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

Optimal. Leaf size=92 \[ -\frac{4 i a^2 \sqrt{a^2 x^2+1}}{a x+i}-\frac{3 i a \sqrt{a^2 x^2+1}}{x}-\frac{\sqrt{a^2 x^2+1}}{2 x^2}+\frac{9}{2} a^2 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

-Sqrt[1 + a^2*x^2]/(2*x^2) - ((3*I)*a*Sqrt[1 + a^2*x^2])/x - ((4*I)*a^2*Sqrt[1 + a^2*x^2])/(I + a*x) + (9*a^2*
ArcTanh[Sqrt[1 + a^2*x^2]])/2

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Rubi [A]  time = 0.601464, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5060, 6742, 266, 51, 63, 208, 264, 651} \[ -\frac{4 i a^2 \sqrt{a^2 x^2+1}}{a x+i}-\frac{3 i a \sqrt{a^2 x^2+1}}{x}-\frac{\sqrt{a^2 x^2+1}}{2 x^2}+\frac{9}{2} a^2 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + a^2*x^2]/(2*x^2) - ((3*I)*a*Sqrt[1 + a^2*x^2])/x - ((4*I)*a^2*Sqrt[1 + a^2*x^2])/(I + a*x) + (9*a^2*
ArcTanh[Sqrt[1 + a^2*x^2]])/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 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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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^3} \, dx &=\int \frac{(1+i a x)^2}{x^3 (1-i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=\int \left (\frac{1}{x^3 \sqrt{1+a^2 x^2}}+\frac{3 i a}{x^2 \sqrt{1+a^2 x^2}}-\frac{4 a^2}{x \sqrt{1+a^2 x^2}}+\frac{4 a^3}{(i+a x) \sqrt{1+a^2 x^2}}\right ) \, dx\\ &=(3 i a) \int \frac{1}{x^2 \sqrt{1+a^2 x^2}} \, dx-\left (4 a^2\right ) \int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx+\left (4 a^3\right ) \int \frac{1}{(i+a x) \sqrt{1+a^2 x^2}} \, dx+\int \frac{1}{x^3 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{3 i a \sqrt{1+a^2 x^2}}{x}-\frac{4 i a^2 \sqrt{1+a^2 x^2}}{i+a x}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+a^2 x}} \, dx,x,x^2\right )-\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{2 x^2}-\frac{3 i a \sqrt{1+a^2 x^2}}{x}-\frac{4 i a^2 \sqrt{1+a^2 x^2}}{i+a x}-4 \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )-\frac{1}{4} a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{2 x^2}-\frac{3 i a \sqrt{1+a^2 x^2}}{x}-\frac{4 i a^2 \sqrt{1+a^2 x^2}}{i+a x}+4 a^2 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{2 x^2}-\frac{3 i a \sqrt{1+a^2 x^2}}{x}-\frac{4 i a^2 \sqrt{1+a^2 x^2}}{i+a x}+\frac{9}{2} a^2 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0766212, size = 79, normalized size = 0.86 \[ \sqrt{a^2 x^2+1} \left (-\frac{4 i a^2}{a x+i}-\frac{3 i a}{x}-\frac{1}{2 x^2}\right )+\frac{9}{2} a^2 \log \left (\sqrt{a^2 x^2+1}+1\right )-\frac{9}{2} a^2 \log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[1 + a^2*x^2]*(-1/(2*x^2) - ((3*I)*a)/x - ((4*I)*a^2)/(I + a*x)) - (9*a^2*Log[x])/2 + (9*a^2*Log[1 + Sqrt[
1 + a^2*x^2]])/2

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Maple [A]  time = 0.072, size = 105, normalized size = 1.1 \begin{align*}{-i{a}^{3}x{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{1}{2\,{x}^{2}}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{9\,{a}^{2}}{2} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}-{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \right ) }+3\,ia \left ( -{\frac{1}{x}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-2\,{\frac{{a}^{2}x}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a^3*x/(a^2*x^2+1)^(1/2)-1/2/x^2/(a^2*x^2+1)^(1/2)-9/2*a^2*(1/(a^2*x^2+1)^(1/2)-arctanh(1/(a^2*x^2+1)^(1/2))
)+3*I*a*(-1/x/(a^2*x^2+1)^(1/2)-2*a^2*x/(a^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.05043, size = 112, normalized size = 1.22 \begin{align*} -\frac{7 i \, a^{3} x}{\sqrt{a^{2} x^{2} + 1}} + \frac{9}{2} \, a^{2} \operatorname{arsinh}\left (\frac{1}{\sqrt{a^{2}}{\left | x \right |}}\right ) - \frac{9 \, a^{2}}{2 \, \sqrt{a^{2} x^{2} + 1}} - \frac{3 i \, a}{\sqrt{a^{2} x^{2} + 1} x} - \frac{1}{2 \, \sqrt{a^{2} x^{2} + 1} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*I*a^3*x/sqrt(a^2*x^2 + 1) + 9/2*a^2*arcsinh(1/(sqrt(a^2)*abs(x))) - 9/2*a^2/sqrt(a^2*x^2 + 1) - 3*I*a/(sqrt
(a^2*x^2 + 1)*x) - 1/2/(sqrt(a^2*x^2 + 1)*x^2)

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Fricas [A]  time = 1.71156, size = 294, normalized size = 3.2 \begin{align*} \frac{-14 i \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 9 \,{\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) - 9 \,{\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) + \sqrt{a^{2} x^{2} + 1}{\left (-14 i \, a^{2} x^{2} + 5 \, a x - i\right )}}{2 \,{\left (a x^{3} + i \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-14*I*a^3*x^3 + 14*a^2*x^2 + 9*(a^3*x^3 + I*a^2*x^2)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 9*(a^3*x^3 + I*a
^2*x^2)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + sqrt(a^2*x^2 + 1)*(-14*I*a^2*x^2 + 5*a*x - I))/(a*x^3 + I*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

undef