*sqrt(3)*x^2 + x^2)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)
/(x + I))^(1/3) + 1/2) + (-I*sqrt(3)*x^2 - x^2)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) -
6*(4*x^2 + I*x + 3)*(I*sqrt(x^2 + 1)/(x + I))^(1/3))/x^2

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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