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

Optimal. Leaf size=25 \[ -\tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )+i \sinh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5060, 844, 215, 266, 63, 208} \[ -\tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )+i \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

I*ArcSinh[a*x] - 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 215

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 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 844

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 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} \, dx &=\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 \sinh ^{-1}(a x)+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,x^2\right )\\ &=i \sinh ^{-1}(a x)+\frac {\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^2}\\ &=i \sinh ^{-1}(a x)-\tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.16 \[ -\log \left (\sqrt {a^2 x^2+1}+1\right )+i \sinh ^{-1}(a x)+\log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 58, normalized size = 2.32 \[ -\log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.14, size = 69, normalized size = 2.76 \[ -\frac {a i \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{{\left | a \right |}} - \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*i*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))

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maple [B]  time = 0.16, size = 48, normalized size = 1.92 \[ \frac {i a \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-\arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 18, normalized size = 0.72 \[ i \, \operatorname {arsinh}\left (a x\right ) - \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*arcsinh(a*x) - arcsinh(1/(a*abs(x)))

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mupad [B]  time = 0.04, size = 32, normalized size = 1.28 \[ -\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )+\frac {a\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,1{}\mathrm {i}}{\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 3.35, size = 53, normalized size = 2.12 \[ i a \left (\begin {cases} \sqrt {- \frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \\\sqrt {\frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \end {cases}\right ) - \operatorname {asinh}{\left (\frac {1}{a x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*Piecewise((sqrt(-1/a**2)*asin(x*sqrt(-a**2)), a**2 < 0), (sqrt(a**(-2))*asinh(x*sqrt(a**2)), a**2 > 0)) -
asinh(1/(a*x))

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