\(\int \frac {e^{-i \arctan (a+b x)}}{x} \, dx\) [194]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5203, 132, 55, 633, 221, 12, 95, 214} \[ \int \frac {e^{-i \arctan (a+b x)}}{x} \, dx=-i \text {arcsinh}(a+b x)-\frac {2 \sqrt {a+i} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i}} \]

[In]

Int[1/(E^(I*ArcTan[a + b*x])*x),x]

[Out]

(-I)*ArcSinh[a + b*x] - (2*Sqrt[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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 55

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[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)*ExpandTo
Sum[(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 214

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 221

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 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[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 5203

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1-i a-i b x}}{x \sqrt {1+i a+i b x}} \, dx \\ & = -\left ((i b) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx\right )+\int \frac {1-i a}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx \\ & = (1-i a) \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx-(i b) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx \\ & = (2 (1-i a)) \text {Subst}\left (\int \frac {1}{-1-i a-(-1+i a) x^2} \, dx,x,\frac {\sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}}\right )-\frac {i \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b} \\ & = -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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {e^{-i \arctan (a+b x)}}{x} \, dx=\frac {2 \sqrt [4]{-1} (-i b)^{3/2} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{3/2}}-\frac {2 \sqrt {-1+i a} \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (68 ) = 136\).

Time = 0.46 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.92

method	result	size
default	\(\frac {i \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{i-a}-\frac {i \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{i-a}\)	\(260\)

[In]

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

[Out]

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

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-i \arctan (a+b x)}}{x} \, dx=-\sqrt {-\frac {a + i}{a - i}} \log \left (-b x + {\left (i \, a + 1\right )} \sqrt {-\frac {a + i}{a - i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt {-\frac {a + i}{a - i}} \log \left (-b x + {\left (-i \, a - 1\right )} \sqrt {-\frac {a + i}{a - i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + i \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \]

[In]

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

[Out]

-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)) + sqr
t(-(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]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{-i \arctan (a+b x)}}{x} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-i \arctan (a+b x)}}{x} \, dx=-\frac {{\left (-i \, a + 1\right )} \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1}} + \frac {i \, b \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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