\(\int \frac {e^{-i \arctan (a+b x)}}{x^3} \, dx\) [210]

Optimal result

Integrand size = 16, antiderivative size = 201 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^3} \, dx=\frac {(1-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 (i-a)^2 (i+a) x}-\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 \left (1+a^2\right ) x^2}+\frac {(1-2 i a) b^2 \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{5/2} (i+a)^{3/2}} \] Output:

1/2*(1-2*I*a)*b*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^2/(I+a)/x-1/ 
2*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/(a^2+1)/x^2+(1-2*I*a)*b^2*arctan 
h((I+a)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^(1/2)/(1-I*a-I*b*x)^(1/2))/(I-a)^( 
5/2)/(I+a)^(3/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^3} \, dx=\frac {\frac {i \left (1+a^2-2 i b x-a b x\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}}{x^2}+\frac {2 (i+2 a) b^2 \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a} \sqrt {-1+i a}}}{2 (-i+a)^2 (i+a)} \] Input:

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

Output:

((I*(1 + a^2 - (2*I)*b*x - a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/x^2 + 
 (2*(I + 2*a)*b^2*ArcTanh[(Sqrt[-1 - I*a]*Sqrt[(-I)*(I + a + b*x)])/(Sqrt[ 
-1 + I*a]*Sqrt[1 + I*a + I*b*x])])/(Sqrt[-1 - I*a]*Sqrt[-1 + I*a]))/(2*(-I 
 + a)^2*(I + a))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5618, 107, 105, 104, 221}

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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(2 a+i) b \left (-\frac {2 i b \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{3/2} \sqrt {a+i}}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1+i a) x}\right )}{2 \left (a^2+1\right )}-\frac {\sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{2 \left (a^2+1\right ) x^2}\)

Input:

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

Output:

-1/2*((1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/((1 + a^2)*x^2) - ((I 
 + 2*a)*b*(-((Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/((1 + I*a)*x)) 
- ((2*I)*b*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 
 - I*a - I*b*x])])/((I - a)^(3/2)*Sqrt[I + a])))/(2*(1 + a^2))
 

Defintions of rubi rules used

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 5618

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), 
 x_Symbol] :> Int[(d + e*x)^m*((1 - I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c + 
 I*b*c*x)^(I*(n/2))), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {i \left (-a \,b^{3} x^{3}-2 i b^{3} x^{3}-a^{2} b^{2} x^{2}-4 i a \,b^{2} x^{2}+a^{3} b x -2 i a^{2} b x +a^{4}+b^{2} x^{2}+a b x -2 i b x +2 a^{2}+1\right )}{2 x^{2} \left (i+a \right ) \left (-i+a \right )^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}-\frac {b^{2} \left (i+2 a \right ) \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}} \left (-i+a \right )}\)	\(187\)
default	\(\frac {i \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (a^{2}+1\right ) x^{2}}-\frac {a b \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {a b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{a^{2}+1}+\frac {2 b^{2} \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{a^{2}+1}\right )}{2 \left (a^{2}+1\right )}+\frac {b^{2} \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{2 a^{2}+2}\right )}{i-a}-\frac {i b^{2} \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{\left (i-a \right )^{3}}+\frac {i b \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {a b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{a^{2}+1}+\frac {2 b^{2} \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{a^{2}+1}\right )}{\left (i-a \right )^{2}}+\frac {i b^{2} \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{\left (i-a \right )^{3}}\)	\(1008\)

Input:

int(1/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

1/2*I*(-a*b^3*x^3-a^2*b^2*x^2+a^3*b*x-2*I*b^3*x^3+a^4+b^2*x^2-4*I*a*b^2*x^ 
2+a*b*x-2*I*a^2*b*x+2*a^2-2*I*b*x+1)/x^2/(I+a)/(-I+a)^2/(b^2*x^2+2*a*b*x+a 
^2+1)^(1/2)-1/2*b^2*(I+2*a)/(a^2+1)^(3/2)/(-I+a)*ln((2*a^2+2+2*a*b*x+2*(a^ 
2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (135) = 270\).

Time = 0.15 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.25 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^3} \, dx=\frac {{\left (-i \, a + 2\right )} b^{2} x^{2} + \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}} {\left (a^{3} - i \, a^{2} + a - i\right )} x^{2} \log \left (-\frac {{\left (2 \, a + i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + i\right )} b^{2} + {\left (a^{5} - i \, a^{4} + 2 \, a^{3} - 2 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}}}{{\left (2 \, a + i\right )} b^{2}}\right ) - \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}} {\left (a^{3} - i \, a^{2} + a - i\right )} x^{2} \log \left (-\frac {{\left (2 \, a + i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + i\right )} b^{2} - {\left (a^{5} - i \, a^{4} + 2 \, a^{3} - 2 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 4 i \, a - 1\right )} b^{4}}{a^{8} - 2 i \, a^{7} + 2 \, a^{6} - 6 i \, a^{5} - 6 i \, a^{3} - 2 \, a^{2} - 2 i \, a - 1}}}{{\left (2 \, a + i\right )} b^{2}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (-i \, a + 2\right )} b x + i \, a^{2} + i\right )}}{2 \, {\left (a^{3} - i \, a^{2} + a - i\right )} x^{2}} \] Input:

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

Output:

1/2*((-I*a + 2)*b^2*x^2 + sqrt((4*a^2 + 4*I*a - 1)*b^4/(a^8 - 2*I*a^7 + 2* 
a^6 - 6*I*a^5 - 6*I*a^3 - 2*a^2 - 2*I*a - 1))*(a^3 - I*a^2 + a - I)*x^2*lo 
g(-((2*a + I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + I)*b^2 + (a 
^5 - I*a^4 + 2*a^3 - 2*I*a^2 + a - I)*sqrt((4*a^2 + 4*I*a - 1)*b^4/(a^8 - 
2*I*a^7 + 2*a^6 - 6*I*a^5 - 6*I*a^3 - 2*a^2 - 2*I*a - 1)))/((2*a + I)*b^2) 
) - sqrt((4*a^2 + 4*I*a - 1)*b^4/(a^8 - 2*I*a^7 + 2*a^6 - 6*I*a^5 - 6*I*a^ 
3 - 2*a^2 - 2*I*a - 1))*(a^3 - I*a^2 + a - I)*x^2*log(-((2*a + I)*b^3*x - 
sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + I)*b^2 - (a^5 - I*a^4 + 2*a^3 - 2 
*I*a^2 + a - I)*sqrt((4*a^2 + 4*I*a - 1)*b^4/(a^8 - 2*I*a^7 + 2*a^6 - 6*I* 
a^5 - 6*I*a^3 - 2*a^2 - 2*I*a - 1)))/((2*a + I)*b^2)) + sqrt(b^2*x^2 + 2*a 
*b*x + a^2 + 1)*((-I*a + 2)*b*x + I*a^2 + I))/((a^3 - I*a^2 + a - I)*x^2)
 

Sympy [F]

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

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

Output:

-I*Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a*x**3 + b*x**4 - I*x**3 
), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (135) = 270\).

Time = 0.18 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.34 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^3} \, dx=-\frac {{\left (2 \, a b^{2} + i \, b^{2}\right )} \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{2 \, {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {a^{2} + 1}} - \frac {4 \, {\left (i \, x {\left | b \right |} - i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{4} b^{2} + 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{3} b {\left | b \right |} + 2 i \, a^{5} b {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} a b^{2} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{3} b^{2} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a^{2} b {\left | b \right |} - 2 \, a^{4} b {\left | b \right |} + i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} b^{2} + 5 \, {\left (i \, x {\left | b \right |} - i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} b^{2} + 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b {\left | b \right |} + 4 i \, a^{3} b {\left | b \right |} - 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b^{2} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b {\left | b \right |} - 4 \, a^{2} b {\left | b \right |} - {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} b^{2} + 2 i \, a b {\left | b \right |} - 2 \, b {\left | b \right |}}{{\left (a^{3} - i \, a^{2} + a - i\right )} {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )}^{2}} \] Input:

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

Output:

-1/2*(2*a*b^2 + I*b^2)*log(abs(2*x*abs(b) - 2*sqrt((b*x + a)^2 + 1) - 2*sq 
rt(a^2 + 1))/abs(2*x*abs(b) - 2*sqrt((b*x + a)^2 + 1) + 2*sqrt(a^2 + 1)))/ 
((a^3 - I*a^2 + a - I)*sqrt(a^2 + 1)) - (4*(I*x*abs(b) - I*sqrt((b*x + a)^ 
2 + 1))*a^4*b^2 + 2*I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a^3*b*abs(b) + 
2*I*a^5*b*abs(b) + 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a*b^2 - 2*(x*abs 
(b) - sqrt((b*x + a)^2 + 1))*a^3*b^2 + 2*(x*abs(b) - sqrt((b*x + a)^2 + 1) 
)^2*a^2*b*abs(b) - 2*a^4*b*abs(b) + I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3 
*b^2 + 5*(I*x*abs(b) - I*sqrt((b*x + a)^2 + 1))*a^2*b^2 + 2*I*(x*abs(b) - 
sqrt((b*x + a)^2 + 1))^2*a*b*abs(b) + 4*I*a^3*b*abs(b) - 2*(x*abs(b) - sqr 
t((b*x + a)^2 + 1))*a*b^2 + 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b*abs(b 
) - 4*a^2*b*abs(b) - (-I*x*abs(b) + I*sqrt((b*x + a)^2 + 1))*b^2 + 2*I*a*b 
*abs(b) - 2*b*abs(b))/((a^3 - I*a^2 + a - I)*((x*abs(b) - sqrt((b*x + a)^2 
 + 1))^2 - a^2 - 1)^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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