3.56 \(\int \frac{e^{-3 i \tan ^{-1}(a x)}}{x} \, dx\)

Optimal. Leaf size=52 \[ \frac{4 i \sqrt{a^2 x^2+1}}{-a x+i}-\tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right )+i \sinh ^{-1}(a x) \]

[Out]

((4*I)*Sqrt[1 + a^2*x^2])/(I - a*x) + I*ArcSinh[a*x] - ArcTanh[Sqrt[1 + a^2*x^2]]

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Rubi [A]  time = 0.572048, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5060, 6742, 215, 266, 63, 208, 651} \[ \frac{4 i \sqrt{a^2 x^2+1}}{-a x+i}-\tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right )+i \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*x),x]

[Out]

((4*I)*Sqrt[1 + a^2*x^2])/(I - a*x) + I*ArcSinh[a*x] - ArcTanh[Sqrt[1 + a^2*x^2]]

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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 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 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

\begin{align*} \int \frac{e^{-3 i \tan ^{-1}(a x)}}{x} \, dx &=\int \frac{(1-i a x)^2}{x (1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=\int \left (\frac{i a}{\sqrt{1+a^2 x^2}}+\frac{1}{x \sqrt{1+a^2 x^2}}-\frac{4 a}{(-i+a x) \sqrt{1+a^2 x^2}}\right ) \, dx\\ &=(i a) \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx-(4 a) \int \frac{1}{(-i+a x) \sqrt{1+a^2 x^2}} \, dx+\int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx\\ &=\frac{4 i \sqrt{1+a^2 x^2}}{i-a x}+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 )\\ &=\frac{4 i \sqrt{1+a^2 x^2}}{i-a x}+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}\\ &=\frac{4 i \sqrt{1+a^2 x^2}}{i-a x}+i \sinh ^{-1}(a x)-\tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.040163, size = 55, normalized size = 1.06 \[ -\frac{4 i \sqrt{a^2 x^2+1}}{a x-i}-\log \left (\sqrt{a^2 x^2+1}+1\right )+i \sinh ^{-1}(a x)+\log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*x),x]

[Out]

((-4*I)*Sqrt[1 + a^2*x^2])/(-I + a*x) + I*ArcSinh[a*x] + Log[x] - Log[1 + Sqrt[1 + a^2*x^2]]

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Maple [B]  time = 0.084, size = 257, normalized size = 4.9 \begin{align*}{\frac{i}{{a}^{3}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-3}}-{\frac{1}{{a}^{2}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-2}}+{\frac{2}{3} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{3}{2}}}}+ia\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }x+{ia\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+\sqrt{{a}^{2}{x}^{2}+1}-{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x,x)

[Out]

I/a^3/(x-I/a)^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)-1/a^2/(x-I/a)^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)+2/3*(a
^2*(x-I/a)^2+2*I*a*(x-I/a))^(3/2)+I*a*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)*x+I*a*ln((I*a+a^2*(x-I/a))/(a^2)^(1/
2)+(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2))/(a^2)^(1/2)+1/3*(a^2*x^2+1)^(3/2)+(a^2*x^2+1)^(1/2)-arctanh(1/(a^2*x^2
+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (i \, a x + 1\right )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x,x, algorithm="maxima")

[Out]

integrate((a^2*x^2 + 1)^(3/2)/((I*a*x + 1)^3*x), x)

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Fricas [B]  time = 1.73555, size = 252, normalized size = 4.85 \begin{align*} \frac{-4 i \, a x -{\left (a x - i\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) +{\left (-i \, a x - 1\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) +{\left (a x - i\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) - 4 i \, \sqrt{a^{2} x^{2} + 1} - 4}{a x - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x,x, algorithm="fricas")

[Out]

(-4*I*a*x - (a*x - I)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) + (-I*a*x - 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + (a*x -
I)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) - 4*I*sqrt(a^2*x^2 + 1) - 4)/(a*x - I)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}{x \left (i a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)**3*(a**2*x**2+1)**(3/2)/x,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x,x, algorithm="giac")

[Out]

undef