\(\int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx\) [109]

Optimal result

Integrand size = 16, antiderivative size = 132 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:

1/4*I*a*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x-1/2*(1-I*a*x)^(5/4)*(1+I*a*x)^(3 
/4)/x^2-1/4*a^2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+1/4*a^2*arctanh((1 
+I*a*x)^(1/4)/(1-I*a*x)^(1/4))
 

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {\sqrt [4]{1-i a x} \left (-2+i a x-3 a^2 x^2+2 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{4 x^2 \sqrt [4]{1+i a x}} \] Input:

Integrate[1/(E^((I/2)*ArcTan[a*x])*x^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5585, 107, 105, 104, 25, 827, 216, 219}

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.

\(\displaystyle \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx\)

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 107

\(\displaystyle -\frac {1}{4} i a \int \frac {\sqrt [4]{1-i a x}}{x^2 \sqrt [4]{i a x+1}}dx-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {1}{4} i a \left (-\frac {1}{2} i a \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {1}{4} i a \left (-2 i a \int -\frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{4} i a \left (2 i a \int \frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\)

Input:

Int[1/(E^((I/2)*ArcTan[a*x])*x^3),x]
 

Output:

-1/2*((1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/x^2 - (I/4)*a*(-(((1 - I*a*x)^( 
1/4)*(1 + I*a*x)^(3/4))/x) - (2*I)*a*(ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x) 
^(1/4)]/2 - ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 105

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[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

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

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 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]

\[\int \frac {1}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{3}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-3 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \] Input:

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

Output:

1/8*(a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - I*a^2*x^2*log( 
sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + I*a^2*x^2*log(sqrt(I*sqrt(a^2*x 
^2 + 1)/(a*x + I)) - I) - a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) 
- 1) - 2*sqrt(a^2*x^2 + 1)*(-3*I*a*x + 2)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + 
I)))/x^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {6 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}} a i x -4 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}}-\left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}}}{a^{2} x^{3}+x}d x \right ) a^{2} x^{2}}{8 x^{2}} \] Input:

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

Output:

(6*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(1/4)*a*i*x - 4*sqrt(a*i*x + 1)*(a**2* 
x**2 + 1)**(1/4) - int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(1/4))/(a**2*x**3 
 + x),x)*a**2*x**2)/(8*x**2)