\(\int \frac {e^{-i \arctan (a x)}}{x^2} \, dx\) [40]

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[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 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[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 [A] (verified)

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {\sqrt {a^{2} x^{2}+1}}{x}+i a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\)	\(34\)
default	\(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+2 a^{2} \left (\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}\right )-i a \left (\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )+i a \left (\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}\right )\)	\(195\)

[In]

int(1/(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 = 0.26 (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/(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]

\[ \int \frac {e^{-i \arctan (a x)}}{x^2} \, dx=- i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{3} - i x^{2}}\, dx \]

[In]

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

[Out]

-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**3 - I*x**2), x)

Maxima [F]

\[ \int \frac {e^{-i \arctan (a x)}}{x^2} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{2}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-i \arctan (a x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

[Out]

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