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]

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

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.90 \[ \frac {\sinh ^{-1}(a x)+i \sqrt {a^2 x^2+1}}{a} \]

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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fricas [A]  time = 0.45, size = 37, normalized size = 1.28 \[ \frac {i \, \sqrt {a^{2} x^{2} + 1} - \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{a} \]

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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giac [A]  time = 0.13, size = 41, normalized size = 1.41 \[ \frac {\sqrt {a^{2} x^{2} + 1} i}{a} - \frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} \]

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)

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maple [A]  time = 0.10, size = 48, normalized size = 1.66 \[ \frac {\ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {i \sqrt {a^{2} x^{2}+1}}{a} \]

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(x*a^2/(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 = 0.32, size = 25, normalized size = 0.86 \[ \frac {\operatorname {arsinh}\left (a x\right )}{a} + \frac {i \, \sqrt {a^{2} x^{2} + 1}}{a} \]

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*x)/a + I*sqrt(a^2*x^2 + 1)/a

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mupad [B]  time = 0.04, size = 32, normalized size = 1.10 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}+\frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.25, size = 68, normalized size = 2.34 \[ i a \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\\frac {\sqrt {a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + \begin {cases} \sqrt {- \frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \\\sqrt {\frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \end {cases} \]

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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