\(\int \frac {e^{i \arctan (a+b x)}}{x} \, dx\) [181]

Optimal result

Integrand size = 16, antiderivative size = 89 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=i \text {arcsinh}(a+b x)-\frac {2 \sqrt {i-a} \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{\sqrt {i+a}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\frac {2 (-1)^{3/4} \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{\sqrt {b}}-\frac {2 \sqrt {-1-i a} \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1+i a}} \] Input:

Integrate[E^(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*Sqrt[-1 - 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]
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.25, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5618, 140, 27, 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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 140

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

I*ArcSinh[(2*a*b + 2*b^2*x)/(2*b)] - ((2*I)*(1 + I*a)*ArcTanh[(Sqrt[I + a] 
*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/(Sqrt[I - a] 
*Sqrt[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 140

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

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {i b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {\left (i a +1\right ) \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\sqrt {a^{2}+1}}\)	\(107\)

Input:

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

Output:

I*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)- 
(1+I*a)/(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. 144 vs. \(2 (59) = 118\).

Time = 0.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.62 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\sqrt {-\frac {a - i}{a + i}} \log \left (-b x + {\left (i \, a - 1\right )} \sqrt {-\frac {a - i}{a + i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt {-\frac {a - i}{a + i}} \log \left (-b x + {\left (-i \, a + 1\right )} \sqrt {-\frac {a - i}{a + i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - i \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (59) = 118\).

Time = 0.03 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.62 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=-\frac {i \, a \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {\operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} + i \, \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right ) \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.27 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=-\frac {{\left (-i \, a - 1\right )} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {i \, b \log \left ({\left | -a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |} \right |}\right )}{{\left | b \right |}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 23.89 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.33 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\mathrm {asinh}\left (a+b\,x\right )\,1{}\mathrm {i}-\frac {\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )}{\sqrt {a^2+1}}-\frac {a\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )\,1{}\mathrm {i}}{\sqrt {a^2+1}} \] Input:

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

Output:

asinh(a + b*x)*1i - log(a*b + (a^2 + 1)/x + ((a^2 + 1)^(1/2)*(a^2 + b^2*x^ 
2 + 2*a*b*x + 1)^(1/2))/x)/(a^2 + 1)^(1/2) - (a*log(a*b + (a^2 + 1)/x + (( 
a^2 + 1)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2))/x)*1i)/(a^2 + 1)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.71 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\frac {-2 \sqrt {a^{2}+1}\, \mathit {atan} \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, i +b i x}{\sqrt {a^{2}+1}}\right ) a +2 \sqrt {a^{2}+1}\, \mathit {atan} \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, i +b i x}{\sqrt {a^{2}+1}}\right ) i +\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{2} i +\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) i}{a^{2}+1} \] Input:

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

Output:

( - 2*sqrt(a**2 + 1)*atan((sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*i + b*i*x) 
/sqrt(a**2 + 1))*a + 2*sqrt(a**2 + 1)*atan((sqrt(a**2 + 2*a*b*x + b**2*x** 
2 + 1)*i + b*i*x)/sqrt(a**2 + 1))*i + log(sqrt(a**2 + 2*a*b*x + b**2*x**2 
+ 1) + a + b*x)*a**2*i + log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b* 
x)*i)/(a**2 + 1)