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

Optimal. Leaf size=42 \[ \frac{\sqrt{a^2 x^2+1} (2-i a x)}{2 a^2}+\frac{i \sinh ^{-1}(a x)}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0199573, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5060, 780, 215} \[ \frac{\sqrt{a^2 x^2+1} (2-i a x)}{2 a^2}+\frac{i \sinh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5060

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((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, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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)} x \, dx &=\int \frac{x (1-i a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{(2-i a x) \sqrt{1+a^2 x^2}}{2 a^2}+\frac{i \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{2 a}\\ &=\frac{(2-i a x) \sqrt{1+a^2 x^2}}{2 a^2}+\frac{i \sinh ^{-1}(a x)}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0261351, size = 38, normalized size = 0.9 \[ \frac{\sqrt{a^2 x^2+1} (2-i a x)+i \sinh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.069, size = 152, normalized size = 3.6 \begin{align*}{\frac{-{\frac{i}{2}}x}{a}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{{\frac{i}{2}}}{a}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{{a}^{2}}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}+{\frac{i}{a}\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(x/(1+I*a*x)*(a^2*x^2+1)^(1/2),x)

[Out]

-1/2*I/a*x*(a^2*x^2+1)^(1/2)-1/2*I/a*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)+1/a^2*(a^2*(x-I/a)^2+
2*I*a*(x-I/a))^(1/2)+I/a*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 = 1.56325, size = 57, normalized size = 1.36 \begin{align*} -\frac{i \, \sqrt{a^{2} x^{2} + 1} x}{2 \, a} + \frac{i \, \operatorname{arsinh}\left (a x\right )}{2 \, a^{2}} + \frac{\sqrt{a^{2} x^{2} + 1}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76266, size = 103, normalized size = 2.45 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1}{\left (-i \, a x + 2\right )} - i \, \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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(x/(1+I*a*x)*(a**2*x**2+1)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.13438, size = 73, normalized size = 1.74 \begin{align*} -\frac{1}{2} \, \sqrt{a^{2} x^{2} + 1}{\left (\frac{i x}{a} - \frac{2}{a^{2}}\right )} - \frac{i \log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{2 \, a{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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