\(\int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx\) [209]

Optimal result

Integrand size = 16, antiderivative size = 130 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=-\frac {\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{(1+i a) x}-\frac {2 i b \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{3/2} \sqrt {i+a}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=i \left (\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{(-i+a) x}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{(-1-i a)^{3/2} \sqrt {-1+i a}}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5618, 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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((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])
 

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 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 = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (-i+a \right ) x}+\frac {b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (-i+a \right ) \sqrt {a^{2}+1}}\)	\(93\)
default	\(\frac {i \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {a b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{a^{2}+1}+\frac {2 b^{2} \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{a^{2}+1}\right )}{i-a}-\frac {i b \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{\left (i-a \right )^{2}}+\frac {i b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{\left (i-a \right )^{2}}\)	\(546\)

Input:

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

Output:

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

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. 224 vs. \(2 (86) = 172\).

Time = 0.14 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.72 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=-\frac {{\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b + {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - {\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b - {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - i \, b x - i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a - i\right )} x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=\frac {b \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a - i\right )}} - \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b + a^{2} {\left | b \right |} + {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )} {\left (-i \, a - 1\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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