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]

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

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.03, 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]]

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fricas [B]  time = 0.45, size = 66, normalized size = 1.74 \[ \frac {-i \, a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + i \, a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - a x - \sqrt {a^{2} x^{2} + 1}}{x} \]

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

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giac [B]  time = 0.14, size = 76, normalized size = 2.00 \[ -a i \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + a i \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) + \frac {2 \, {\left | a \right |}}{{\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1} \]

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)

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maple [A]  time = 0.16, size = 34, normalized size = 0.89 \[ -i a \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )-\frac {\sqrt {a^{2} x^{2}+1}}{x} \]

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]

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

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maxima [A]  time = 0.34, size = 29, normalized size = 0.76 \[ -i \, a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\sqrt {a^{2} x^{2} + 1}}{x} \]

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/(a*abs(x))) - sqrt(a^2*x^2 + 1)/x

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mupad [B]  time = 0.04, size = 33, normalized size = 0.87 \[ -\frac {\sqrt {a^2\,x^2+1}}{x}-a\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- a*atanh((a^2*x^2 + 1)^(1/2))*1i - (a^2*x^2 + 1)^(1/2)/x

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sympy [A]  time = 2.30, size = 26, normalized size = 0.68 \[ - a \sqrt {1 + \frac {1}{a^{2} x^{2}}} - i a \operatorname {asinh}{\left (\frac {1}{a x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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