\(\int \frac {e^{i \arctan (a x)}}{x} \, dx\) [6]

Optimal result

Integrand size = 14, antiderivative size = 25 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=i \text {arcsinh}(a x)-\text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5168, 858, 221, 272, 65, 214} \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=-\text {arctanh}\left (\sqrt {a^2 x^2+1}\right )+i \text {arcsinh}(a x) \]

[In]

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

[Out]

I*ArcSinh[a*x] - 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 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[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 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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 \sqrt {1+a^2 x^2}} \, dx \\ & = (i a) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx+\int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx \\ & = i \text {arcsinh}(a x)+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right ) \\ & = i \text {arcsinh}(a x)+\frac {\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^2} \\ & = i \text {arcsinh}(a x)-\text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=i \text {arcsinh}(a x)+\log (x)-\log \left (1+\sqrt {1+a^2 x^2}\right ) \]

[In]

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

[Out]

I*ArcSinh[a*x] + Log[x] - Log[1 + Sqrt[1 + a^2*x^2]]

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22 ) = 44\).

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92

method	result	size
default	\(-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {i a \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}\)	\(48\)
meijerg	\(\frac {-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2 \sqrt {\pi }}+i \operatorname {arcsinh}\left (a x \right )\)	\(53\)

[In]

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

[Out]

-arctanh(1/(a^2*x^2+1)^(1/2))+I*a*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(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. 58 vs. \(2 (21) = 42\).

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=-\log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=i a \left (\begin {cases} \frac {\log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{\sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) - \operatorname {asinh}{\left (\frac {1}{a x} \right )} \]

[In]

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

[Out]

I*a*Piecewise((log(2*a**2*x + 2*sqrt(a**2*x**2 + 1)*sqrt(a**2))/sqrt(a**2), Ne(a**2, 0)), (x, True)) - asinh(1
/(a*x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=i \, \operatorname {arsinh}\left (a x\right ) - \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) \]

[In]

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

[Out]

I*arcsinh(a*x) - arcsinh(1/(a*abs(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. 68 vs. \(2 (21) = 42\).

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=-\frac {i \, a \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} - \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{i \arctan (a x)}}{x} \, dx=-\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )+\frac {a\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,1{}\mathrm {i}}{\sqrt {a^2}} \]

[In]

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

[Out]

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