\(\int e^{i \arctan (a x)} \, dx\) [5]

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 = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41

method	result	size
meijerg	\(\frac {\operatorname {arcsinh}\left (a x \right )}{a}+\frac {i \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\right )}{2 a \sqrt {\pi }}\)	\(41\)
default	\(\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {i \sqrt {a^{2} x^{2}+1}}{a}\)	\(48\)
risch	\(\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {i \sqrt {a^{2} x^{2}+1}}{a}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (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+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 [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (20) = 40\).

Time = 0.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10 \[ \int e^{i \arctan (a x)} \, dx=\begin {cases} \frac {\log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{\sqrt {a^{2}}} + \frac {i \sqrt {a^{2} x^{2} + 1}}{a} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{2}}{2} + x & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(2*a**2*x + 2*sqrt(a**2*x**2 + 1)*sqrt(a**2))/sqrt(a**2) + I*sqrt(a**2*x**2 + 1)/a, Ne(a**2, 0))
, (I*a*x**2/2 + x, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (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+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+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 = 0.04 (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*x*1i + 1)/(a^2*x^2 + 1)^(1/2),x)

[Out]

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