\(\int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx\) [137]

Optimal result

Integrand size = 14, antiderivative size = 177 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=-\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x+\frac {22 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x) \] Output:

-11/27*I*(1-I*x)^(2/3)*(1+I*x)^(1/3)-1/9*I*(1-I*x)^(2/3)*(1+I*x)^(4/3)+1/3 
*(1-I*x)^(2/3)*(1+I*x)^(4/3)*x+22/81*I*arctan(1/3*3^(1/2)-2/3*(1-I*x)^(1/3 
)*3^(1/2)/(1+I*x)^(1/3))*3^(1/2)+11/27*I*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3)) 
+11/81*I*ln(1+I*x)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\frac {1}{18} (1-i x)^{2/3} \left (2 \sqrt [3]{1+i x} \left (-i+4 x+3 i x^2\right )-11 i \sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )\right ) \] Input:

Integrate[E^(((2*I)/3)*ArcTan[x])*x^2,x]
 

Output:

((1 - I*x)^(2/3)*(2*(1 + I*x)^(1/3)*(-I + 4*x + (3*I)*x^2) - (11*I)*2^(1/3 
)*Hypergeometric2F1[-1/3, 2/3, 5/3, 1/2 - (I/2)*x]))/18
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5585, 101, 27, 90, 60, 72}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 e^{\frac {2}{3} i \arctan (x)} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {\sqrt [3]{1+i x} x^2}{\sqrt [3]{1-i x}}dx\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {1}{3} \int -\frac {\sqrt [3]{i x+1} (2 i x+3)}{3 \sqrt [3]{1-i x}}dx+\frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x-\frac {1}{9} \int \frac {\sqrt [3]{i x+1} (2 i x+3)}{\sqrt [3]{1-i x}}dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {1}{9} \left (-\frac {11}{3} \int \frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}dx-i (1-i x)^{2/3} (1+i x)^{4/3}\right )+\frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{9} \left (-\frac {11}{3} \left (\frac {2}{3} \int \frac {1}{\sqrt [3]{1-i x} (i x+1)^{2/3}}dx+i (1-i x)^{2/3} \sqrt [3]{1+i x}\right )-i (1-i x)^{2/3} (1+i x)^{4/3}\right )+\frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}\)

\(\Big \downarrow \) 72

\(\displaystyle \frac {1}{9} \left (-\frac {11}{3} \left (\frac {2}{3} \left (-i \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )-\frac {3}{2} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{2} i \log (1+i x)\right )+i (1-i x)^{2/3} \sqrt [3]{1+i x}\right )-i (1-i x)^{2/3} (1+i x)^{4/3}\right )+\frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}\)

Input:

Int[E^(((2*I)/3)*ArcTan[x])*x^2,x]
 

Output:

((1 - I*x)^(2/3)*(1 + I*x)^(4/3)*x)/3 + ((-I)*(1 - I*x)^(2/3)*(1 + I*x)^(4 
/3) - (11*(I*(1 - I*x)^(2/3)*(1 + I*x)^(1/3) + (2*((-I)*Sqrt[3]*ArcTan[1/S 
qrt[3] - (2*(1 - I*x)^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))] - ((3*I)/2)*Log[1 
+ (1 - I*x)^(1/3)/(1 + I*x)^(1/3)] - (I/2)*Log[1 + I*x]))/3))/3)/9
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 72

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> 
 With[{q = Rt[-d/b, 3]}, Simp[Sqrt[3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a + b* 
x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 5585

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] &&  !Intege 
rQ[(I*n - 1)/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.66 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=-\frac {11}{81} \, {\left (\sqrt {3} + i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {11}{81} \, {\left (\sqrt {3} - i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{27} \, {\left (9 \, x^{3} - 3 i \, x^{2} - 2 \, x - 14 i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {22}{81} i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \] Input:

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

Output:

-11/81*(sqrt(3) + I)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 
 1/2) + 11/81*(sqrt(3) - I)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sq 
rt(3) - 1/2) + 1/27*(9*x^3 - 3*I*x^2 - 2*x - 14*I)*(I*sqrt(x^2 + 1)/(x + I 
))^(2/3) + 22/81*I*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)
 

Sympy [F(-1)]

Timed out. \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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