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

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 = 0.0094273, 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 = 0.0129128, 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 [B]  time = 0.058, size = 97, normalized size = 3.3 \begin{align*}{\frac{-i}{a}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}+{\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56001, size = 34, normalized size = 1.17 \begin{align*} \frac{\operatorname{arsinh}\left (a x\right )}{a} - \frac{i \, \sqrt{a^{2} x^{2} + 1}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.6649, size = 78, normalized size = 2.69 \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/(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{i a x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10608, size = 57, normalized size = 1.97 \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/(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)