3.2.21 \(\int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx\) [121]

3.2.21.1 Optimal result
3.2.21.2 Mathematica [C] (verified)
3.2.21.3 Rubi [A] (verified)
3.2.21.4 Maple [F]
3.2.21.5 Fricas [A] (verification not implemented)
3.2.21.6 Sympy [F(-1)]
3.2.21.7 Maxima [F]
3.2.21.8 Giac [F]
3.2.21.9 Mupad [F(-1)]

3.2.21.1 Optimal result

Integrand size = 14, antiderivative size = 319 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{18 x^2}+\frac {11 (1-i x)^{5/6} \sqrt [6]{1+i x}}{27 x}-\frac {19 i \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19 i \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{54 \sqrt {3}}+\frac {19}{81} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {19}{324} i \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\frac {19}{324} i \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \]

output
-1/3*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x^3-7/18*I*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x^ 
2+11/27*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x+19/81*I*arctanh((1+I*x)^(1/6)/(1-I*x 
)^(1/6))-19/324*I*ln(1-(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^( 
1/3))+19/324*I*ln(1+(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^(1/3 
))-19/162*I*arctan(1/3*(1-2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)+ 
19/162*I*arctan(1/3*(1+2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)
 
3.2.21.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.25 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\frac {(1-i x)^{5/6} \left (5 \left (-18-39 i x+43 x^2+22 i x^3\right )+38 i x^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{270 (1+i x)^{5/6} x^3} \]

input
Integrate[E^((I/3)*ArcTan[x])/x^4,x]
 
output
((1 - I*x)^(5/6)*(5*(-18 - (39*I)*x + 43*x^2 + (22*I)*x^3) + (38*I)*x^3*Hy 
pergeometric2F1[5/6, 1, 11/6, (I + x)/(I - x)]))/(270*(1 + I*x)^(5/6)*x^3)
 
3.2.21.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {5585, 110, 27, 168, 27, 168, 27, 104, 754, 27, 219, 1142, 25, 1083, 217, 1103}

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.

\(\displaystyle \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x^4}dx\)

\(\Big \downarrow \) 110

\(\displaystyle \frac {1}{3} \int \frac {7 i-6 x}{3 \sqrt [6]{1-i x} (i x+1)^{5/6} x^3}dx-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \frac {7 i-6 x}{\sqrt [6]{1-i x} (i x+1)^{5/6} x^3}dx-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{9} \left (-\frac {1}{2} \int \frac {21 i x+22}{3 \sqrt [6]{1-i x} (i x+1)^{5/6} x^2}dx-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (-\frac {1}{6} \int \frac {21 i x+22}{\sqrt [6]{1-i x} (i x+1)^{5/6} x^2}dx-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\int -\frac {19 i}{3 \sqrt [6]{1-i x} (i x+1)^{5/6} x}dx+\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-\frac {19}{3} i \int \frac {1}{\sqrt [6]{1-i x} (i x+1)^{5/6} x}dx\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \int \frac {1}{\frac {i x+1}{1-i x}-1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 754

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (\frac {1}{6} \left (\frac {1}{2} \int -\frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-1\right )-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (\frac {1}{6} \left (-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{9} \left (\frac {1}{6} \left (\frac {22 (1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-38 i \left (\frac {1}{6} \left (\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )\right )-\frac {7 i (1-i x)^{5/6} \sqrt [6]{1+i x}}{2 x^2}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{3 x^3}\)

input
Int[E^((I/3)*ArcTan[x])/x^4,x]
 
output
-1/3*((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x^3 + ((((-7*I)/2)*(1 - I*x)^(5/6)* 
(1 + I*x)^(1/6))/x^2 + ((22*(1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x - (38*I)*(- 
1/3*ArcTanh[(1 + I*x)^(1/6)/(1 - I*x)^(1/6)] + (-(Sqrt[3]*ArcTan[(-1 + (2* 
(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]]) + Log[1 - (1 + I*x)^(1/6)/(1 - 
 I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]/2)/6 + (-(Sqrt[3]*ArcTan[(1 
 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]]) - Log[1 + (1 + I*x)^(1/6 
)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]/2)/6))/6)/9
 

3.2.21.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 110
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 217
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 219
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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 754
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*Cos[(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] + 2*(r/(a*n))   Sum[u, {k, 1, (n - 2)/4}], x]] / 
; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

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]
 
3.2.21.4 Maple [F]

\[\int \frac {{\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )}^{\frac {1}{3}}}{x^{4}}d x\]

input
int(((1+I*x)/(x^2+1)^(1/2))^(1/3)/x^4,x)
 
output
int(((1+I*x)/(x^2+1)^(1/2))^(1/3)/x^4,x)
 
3.2.21.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\frac {38 i \, x^{3} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) - 38 i \, x^{3} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) - 19 \, {\left (\sqrt {3} x^{3} - i \, x^{3}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) - 19 \, {\left (\sqrt {3} x^{3} + i \, x^{3}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) + 19 \, {\left (\sqrt {3} x^{3} + i \, x^{3}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + 19 \, {\left (\sqrt {3} x^{3} - i \, x^{3}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - 6 \, {\left (22 i \, x^{3} - x^{2} + 3 i \, x + 18\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}}}{324 \, x^{3}} \]

input
integrate(((1+I*x)/(x^2+1)^(1/2))^(1/3)/x^4,x, algorithm="fricas")
 
output
1/324*(38*I*x^3*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1) - 38*I*x^3*log((I 
*sqrt(x^2 + 1)/(x + I))^(1/3) - 1) - 19*(sqrt(3)*x^3 - I*x^3)*log(1/2*I*sq 
rt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) - 19*(sqrt(3)*x^3 + I*x^3)* 
log(1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) + 19*(sqrt(3)*x 
^3 + I*x^3)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + 
19*(sqrt(3)*x^3 - I*x^3)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1 
/3) - 1/2) - 6*(22*I*x^3 - x^2 + 3*I*x + 18)*(I*sqrt(x^2 + 1)/(x + I))^(1/ 
3))/x^3
 
3.2.21.6 Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\text {Timed out} \]

input
integrate(((1+I*x)/(x**2+1)**(1/2))**(1/3)/x**4,x)
 
output
Timed out
 
3.2.21.7 Maxima [F]

\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{4}} \,d x } \]

input
integrate(((1+I*x)/(x^2+1)^(1/2))^(1/3)/x^4,x, algorithm="maxima")
 
output
integrate(((I*x + 1)/sqrt(x^2 + 1))^(1/3)/x^4, x)
 
3.2.21.8 Giac [F]

\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{4}} \,d x } \]

input
integrate(((1+I*x)/(x^2+1)^(1/2))^(1/3)/x^4,x, algorithm="giac")
 
output
integrate(((I*x + 1)/sqrt(x^2 + 1))^(1/3)/x^4, x)
 
3.2.21.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^4} \, dx=\int \frac {{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{1/3}}{x^4} \,d x \]

input
int(((x*1i + 1)/(x^2 + 1)^(1/2))^(1/3)/x^4,x)
 
output
int(((x*1i + 1)/(x^2 + 1)^(1/2))^(1/3)/x^4, x)