3.57 \(\int \frac{e^{-3 i \tan ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=64 \[ \frac{4 a \sqrt{a^2 x^2+1}}{-a x+i}-\frac{\sqrt{a^2 x^2+1}}{x}+3 i a \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.552167, antiderivative size = 64, 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, 264, 266, 63, 208, 651} \[ \frac{4 a \sqrt{a^2 x^2+1}}{-a x+i}-\frac{\sqrt{a^2 x^2+1}}{x}+3 i a \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*x^2),x]

[Out]

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

Mathematica [A]  time = 0.0475849, size = 61, normalized size = 0.95 \[ \sqrt{a^2 x^2+1} \left (-\frac{1}{x}-\frac{4 a}{a x-i}\right )+3 i a \log \left (\sqrt{a^2 x^2+1}+1\right )-3 i a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*x^2),x]

[Out]

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

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Maple [B]  time = 0.086, size = 305, normalized size = 4.8 \begin{align*} ia \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{3}{2}}}-ia \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}-{\frac{3\,{a}^{2}x}{2}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}-{\frac{3\,{a}^{2}}{2}\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}{{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 ) ^{-3}}+3\,ia{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -3\,ia\sqrt{{a}^{2}{x}^{2}+1}-{\frac{1}{x} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{a}^{2}x \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}+{\frac{3\,{a}^{2}x}{2}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{3\,{a}^{2}}{2}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \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^2,x)

[Out]

I*a*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(3/2)-I*a*(a^2*x^2+1)^(3/2)-3/2*a^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)*x-3/
2*a^2*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/a^2/(x-I/a)^3*(a^2*(
x-I/a)^2+2*I*a*(x-I/a))^(5/2)+3*I*a*arctanh(1/(a^2*x^2+1)^(1/2))-3*I*a*(a^2*x^2+1)^(1/2)-1/x*(a^2*x^2+1)^(5/2)
+a^2*x*(a^2*x^2+1)^(3/2)+3/2*a^2*x*(a^2*x^2+1)^(1/2)+3/2*a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(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^{2}}\,{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^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.64274, size = 247, normalized size = 3.86 \begin{align*} -\frac{5 \, a^{2} x^{2} - 5 i \, a x + 3 \,{\left (-i \, a^{2} x^{2} - a x\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) + 3 \,{\left (i \, a^{2} x^{2} + a x\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) + \sqrt{a^{2} x^{2} + 1}{\left (5 \, a x - i\right )}}{a x^{2} - i \, 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^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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**2,x)

[Out]

Timed out

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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^2,x, algorithm="giac")

[Out]

undef