\(\int \frac {e^{i \arctan (a x)}}{x^2} \, dx\) [7]

Optimal result

Integrand size = 14, antiderivative size = 38 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-\frac {\sqrt {1+a^2 x^2}}{x}-i a \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5168, 821, 272, 65, 214} \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-\frac {\sqrt {a^2 x^2+1}}{x}-i a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right ) \]

[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 65

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 214

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

Rule 272

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 821

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 5168

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*} \text {integral}& = \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) \text {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 \text {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 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-\frac {\sqrt {1+a^2 x^2}}{x}+i a \log (x)-i a \log \left (1+\sqrt {1+a^2 x^2}\right ) \]

[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] (verified)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sqrt {a^{2} x^{2}+1}}{x}-i a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\)	\(34\)
risch	\(-\frac {\sqrt {a^{2} x^{2}+1}}{x}-i a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\)	\(34\)
meijerg	\(-\frac {\sqrt {a^{2} x^{2}+1}}{x}+\frac {i a \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }\right )}{2 \sqrt {\pi }}\)	\(64\)

[In]

int((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (32) = 64\).

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.74 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=\frac {-i \, a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + i \, a x \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - a x - \sqrt {a^{2} x^{2} + 1}}{x} \]

[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] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=- a \sqrt {1 + \frac {1}{a^{2} x^{2}}} - i a \operatorname {asinh}{\left (\frac {1}{a x} \right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-i \, a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\sqrt {a^{2} x^{2} + 1}}{x} \]

[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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (32) = 64\).

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.97 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-i \, a \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + i \, a \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) + \frac {2 \, {\left | a \right |}}{{\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {e^{i \arctan (a x)}}{x^2} \, dx=-\frac {\sqrt {a^2\,x^2+1}}{x}-a\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )\,1{}\mathrm {i} \]

[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