\(\int e^{\frac {1}{3} i \arctan (x)} \, dx\) [117]

Optimal result

Integrand size = 10, antiderivative size = 262 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=i (1-i x)^{5/6} \sqrt [6]{1+i x}-\frac {1}{3} i \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{3} i \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}}-\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}} \]

[Out]

I*(1-I*x)^(5/6)*(1+I*x)^(1/6)+2/3*I*arctan((1-I*x)^(1/6)/(1+I*x)^(1/6))+1/3*I*arctan(2*(1-I*x)^(1/6)/(1+I*x)^(
1/6)-3^(1/2))+1/3*I*arctan(2*(1-I*x)^(1/6)/(1+I*x)^(1/6)+3^(1/2))+1/6*I*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3)-(1-I*
x)^(1/6)*3^(1/2)/(1+I*x)^(1/6))*3^(1/2)-1/6*I*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3)+(1-I*x)^(1/6)*3^(1/2)/(1+I*x)^(
1/6))*3^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5169, 52, 65, 338, 301, 648, 632, 210, 642, 209} \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=-\frac {1}{3} i \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{3} i \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+i (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac {i \log \left (\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{2 \sqrt {3}}-\frac {i \log \left (\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{2 \sqrt {3}} \]

[In]

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

[Out]

I*(1 - I*x)^(5/6)*(1 + I*x)^(1/6) - (I/3)*ArcTan[Sqrt[3] - (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] + (I/3)*ArcTan
[Sqrt[3] + (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] + ((2*I)/3)*ArcTan[(1 - I*x)^(1/6)/(1 + I*x)^(1/6)] + ((I/2)*L
og[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3) - (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)])/Sqrt[3] - ((I/2)*Log[1 +
(1 - I*x)^(1/3)/(1 + I*x)^(1/3) + (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)])/Sqrt[3]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 301

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

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 5169

Int[E^(ArcTan[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2)), x] /; FreeQ[{a
, 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}} \, dx \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac {1}{3} \int \frac {1}{\sqrt [6]{1-i x} (1+i x)^{5/6}} \, dx \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+2 i \text {Subst}\left (\int \frac {x^4}{\left (2-x^6\right )^{5/6}} \, dx,x,\sqrt [6]{1-i x}\right ) \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+2 i \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac {2}{3} 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 {2}{3} i \text {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {2}{3} i \text {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{6} i \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{6} i \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {i \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}}-\frac {i \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}} \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}}-\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}}-\frac {1}{3} i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\frac {1}{3} i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = i (1-i x)^{5/6} \sqrt [6]{1+i x}-\frac {1}{3} i \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{3} i \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}}-\frac {i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )}{2 \sqrt {3}} \\ \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 = 34, normalized size of antiderivative = 0.13 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=-\frac {12}{7} i e^{\frac {7}{3} i \arctan (x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{6},2,\frac {13}{6},-e^{2 i \arctan (x)}\right ) \]

[In]

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

[Out]

((-12*I)/7)*E^(((7*I)/3)*ArcTan[x])*Hypergeometric2F1[7/6, 2, 13/6, -E^((2*I)*ArcTan[x])]

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.74 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=\frac {1}{6} \, {\left (-i \, \sqrt {3} + 1\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (i \, \sqrt {3} + 1\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + {\left (x + i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{3} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i\right ) - \frac {1}{3} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - i\right ) \]

[In]

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

[Out]

1/6*(-I*sqrt(3) + 1)*log(1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2*I) + 1/6*(-I*sqrt(3) - 1)*log(1/2
*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2*I) + 1/6*(I*sqrt(3) + 1)*log(-1/2*sqrt(3) + (I*sqrt(x^2 + 1)/
(x + I))^(1/3) + 1/2*I) + 1/6*(I*sqrt(3) - 1)*log(-1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2*I) + (x
+ I)*(I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/3*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + I) - 1/3*log((I*sqrt(x^2 + 1
)/(x + I))^(1/3) - I)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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