3.122 \(\int e^{\frac {2}{3} i \tan ^{-1}(x)} x^2 \, dx\)

Optimal. Leaf size=177 \[ \frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}-\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x)+\frac {22 i \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5062, 90, 80, 50, 60} \[ \frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}-\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x)+\frac {22 i \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*I)/27)*(1 - I*x)^(2/3)*(1 + I*x)^(1/3) - (I/9)*(1 - I*x)^(2/3)*(1 + I*x)^(4/3) + ((1 - I*x)^(2/3)*(1 + I
*x)^(4/3)*x)/3 + (((22*I)/27)*ArcTan[1/Sqrt[3] - (2*(1 - I*x)^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))])/Sqrt[3] + ((1
1*I)/27)*Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3)] + ((11*I)/81)*Log[1 + I*x]

Rule 50

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 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[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 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]

Rubi steps

\begin {align*} \int e^{\frac {2}{3} i \tan ^{-1}(x)} x^2 \, dx &=\int \frac {\sqrt [3]{1+i x} x^2}{\sqrt [3]{1-i x}} \, dx\\ &=\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x+\frac {1}{3} \int \frac {\left (-1-\frac {2 i x}{3}\right ) \sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx\\ &=-\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 {11}{27} \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, 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}{81} \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, 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 \tan ^{-1}\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)\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 73, normalized size = 0.41 \[ \frac {1}{18} (1-i x)^{2/3} \left (2 \sqrt [3]{1+i x} \left (3 i x^2+4 x-i\right )-11 i \sqrt [3]{2} \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};\frac {1}{2}-\frac {i x}{2}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((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

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fricas [A]  time = 1.54, size = 121, normalized size = 0.68 \[ -\frac {1}{81} \, {\left (11 \, \sqrt {3} + 11 i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{81} \, {\left (11 \, \sqrt {3} - 11 i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{81} \, {\left (27 \, x^{3} - 9 i \, x^{2} - 6 \, x - 42 i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {22}{81} i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/81*(11*sqrt(3) + 11*I)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 1/2) + 1/81*(11*sqrt(3) - 11*I
)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sqrt(3) - 1/2) + 1/81*(27*x^3 - 9*I*x^2 - 6*x - 42*I)*(I*sqrt(x^
2 + 1)/(x + I))^(2/3) + 22/81*I*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )^{\frac {2}{3}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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