Integrand size = 12, antiderivative size = 140 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\frac {1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}-\frac {1}{3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{9} \log (1+i x) \] Output:
1/3*(1-I*x)^(2/3)*(1+I*x)^(1/3)+1/2*(1-I*x)^(2/3)*(1+I*x)^(4/3)-2/9*arctan (1/3*3^(1/2)-2/3*(1-I*x)^(1/3)*3^(1/2)/(1+I*x)^(1/3))*3^(1/2)-1/3*ln(1+(1- I*x)^(1/3)/(1+I*x)^(1/3))-1/9*ln(1+I*x)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\frac {1}{2} (1-i x)^{2/3} \left ((1+i x)^{4/3}+\sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )\right ) \] Input:
Integrate[E^(((2*I)/3)*ArcTan[x])*x,x]
Output:
((1 - I*x)^(2/3)*((1 + I*x)^(4/3) + 2^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5 /3, 1/2 - (I/2)*x]))/2
Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5585, 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.
\(\displaystyle \int x e^{\frac {2}{3} i \arctan (x)} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {\sqrt [3]{1+i x} x}{\sqrt [3]{1-i x}}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {1}{3} i \int \frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {1}{3} i \left (\frac {2}{3} \int \frac {1}{\sqrt [3]{1-i x} (i x+1)^{2/3}}dx+i (1-i x)^{2/3} \sqrt [3]{1+i x}\right )\) |
\(\Big \downarrow \) 72 |
\(\displaystyle \frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {1}{3} i \left (\frac {2}{3} \left (-i \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )-\frac {3}{2} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{2} i \log (1+i x)\right )+i (1-i x)^{2/3} \sqrt [3]{1+i x}\right )\) |
Input:
Int[E^(((2*I)/3)*ArcTan[x])*x,x]
Output:
((1 - I*x)^(2/3)*(1 + I*x)^(4/3))/2 - (I/3)*(I*(1 - I*x)^(2/3)*(1 + I*x)^( 1/3) + (2*((-I)*Sqrt[3]*ArcTan[1/Sqrt[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)
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]
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]
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]
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]
\[\int {\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )}^{\frac {2}{3}} x d x\]
Input:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x)
Output:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x)
Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=-\frac {1}{9} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) - \frac {1}{9} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{6} \, {\left (3 \, x^{2} - 2 i \, x + 5\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {2}{9} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x, algorithm="fricas")
Output:
-1/9*(I*sqrt(3) - 1)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 1/2) - 1/9*(-I*sqrt(3) - 1)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*s qrt(3) - 1/2) + 1/6*(3*x^2 - 2*I*x + 5)*(I*sqrt(x^2 + 1)/(x + I))^(2/3) - 2/9*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)
Timed out. \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\text {Timed out} \] Input:
integrate(((1+I*x)/(x**2+1)**(1/2))**(2/3)*x,x)
Output:
Timed out
\[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\int { x \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}} \,d x } \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x, algorithm="maxima")
Output:
integrate(x*((I*x + 1)/sqrt(x^2 + 1))^(2/3), x)
\[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\int { x \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}} \,d x } \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x, algorithm="giac")
Output:
integrate(x*((I*x + 1)/sqrt(x^2 + 1))^(2/3), x)
Timed out. \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\int x\,{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{2/3} \,d x \] Input:
int(x*((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3),x)
Output:
int(x*((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3), x)
\[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\int \frac {\left (i x +1\right )^{\frac {2}{3}} x}{\left (x^{2}+1\right )^{\frac {1}{3}}}d x \] Input:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x,x)
Output:
int(((i*x + 1)**(2/3)*x)/(x**2 + 1)**(1/3),x)