\(\int \frac {e^{-3 i \arctan (a+b x)}}{x} \, dx\) [226]

Optimal result

Integrand size = 16, antiderivative size = 134 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x} \, dx=\frac {4 \sqrt {1-i a-i b x}}{(1+i a) \sqrt {1+i a+i b x}}+i \text {arcsinh}(a+b x)-\frac {2 (i+a)^{3/2} \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x} \, dx=\frac {2 (-1)^{3/4} \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{\sqrt {b}}+\frac {2 \left (-\frac {2 \sqrt {1+a^2+2 a b x+b^2 x^2}}{-i+a+b x}+\frac {\sqrt {-1+i a} (i+a) \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a}}\right )}{-i+a} \] Input:

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

Output:

(2*(-1)^(3/4)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + 
 b*x)])/Sqrt[(-I)*b]])/Sqrt[b] + (2*((-2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] 
)/(-I + a + b*x) + (Sqrt[-1 + I*a]*(I + a)*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] 
))/(-I + a)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5618, 109, 27, 175, 62, 104, 221, 1090, 222}

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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {2 \int -\frac {b \left (i (a+i)^2+(i a+1) b x\right )}{2 x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{(-a+i) b}+\frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {\int \frac {i (a+i)^2+(i a+1) b x}{x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{-a+i}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {i (a+i)^2 \int \frac {1}{x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx+(1+i a) b \int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{-a+i}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {(1+i a) b \int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx+i (a+i)^2 \int \frac {1}{x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{-a+i}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {(1+i a) b \int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx+2 i (a+i)^2 \int \frac {1}{-i a+\frac {(1-i a) (i a+i b x+1)}{-i a-i b x+1}-1}d\frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}{-a+i}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {(1+i a) b \int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx+\frac {2 (a+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i}}}{-a+i}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {\frac {(1+i a) \int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b}+\frac {2 (a+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i}}}{-a+i}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {4 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {(1+i a) \text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )+\frac {2 (a+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i}}}{-a+i}\)

Input:

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

Output:

(4*Sqrt[1 - I*a - I*b*x])/((1 + I*a)*Sqrt[1 + I*a + I*b*x]) - ((1 + I*a)*A 
rcSinh[(2*a*b + 2*b^2*x)/(2*b)] + (2*(I + a)^(3/2)*ArcTanh[(Sqrt[I + a]*Sq 
rt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/Sqrt[I - a])/(I 
 - a)
 

Defintions of rubi rules used

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 62

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 
1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b 
+ d, 0] && GtQ[a + c, 0]
 

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 109

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

rule 175

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

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 222

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

rule 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

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 [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1066 vs. \(2 (104 ) = 208\).

Time = 0.79 (sec) , antiderivative size = 1067, normalized size of antiderivative = 7.96

method	result	size
default	\(-\frac {i \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3}+a b \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 )+\left (a^{2}+1\right ) \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 )\right )}{\left (i-a \right )^{3}}-\frac {i \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )}{\left (i-a \right )^{2} b}+\frac {i \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )}{\left (i-a \right )^{3}}+\frac {i \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{\left (i-a \right ) b^{2}}\)	\(1067\)

Input:

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

Output:

-I/(I-a)^3*(1/3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+a*b*(1/4*(2*b^2*x+2*a*b)/b^2 
*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/8*(4*b^2*(a^2+1)-4*a^2*b^2)/b^2*ln((b^2*x 
+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2))+(a^2+1)*((b^ 
2*x^2+2*a*b*x+a^2+1)^(1/2)+a*b*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x 
+a^2+1)^(1/2))/(b^2)^(1/2)-(a^2+1)^(1/2)*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)))-I/(I-a)^2/b*(-I/b/(x-(I-a)/b)^2*((x 
-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)+3*I*b*(1/3*((x-(I-a)/b)^2*b^2+2*I 
*b*(x-(I-a)/b))^(3/2)+I*b*(1/4*(2*(x-(I-a)/b)*b^2+2*I*b)/b^2*((x-(I-a)/b)^ 
2*b^2+2*I*b*(x-(I-a)/b))^(1/2)+1/2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+(( 
x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))/(b^2)^(1/2))))+I/(I-a)^3*(1/3*( 
(x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)+I*b*(1/4*(2*(x-(I-a)/b)*b^2+2*I 
*b)/b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)+1/2*ln((I*b+(x-(I-a)/b 
)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))/(b^2)^(1/2 
)))+I/(I-a)/b^2*(I/b/(x-(I-a)/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^( 
5/2)-2*I*b*(-I/b/(x-(I-a)/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2) 
+3*I*b*(1/3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)+I*b*(1/4*(2*(x-(I- 
a)/b)*b^2+2*I*b)/b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)+1/2*ln((I 
*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2 
))/(b^2)^(1/2)))))
 

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. 356 vs. \(2 (90) = 180\).

Time = 0.15 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.66 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x} \, dx=\frac {{\left ({\left (a - i\right )} b x + a^{2} - 2 i \, a - 1\right )} \sqrt {-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}} \log \left (-\frac {{\left (a + i\right )} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a + i\right )} - {\left (i \, a^{2} + 2 \, a - i\right )} \sqrt {-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}}}{a + i}\right ) - {\left ({\left (a - i\right )} b x + a^{2} - 2 i \, a - 1\right )} \sqrt {-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}} \log \left (-\frac {{\left (a + i\right )} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a + i\right )} - {\left (-i \, a^{2} - 2 \, a + i\right )} \sqrt {-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}}}{a + i}\right ) - 4 \, b x - {\left ({\left (i \, a + 1\right )} b x + i \, a^{2} + 2 \, a - i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, a - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 4 i}{{\left (a - i\right )} b x + a^{2} - 2 i \, a - 1} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x + 3*a**2*b*x**2 - 
 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 - 3*a*x + b**3*x**4 - 3*I*b**2* 
x**3 - 3*b*x**2 + I*x), x) + Integral(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 
 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 
- 3*a*x + b**3*x**4 - 3*I*b**2*x**3 - 3*b*x**2 + I*x), x) + Integral(b**2* 
x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x + 3*a**2*b*x**2 - 3*I*a* 
*2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 - 3*a*x + b**3*x**4 - 3*I*b**2*x**3 - 
3*b*x**2 + I*x), x) + Integral(2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1 
)/(a**3*x + 3*a**2*b*x**2 - 3*I*a**2*x + 3*a*b**2*x**3 - 6*I*a*b*x**2 - 3* 
a*x + b**3*x**4 - 3*I*b**2*x**3 - 3*b*x**2 + I*x), x))
 

Maxima [F]

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

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

Output:

integrate(((b*x + a)^2 + 1)^(3/2)/((I*b*x + I*a + 1)^3*x), 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. 252 vs. \(2 (90) = 180\).

Time = 0.25 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x} \, dx=-\frac {i \, b \log \left (-3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b - a^{3} b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} + 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 i \, a^{2} b - 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} + a b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{3 \, {\left | b \right |}} - \frac {{\left (-i \, a^{2} + 2 \, a + i\right )} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a - i\right )}} \] Input:

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

Output:

-1/3*I*b*log(-3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b - a^3*b - (x*abs( 
b) - sqrt((b*x + a)^2 + 1))^3*abs(b) - 3*(x*abs(b) - sqrt((b*x + a)^2 + 1) 
)*a^2*abs(b) + 2*I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b + 2*I*a^2*b - 4* 
(-I*x*abs(b) + I*sqrt((b*x + a)^2 + 1))*a*abs(b) + a*b + (x*abs(b) - sqrt( 
(b*x + a)^2 + 1))*abs(b))/abs(b) - (-I*a^2 + 2*a + I)*log(abs(-2*x*abs(b) 
+ 2*sqrt((b*x + a)^2 + 1) - 2*sqrt(a^2 + 1))/abs(-2*x*abs(b) + 2*sqrt((b*x 
 + a)^2 + 1) + 2*sqrt(a^2 + 1)))/(sqrt(a^2 + 1)*(a - I))
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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