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

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 = 0.0369982, 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[1/(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 = 0.0262643, 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[1/(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 [B]  time = 0.095, size = 194, normalized size = 5.1 \begin{align*} -{\frac{1}{x} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{a}^{2}x\sqrt{{a}^{2}{x}^{2}+1}+{{a}^{2}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+ia\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }-{{a}^{2}\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}}}}}+ia{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -ia\sqrt{{a}^{2}{x}^{2}+1} \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^2,x)

[Out]

-1/x*(a^2*x^2+1)^(3/2)+a^2*x*(a^2*x^2+1)^(1/2)+a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)+I*a*(a^
2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)-a^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)+I*a*arctanh(1/(a^2*x^2+1)^(1/2))-I*a*(a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{2}}\,{d x} \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^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.66206, size = 153, normalized size = 4.03 \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/(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{x^{2} \left (i a x + 1\right )}\, 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**2,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \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^2,x, algorithm="giac")

[Out]

undef