3.5 \(\int e^{i \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=29 \[ \frac{\sinh ^{-1}(a x)}{a}+\frac{i \sqrt{a^2 x^2+1}}{a} \]

[Out]

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

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Rubi [A]  time = 0.0090885, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5059, 641, 215} \[ \frac{\sinh ^{-1}(a x)}{a}+\frac{i \sqrt{a^2 x^2+1}}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5059

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]

Rule 641

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 215

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]

Rubi steps

\begin{align*} \int e^{i \tan ^{-1}(a x)} \, dx &=\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{\sinh ^{-1}(a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0147939, size = 26, normalized size = 0.9 \[ \frac{\sinh ^{-1}(a x)+i \sqrt{a^2 x^2+1}}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 48, normalized size = 1.7 \begin{align*}{\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{i}{a}\sqrt{{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02399, size = 46, normalized size = 1.59 \begin{align*} \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} + \frac{i \, \sqrt{a^{2} x^{2} + 1}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61595, size = 77, normalized size = 2.66 \begin{align*} \frac{i \, \sqrt{a^{2} x^{2} + 1} - \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 1.3362, size = 68, normalized size = 2.34 \begin{align*} i a \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: a^{2} = 0 \\\frac{\sqrt{a^{2} x^{2} + 1}}{a^{2}} & \text{otherwise} \end{cases}\right ) + \begin{cases} \sqrt{- \frac{1}{a^{2}}} \operatorname{asin}{\left (x \sqrt{- a^{2}} \right )} & \text{for}\: a^{2} < 0 \\\sqrt{\frac{1}{a^{2}}} \operatorname{asinh}{\left (x \sqrt{a^{2}} \right )} & \text{for}\: a^{2} > 0 \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*Piecewise((x**2/2, Eq(a**2, 0)), (sqrt(a**2*x**2 + 1)/a**2, True)) + Piecewise((sqrt(-1/a**2)*asin(x*sqrt(
-a**2)), a**2 < 0), (sqrt(a**(-2))*asinh(x*sqrt(a**2)), a**2 > 0))

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Giac [A]  time = 1.11413, size = 55, normalized size = 1.9 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1} i}{a} - \frac{\log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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