3.7 \(\int \frac{e^{i \tan ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=38 \[ -\frac{\sqrt{a^2 x^2+1}}{x}-i a \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0382376, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5060, 807, 266, 63, 208} \[ -\frac{\sqrt{a^2 x^2+1}}{x}-i a \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a*x])/x^2,x]

[Out]

-(Sqrt[1 + a^2*x^2]/x) - I*a*ArcTanh[Sqrt[1 + a^2*x^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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{e^{i \tan ^{-1}(a x)}}{x^2} \, dx &=\int \frac{1+i a x}{x^2 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{x}+(i a) \int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{x}+\frac{1}{2} (i a) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{x}+\frac{i \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )}{a}\\ &=-\frac{\sqrt{1+a^2 x^2}}{x}-i a \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0295662, size = 47, normalized size = 1.24 \[ -\frac{\sqrt{a^2 x^2+1}}{x}-i a \log \left (\sqrt{a^2 x^2+1}+1\right )+i a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(I*ArcTan[a*x])/x^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.063, size = 34, normalized size = 0.9 \begin{align*} -{\frac{1}{x}\sqrt{{a}^{2}{x}^{2}+1}}-ia{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \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^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00049, size = 42, normalized size = 1.11 \begin{align*} -i \, a \operatorname{arsinh}\left (\frac{1}{\sqrt{a^{2}}{\left | x \right |}}\right ) - \frac{\sqrt{a^{2} x^{2} + 1}}{x} \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^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.75318, size = 154, normalized size = 4.05 \begin{align*} \frac{-i \, a x \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) + i \, a x \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) - a x - \sqrt{a^{2} x^{2} + 1}}{x} \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^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 2.52476, size = 26, normalized size = 0.68 \begin{align*} - a \sqrt{1 + \frac{1}{a^{2} x^{2}}} - i a \operatorname{asinh}{\left (\frac{1}{a x} \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**2,x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.17661, size = 103, normalized size = 2.71 \begin{align*} -a i \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1} + 1 \right |}\right ) + a i \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1} - 1 \right |}\right ) + \frac{2 \,{\left | a \right |}}{{\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}\right )}^{2} - 1} \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^2,x, algorithm="giac")

[Out]

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