\(\int e^{-i \arctan (a x)} \, dx\) [38]

Optimal result

Integrand size = 10, antiderivative size = 29 \[ \int e^{-i \arctan (a x)} \, dx=-\frac {i \sqrt {1+a^2 x^2}}{a}+\frac {\text {arcsinh}(a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5167, 655, 221} \[ \int e^{-i \arctan (a x)} \, dx=\frac {\text {arcsinh}(a x)}{a}-\frac {i \sqrt {a^2 x^2+1}}{a} \]

[In]

Int[E^((-I)*ArcTan[a*x]),x]

[Out]

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

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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 5167

Int[E^(ArcTan[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1
+ a^2*x^2]), x] /; FreeQ[a, x] && IntegerQ[(I*n - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int e^{-i \arctan (a x)} \, dx=\frac {-i \sqrt {1+a^2 x^2}+\text {arcsinh}(a x)}{a} \]

[In]

Integrate[E^((-I)*ArcTan[a*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66

method	result	size
risch	\(-\frac {i \sqrt {a^{2} x^{2}+1}}{a}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}\)	\(48\)
default	\(-\frac {i \left (\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}\right )}{a}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int e^{-i \arctan (a x)} \, dx=\frac {-i \, \sqrt {a^{2} x^{2} + 1} - \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{a} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int e^{-i \arctan (a x)} \, dx=\frac {\operatorname {arsinh}\left (a x\right )}{a} - \frac {i \, \sqrt {a^{2} x^{2} + 1}}{a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int e^{-i \arctan (a x)} \, dx=-\frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} - \frac {i \, \sqrt {a^{2} x^{2} + 1}}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int e^{-i \arctan (a x)} \, dx=\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a} \]

[In]

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

[Out]

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